W hen it is cold, warm clothes act as insulators to reduce heat loss from the body to the environment by conduction and convection. H eat radiation from a campfire can warm you and your clothes. The fire can also transfer energy directly by heat convection and conduction to what you are cooking. Heat, like work, represents a transfer o f energy. H eat is defined as a transfer o f energy due to a difference of temperature. Work is a transfer o f energy by mechanical means, not due to a temperature difference. The first law of thermodynamics links the two in a general statement o f energy conservation: the heat Q added to a system minus the net work W done by the system equals the change in internal energy A £ int o f the system: A £ int = Q - W . Internal energy E int is the sum total o f all the energy of the m olecules of the system.

Heat and the First Law of ThermodynamicsCHAPTER-OPENING QUESTION—Guess now!A 5-kg cube of warm iron (60°C) is put in thermal contact with a 10-kg cube of cold iron (15°C). Which statement is valid:

(a) Heat flows spontaneously from the warm cube to the cold cube until both cubes have the same heat content.

(b) Heat flows spontaneously from the warm cube to the cold cube until both cubes have the same temperature.

(c) Heat can flow spontaneously from the warm cube to the cold cube, but can also flow spontaneously from the cold cube to the warm cube.

(d) Heat can never flow from a cold object or area to a hot object or area.(e) Heat flows from the larger cube to the smaller one because the larger one has

more internal energy.

W hen a pot of cold water is placed on a hot burner of a stove, the temperature of the water increases. We say that heat “flows” from the hot burner to the cold water. When two objects at different temperatures are put in contact, heat spontaneously flows from the hotter one to the

colder one. The spontaneous flow of heat is in the direction tending to equalize the temperature. If the two objects are kept in contact long enough for their temperatures to become equal, the objects are said to be in thermal equilibrium, and there is no further heat flow between them. For example, when a fever thermometer is first placed in your mouth, heat flows from your mouth to the thermometer. When the thermometer reaches the same temoerature as the inside of vour mouth, the

CONTENTS19-1 Heat as Energy Transfer1 9 -2 Internal Energy19-3 Specific Heat1 9 -4 Calorimetry— Solving

Problems

19-5 Latent Heat

1 9 -6 The First Law of Thermodynamics

1 9 -7 The First Lawof Thermodynamics Applied; Calculating the Work

1 9-8 Molar Specific Heats for Gases, and the Equipartition of Energy

19-9 Adiabatic Expansion of a Gas

19-10 Heat Transfer: Conduction, Convection, Radiation

Heat and temperature are often confused. They are very different concepts and we will make the clear distinction between them. We start this Chapter by defining and using the concept of heat. We also begin our discussion of thermodynamics, which is the name we give to the study of processes in which energy is transferred as heat and as work.

19-1 Heat as Energy TransferWe use the term “heat” in everyday life as if we knew what we meant. But the term is often used inconsistently, so it is important for us to define heat clearly, and to clarify the phenomena and concepts related to heat.

We commonly speak of the flow of heat—heat flows from a stove burner to a pot of soup, from the Sun to the Earth, from a person’s mouth into a fever thermometer. Heat flows spontaneously from an object at higher temperature to one at lower temperature. Indeed, an eighteenth-century model of heat pictured heat flow as movement of a fluid substance called caloric. However, the caloric fluid could never be detected. In the nineteenth century, it was found that the various phenomena associated with heat could be described consistently using a new model that views heat as akin to work, as we will discuss in a moment. First we note that a common unit for heat, still in use today, is named after caloric. It is called the calorie (cal) and is defined as the amount o f heat necessary to raise the temperature o f 1 gram o f water by 1 Celsius degree. [To be precise, the particular temperature range from 14.5°C to 15.5°C is specified because the heat required is very slightly different at different temperatures. The difference is less than 1% over the range 0 to 100°C, and we will ignore it for most purposes.] More often used than the calorie is the kilocalorie (kcal), which is 1000 calories. Thus 1 kcal is the heat needed to raise the temperature o f 1 kg o f water by 1 C°. Often a kilocalorie is called a Calorie (with a capital C), and this Calorie (or the kJ) is used to specify the energy value of food. In the British system of units, heat is measured in British thermal units (Btu). One Btu is defined as the heat needed to raise the temperature of 1 lb of water by 1 F°. It can be shown (Problem 4) that 1 Btu = 0.252 kcal = 1056 J.

The idea that heat is related to energy transfer was pursued by a number of scientists in the 1800s, particularly by an English brewer, James Prescott Joule (1818-1889). One of Joule’s experiments is shown (simplified) in Fig. 19-1. The falling weight causes the paddle wheel to turn. The friction between the water and the paddle wheel causes the temperature of the water to rise slightly (barely measurable, in fact, by Joule). In this and many other experiments (some involving electrical energy), Joule determined that a given amount of work done was always equivalent to a particular amount of heat input. Quantitatively, 4.186 joules (J) of work was found to be equivalent to 1 calorie (cal) of heat. This is known as the mechanical equivalent of heat:

4.186 J = leal;4.186 kJ = 1 kcal.

As a result of these and other experiments, scientists came to interpret heat not as a substance, and not exactly as a form of energy. Rather, heat refers to a transfer o f energy: when heat flows from a hot object to a cooler one, it is energy that is being transferred from the hot to the cold object. Thus, heat is energy transferred from one object to another because o f a difference in temperature. In SI units, the unit for heat, as for any form of energy, is the joule. Nonetheless, calories and kcal are still sometimes used. Today the calorie is defined in terms of the joule (via the mechanical equivalent of heat, above), rather than in terms of the properties of water, as given previously. The latter is still handy to remember: 1 cal raises 1 g of water by 1 C°, or 1 kcal raises 1 kg of water by 1 C°.

Joule’s result was crucial because it extended the work-energy principle to include orocesses involving heat. It also led to the establishment of the law of

/? \ CAUTI ONHeat is not a fluid

FIGURE 19-1 Joule’s experiment on the mechanical equivalent of heat.

/ j \ CAUTI ONHeat is energy transferred because o f a A T

@ P H Y S I C S A P P L I E DWorking o ff Calories

/ j \ CAUTI ONDistinguish heat from

internal energy and from temperature

/j\ CAUTION__________Direction o f heat flo w

depends on temperature (not on amount o f internal energy)

EXAMPLE 19-1 ESTIMATE"! Working off the extra calories. Suppose you throw caution to the wind and eat too much ice cream and cake on the order of 500 Calories. To compensate, you want to do an equivalent amount of work climbing stairs or a mountain. How much total height must you climb?APPROACH The work W you need to do in climbing stairs equals the change in gravitational potential energy: W = A PE = mgh, where h is the vertical height climbed. For this estimate, approximate your mass as m ~ 60 kg.SOLUTION 500 food Calories are 500 kcal, which in joules is

(500 kcal)(4.186 X 103J/kcal) = 2.1 X 106J.The work done to climb a vertical height h is W = mgh. We solve for h:

W 2.1 X 106J _h = ---- = --------- ----------- -rr = 3600 m.

mg (60 kg) (9.80 m/s2)

This is a huge elevation change (over 11,000 ft).NOTE The human body does not transform food energy with 100% efficiency—it is more like 20% efficient. As we shall discuss in the next Chapter, some energy is always “wasted,” so you would actually have to climb only about (0.2)(3600 m) « 700 m, which is more reasonable (about 2300 ft of elevation gain).

19—2 Internal EnergyThe sum total of all the energy of all the molecules in an object is called its internal energy. (Sometimes thermal energy is used to mean the same thing.) We introduce the concept of internal energy now since it will help clarify ideas about heat.

Distinguishing Temperature, Heat, and Internal EnergyUsing the kinetic theory, we can make a clear distinction between temperature, heat, and internal energy. Temperature (in kelvins) is a measure of the average kinetic energy of individual molecules. Internal energy refers to the total energy of all the molecules within the object. (Thus two equal-mass hot ingots of iron may have the same temperature, but two of them have twice as much internal energy as one does.) Heat, finally, refers to a transfer of energy from one object to another because of a difference in temperature.

Notice that the direction of heat flow between two objects depends on their temperatures, not on how much internal energy each has. Thus, if 50 g of water at 30°C is placed in contact (or mixed) with 200 g of water at 25°C, heat flows from the water at 30°C to the water at 25°C even though the internal energy of the 25°C water is much greater because there is so much more of it.

EXERCISE A Return to the Chapter-Opening Question, page 496, and answer it again now. Try to explain why you may have answered differently the first time.

Internal Energy of anldeal GasLet us calculate the internal energy of n moles of an ideal monatomic (one atom per molecule) gas. The internal energy, Emu is the sum of the translational kinetic energies of all the atoms.1 This sum is equal to the average kinetic energy per molecule times the total number of molecules, N:

Emt =

Using Eq. 18-4, K = \m vL = \kT , we can write this as

Eint = I nRT, [ideal monatomic gas] (19-lb)

where n is the number of moles. Thus, the internal energy of an ideal gas depends only on temperature and the number of moles of gas.

If the gas molecules contain more than one atom, then the rotational and vibrational energy of the molecules (Fig. 19-2) must also be taken into account. The internal energy will be greater at a given temperature than for a monatomic gas, but it will still be a function only of temperature for an ideal gas.

The internal energy of real gases also depends mainly on temperature, but where real gases deviate from ideal gas behavior, their internal energy depends also somewhat on pressure and volume (due to atomic potential energy).

The internal energy of liquids and solids is quite complicated, for it includes electrical potential energy associated with the forces (or “chemical” bonds) between atoms and molecules.

or (recall Section 17-9)

19—3 Specific HeatIf heat flows into an object, the object’s temperature rises (assuming no phase change). But how much does the temperature rise? That depends. As early as the eighteenth century, experimenters had recognized that the amount of heat Q required to change the temperature of a given material is proportional to the mass m of the material present and to the temperature change AT. This remarkable simplicity in nature can be expressed in the equation

Q = mcAT, (19-2)where c is a quantity characteristic of the material called its specific heat. Because c = Q/mAT, specific heat is specified in units1" of J/kg-C° (the proper SI unit) or kcal/kg-C°. For water at 15°C and a constant pressure of 1 atm, c = 4.19 X 103J/kg*C° or 1.00kcal/kg*C°, since, by definition of the cal and the joule, it takes 1 kcal of heat to raise the temperature of 1 kg of water by 1 C°. Table 19-1 gives the values of specific heat for other solids and liquids at 20°C. The values of c for solids and liquids depend to some extent on temperature (as well as slightly on pressure), but for temperature changes that are not too great, c can often be considered constant.* Gases are more complicated and are treated in Section 19-8.

How heat transferred depends on specific heat, (a) Howmuch heat input is needed to raise the temperature of an empty 20-kg vat made of iron from 10°C to 90°C? (b) What if the vat is filled with 20 kg of water?APPROACH We apply Eq. 19-2 to the different materials involved.SOLUTION (a) Our system is the iron vat alone. From Table 19-1, the specific heat of iron is 450 J/kg-C°. The change in temperature is (90°C - 10°C) = 80 C°. Thus,

Q = mc AT = (20 kg) (450 J/kg • C°) (80 C°) = 7.2 X l t f j = 720 kJ.(b) Our system is the vat plus the water. The water alone would require

Q = mc AT = (20 kg) (4186 J/kg-C°) (80 C°) = 6.7 X 106J = 6700 kJ,or almost 10 times what an equal mass of iron requires. The total, for the vat plus the water, is 720 kJ + 6700 kJ = 7400 kJ.NOTE In (b), the iron vat and the water underwent the same temperature change, AT = 80 C°, but their specific heats are different.

tNote that J/kg - C° means that way). kg-C

and not (J/kg) • C° = J - C°/kg (otherwise we would have written it

*To take into account the dependence of c on T, we can write Eq. 19-2 in differential form: dQ = mc(T) dT, where c(T) means c is a function of temperature T. Then the heat Q required to change the temperature from 7\ to T2 is

/ ' 1 / V

« V ../ 7

Q Q ^ O(b)

FIGURE 1 9 -2 Besides translational kinetic energy, molecules can have(a) rotational kinetic energy, and(b) vibrational energy (both kinetic and potential).

TABLE 19-1 Specific Heats (at 1 atm constant pressure and 20°C unless otherwise stated)

Specific Heat, cSubstance kcal/kg

(= cal/gC° JAg C°)

c°

Aluminum 0.22 900Alcohol

(ethyl) 0.58 2400Copper 0.093 390Glass 0.20 840Iron or steel 0.11 450Lead 0.031 130Marble 0.21 860Mercury 0.033 140Silver 0.056 230Wood 0.4 1700Water

Ice ( —5°C) 0.50 2100Liquid (15°C) 1.00 4186Steam (110°C) 0.48 2010

Human body (average) 0.83 3470

© - P H Y S I C S A P P L I E DPractical effects o f w ater’s

high specific heat

b

<t>)(a)

FIGURE 1 9 -3 Example 19-3.

If the iron vat in part (a) of Example 19-2 had been cooled from 90°C to 10°C, 720 kJ of heat would have flowed out of the iron. In other words, Eq. 19-2 is valid for heat flow either in or out, with a corresponding increase or decrease in tempera­ture. We saw in part (b) that water requires almost 10 times as much heat as an equal mass of iron to make the same temperature change. Water has one of the highest specific heats of all substances, which makes it an ideal substance for hot-water space-heating systems and other uses that require a minimal drop in temperature for a given amount of heat transfer. It is the water content, too, that causes the apples rather than the crust in hot apple pie to burn our tongues, through heat transfer.

19—4 Calorimetry—Solving ProblemsIn discussing heat and thermodynamics, we shall often refer to particular systems. As already mentioned in earlier Chapters, a system is any object or set of objects that we wish to consider. Everything else in the universe we will refer to as its “environment” or the “surroundings.” There are several categories of systems. A closed system is one for which no mass enters or leaves (but energy may be exchanged with the environment). In an open system, mass may enter or leave (as may energy). Many (idealized) systems we study in physics are closed systems. But many systems, including plants and animals, are open systems since they exchange materials (food, oxygen, waste products) with the environment. A closed system is said to be isolated if no energy in any form passes across its boundaries; otherwise it is not isolated.

When different parts of an isolated system are at different temperatures, heat will flow (energy is transferred) from the part at higher temperature to the part at lower temperature—that is, within the system. If the system is truely isolated, no energy is transferred into or out of it. So the conservation o f energy again plays an important role for us: the heat lost by one part of the system is equal to the heat gained by the other part:

heat lost = heat gainedor

energy out of one part = energy into another part.These simple relations are very useful, but depend on the (often very good) approximation that the whole system is isolated (no other energy transfers occur). Let us take an Example.

U 5 E Q H E H S The CUP cools the tea. If 200 cm3 of tea at 95°C is poured into a 150-g glass cup initially at 25°C (Fig. 19-3), what will be the common final temperature T of the tea and cup when equilibrium is reached, assuming no heat flows to the surroundings?APPROACH We apply conservation of energy to our system of tea plus cup, which we are assuming is isolated: all of the heat that leaves the tea flows into the cup. We use the specific heat equation, Eq. 19-2, to determine how the heat flow is related to the temperature changes.SOLUTION Because tea is mainly water, its specific heat is 4186 J/kg • C ° (Table 19-1), and its mass m is its density times its volume (V = 200 cm3 = 200 X 10-6 m3): m = pV = (1.0 X 103kg/m3)(200 X 10“6m3) = 0.20 kg. We use Eq. 19-2, apply conservation of energy, and let T be the as yet unknown final temperature:

heat lost by tea = heat gained by cup rateaCtea (9 5 °C - T) = mcupccup(T - 2 5 ° C ) .

Putting in numbers and using Table 19-1 (ccup = 840 J/kg-C° for glass), we solve for T, and find

(0.20 kg)(4186 J/kg • C°)(95°C - T) =79,500 J - (837J/C°)T =

T =

(0.15 kg)(840 J/kg-C °)(r - 25°C)(126J/C°)T - 3150 J86°C.

The tea drops in temperature by 9 C° by coming into equilibrium with the cup.

NOTE In this calculation, the AT (of Eq. 19-2, Q = mc AT) is a positive quantity on both sides of our conservation of energy equation. On the left is “heat lost” and AT is the initial minus the final temperature (95°C - T), whereas on the right is “heat gained” and AT is the final minus the initial temperature. But consider the following alternate approach.Alternate Solution We can set up this Example (and others) by a different approach. We can write that the total heat transferred into or out of the isolated system is zero:

2 0 = 0.Then each term is written as Q = mc(Tt — 7]), and AT = Tf — 7] is always the final minus the initial temperature, and each AT can be positive or negative. In the present Example:

2 0 = m cupccup( r - 25°C) + mteactea(T - 95°C) = 0.The second term is negative because T will be less than 95°C. Solving the algebra gives the same result.

The exchange of energy, as exemplified in Example 19-3, is the basis for a technique known as calorimetry, which is the quantitative measurement of heat exchange. To make such measurements, a calorimeter is used; a simple water calorimeter is shown in Fig. 19-4. It is very important that the calorimeter be well insulated so that almost no heat is exchanged with the surroundings. One important use of the calorimeter is in the determination of specific heats of substances. In the technique known as the “method of mixtures,” a sample of a substance is heated to a high temperature, which is accurately measured, and then quickly placed in the cool water of the calorimeter. The heat lost by the sample will be gained by the water and the calorimeter cup. By measuring the final temperature of the mixture, the specific heat can be calculated, as illustrated in the following Example.

Unknown specific heat determined by calorimetry.An engineer wishes to determine the specific heat of a new metal alloy. A 0.150-kg sample of the alloy is heated to 540°C. It is then quickly placed in 0.400 kg of water at 10.0°C, which is contained in a 0.200-kg aluminum calorimeter cup. (We do not need to know the mass of the insulating jacket since we assume the air space between it and the cup insulates it well, so that its temperature does not change significantly.) The final temperature of the system is 30.5°C. Calculate the specific heat of the alloy. APPROACH We apply conservation of energy to our system, which we take to be the alloy sample, the water, and the calorimeter cup. We assume this system is isolated, so the energy lost by the hot alloy equals the energy gained by the water and calorimeter cup.SOLUTION The heat lost equals the heat gained:

heat lost j _ / heat gained \ ( heat gained by by alloy J y by water J y calorimeter cup

^a ^-^a ^ c a l ^cal ^-^calwhere the subscripts a, w, and cal refer to the alloy, water, and calorimeter, respectively, and each AT > 0. When we put in values and use Table 19-1, this equation becomes (0.150 kg) (ca) (540°C - 30.5°C) = (0.400 kg) (4186 J/kg-C°)(30.5°C - 10.0°C)

+ (0.200 kg) (900 J/kg-C°)(30.5°C - 10.0°C) (76.4 kg • C°) ca = (34,300 + 3690) J

ca = 497 J/kg • C°.In making this calculation, we have ignored any heat transferred to the thermometer and the stirrer (which is used to quicken the heat transfer process and thus reduce heat loss to the outside). It can be taken into account by adding additional terms to the right side of the above equation and will result in a slight correction to the value of ca .

In all Examples and Problems of this sort, be sure to include all objects that sain or lose heat (within reason). On the “heat loss” side here, it is onlv the hot

/9\ CAUTION______When usingheat lost = heat gained, AT is positive on both sides

i P R O B L E M S O L V I N GAlternate approach: HQ = 0

Tht>nnomoier Slirrcr

jacket tup

FIGURE 1 9 -4 Simple water calorimeter.

I P R O B L E M S O L V I N GBe sure to consider all possible

FIGURE 19-5 Temperature as a function of the heat added to bring 1.0 kg of ice at — 40°C to steam above 100°C.

19—5 Latent HeatWhen a material changes phase from solid to liquid, or from liquid to gas (see also Section 18-3), a certain amount of energy is involved in this change of phase. For example, let us trace what happens when a 1.0-kg block of ice at -40°C is heated at a slow steady rate until all the ice has changed to water, then the (liquid) water is heated to 100°C and changed to steam, and heated further above 100°C, all at 1 atm pressure. As shown in the graph of Fig. 19-5, as the ice is heated starting at -40°C, its temperature rises at a rate of about 2 C°/kcal of heat added (for ice, c ~ 0.50 kcal/kg • C°). However, when 0°C is reached, the temperature stops increasing even though heat is still being added. The ice gradually changes to water in the liquid state, with no change in temperature. After about 40 kcal has been added at 0°C, half the ice remains and half has changed to water. After about 80 kcal, or 330 kJ, has been added, all the ice has changed to water, still at 0°C. Continued addition of heat causes the water’s temperature to again increase, now at a rate of1 C°/kcal. When 100°C is reached, the temperature again remains constant as the heat added changes the liquid water to vapor (steam). About 540 kcal (2260 kJ) is required to change the 1.0 kg of water completely to steam, after which the graph rises again, indicating that the temperature of the steam rises as heat is added.

The heat required to change 1.0 kg of a substance from the solid to the liquid state is called the heat of fusion; it is denoted by LF. The heat of fusion of water is79.7 kcal/kg or, in proper SI units, 333 kJ/kg (= 3.33 X 105 J/kg). The heat required to change a substance from the liquid to the vapor phase is called the heat of vaporization, Lv . For water it is 539 kcal/kg or 2260 kJ/kg. Other substances follow graphs similar to Fig. 19-5, although the melting-point and boiling-point temperatures are different, as are the specific heats and heats of fusion and vaporization. Values for the heats of fusion and vaporization, which are also called the latent heats, are given in Table 19-2 for a number of substances.

The heats of vaporization and fusion also refer to the amount of heat released by a substance when it changes from a gas to a liquid, or from a liquid to a solid. Thus, steam releases 2260 kJ/kg when it changes to water, and water releases 333 kJ/kg when it becomes ice.

The heat involved in a change of phase depends not only on the latent heat but also on the total mass of the substance. That is,

Q = mL, (19-3)where L is the latent heat of the particular process and substance, m is the mass of the

TABLE 1 9 -2 Latent Heats (at l atm)

Melting Point Heat of Fusion Boiling Point Heat of VaporizationSubstance (°C) kcal/kg1 kJ/kg (°C) kcal/kg1 kJ/kg

Oxygen -2 1 8 .8 3.3 14 -1 8 3 51 210Nitrogen -2 1 0 .0 6.1 26 -1 9 5 .8 48 200Ethyl alcohol -1 1 4 25 104 78 204 850Ammonia -7 7 .8 8.0 33 -3 3 .4 33 137Water 0 79.7 333 100 539 2260Lead 327 5.9 25 1750 208 870Silver 961 21 88 2193 558 2300Iron 1808 69.1 289 3023 1520 6340Tungsten 3410 44 184 5900 1150 4800

Numerical values in kcal/kg are the same in cal/g.

EXERCISE B A pot of water is boiling on a gas stove, and then you turn up the heat. What happens? (a) The temperature of the water starts increasing. (b ) There is a tiny decrease in the rate of water loss by evaporation, (c) The rate of water loss by boiling increases. ((d) There is an appreciable increase in both the rate of boiling and the temperature of the water. (e) N one of these.

Calorimetry sometimes involves a change of state, as the following Examples show. Indeed, latent heats are often measured using calorimetry.

EXAMPLE 19-5 Will all the ice melt? A 0.50-kg chunk of ice at — 10°C is placed in 3.0 kg of “iced” tea at 20°C. At what temperature and in what phase will the final mixture be? The tea can be considered as water. Ignore any heat flow to the surroundings, including the container.APPROACH Before we can write down an equation applying conservation of energy, we must first check to see if the final state will be all ice, a mixture of ice and water at 0°C, or all water. To bring the 3.0 kg of water at 20°C down to 0°C would require an energy release of (Eq. 19-2)

rawcw(20°C - 0°C) = (3.0kg)(4186J/k g • C°)(20C°) = 250kJ.On the other hand, to raise the ice from — 10°C to 0°C would require

micecice[0oC - (-10°C)] = (0.50kg)(2100J/kg-C°)(10C°) = 10.5kJ,and to change the ice to water at 0°C would require (Eq. 19-3)

miceLF = (0.50 kg) (333 kJ/kg) = 167 kJ,for a total of 10.5 kJ + 167 kJ = 177 kJ. This is not enough energy to bring the3.0 kg of water at 20°C down to 0°C, so we see that the mixture must end up all water, somewhere between 0°C and 20°C.SOLUTION To determine the final temperature T, we apply conservation of energy and write: heat gain = heat loss,

^heat to raise ̂ ^heat to change^ ( heat to raise ( heat lost by \0.50 kg of ice -L 0.50 kg -L 0.50 kg of water 3.0 kg offrom -10°C l of ice l from 0°C water cooling ̂ to 0°C j ̂ to water J tor j ^from 20°C to TJ

Using some of the results from above, we obtain10.5 kJ + 167kJ + (0.50kg)(4186 J/k g -C°){T - 0°C)

= (3.0kg)(4186J/kg• C°)(20°C - T).Solving for T we obtain

T = 5.0°C.

j P R O B L E M S O L V I N GFirst determine (or estimate) the final state

i P R O B L E M S O L V I N GThen determine the final temperature

Calorimetry1. Be sure you have sufficient information to apply

energy conservation. Ask yourself: is the systemp* isolated (or very nearly so, enough to get a good

estimate)? Do we know or can we calculate all significant sources of energy transfer?

2. Apply conservation of energy:heat gained = heat lost.

For each substance in the system, a heat (energy) term will appear on either the left or right side of this equation. [Alternatively, use 2 0 = 0.]

3. If no phase changes occur, each term in the energy conservation equation (above) will have the form

G(gain) = mc(Tf - 7-)or

G(lost) = mc(Ti - Tf) where Tx and Tf are the initial and final temperatures

of the substance, and m and c are its mass and specific heat, respectively.

4. If phase changes do or might occur, there may be terms in the energy conservation equation of the form Q = mL, where L is the latent heat. But before applying energy conservation, determine (or estimate) in which phase the final state will be, as we did in Example 19-5 by calculating the different contributing values for heat Q.

5. Be sure each term appears on the correct side of the energy equation (heat gained or heat lost) and that each A T is positive.

6. Note that when the system reaches thermal equilibrium, the final temperature of each substance will have the same value. There is only one 7f.

7. Solve your energy equation for the unknown.

Determining a latent heat. The specific heat of liquid mercury is 140 J/kg-C°. When 1.0 kg of solid mercury at its melting point of -39°C is placed in a 0.50-kg aluminum calorimeter filled with 1.2 kg of water at 20.0°C, the mercury melts and the final temperature of the combination is found to be 16.5°C. What is the heat of fusion of mercury in J/kg?APPROACH We follow the Problem Solving Strategy above.SOLUTION1. Is the system isolated? The mercury is placed in a calorimeter, which we

assume is well insulated. Our isolated system is the calorimeter, the water, and the mercury.

2. Conservation of energy. The heat gained by the mercury = the heat lost by the water and calorimeter.

3. and 4. Phase changes. There is a phase change (of mercury), plus we use specific heat equations. The heat gained by the mercury (Hg) includes a term representing the melting of the Hg,

Q(melt solid Hg) = mHgLHg,plus a term representing the heating of the liquid Hg from -39°C to +16.5°C:

Q(heat liquid Hg) = mHgcHg[16.5°C - (-39°C)]= (1.0 kg) (140 J/kg-C°) (55.5 C°) = 7770 J.

All of this heat gained by the mercury is obtained from the water and calorimeter, which cool down:Qcai + Qw = ^caiccai(20.0°C - 16.5°C) + rawcw(20.0°C - 16.5°C)

= (0.50 kg) (900 J/kg-C°) (3.5 C°) + (1.2 kg) (4186 J/kg- C°) (3.5 C°) = 19,200 J.

5. Energy equation. The conservation of energy tells us the heat lost by the water and calorimeter cup must equal the heat gained by the mercury:

<2cai + Qw = <2(melt solid Hg) + Q(heat liquid Hg)or

EXAMPLE 19-6

19,200 J = mHgLHg + 7770 J.

7. Solve. The only unknown in our energy equation (point 5) is LHg, the latent heat of fusion (or melting) of mercury. We solve for it, putting in raH g = 1.0 kg:

^ = 1 9 ,2 0 0 ^ 7 7 7 0 1 = 11,400 J/kg « llk J/k g ,

where we rounded off to 2 significant figures.

EvaporationThe latent heat to change a liquid to a gas is needed not only at the boiling point. Water can change from the liquid to the gas phase even at room temperature. This process is called evaporation (see also Section 18-4). The value of the heat of vaporization of water increases slightly with a decrease in temperature: at 20°C, for example, it is 2450kJ/kg (585kcal/kg) compared to 2260kJ/kg (= 539kcal/kg) at 100°C. When water evaporates, the remaining liquid cools, because the energy required (the latent heat of vaporization) comes from the water itself; so its internal energy, and therefore its temperature, must drop.*

Evaporation of water from the skin is one of the most important methods the body uses to control its temperature. When the temperature of the blood rises slightly above normal, the hypothalamus region of the brain detects this tempera­ture increase and sends a signal to the sweat glands to increase their production. The energy (latent heat) required to vaporize this water comes from the body, and hence the body cools.

Kinetic Theory of Latent HeatsWe can make use of kinetic theory to see why energy is needed to melt or vaporize a substance. At the melting point, the latent heat of fusion does not act to increase the average kinetic energy (and the temperature) of the molecules in the solid, but instead is used to overcome the potential energy associated with the forces between the molecules. That is, work must be done against these attractive forces to break the molecules loose from their relatively fixed positions in the solid so they can freely roll over one another in the liquid phase. Similarly, energy is required for molecules held close together in the liquid phase to escape into the gaseous phase. This process is a more violent reorganization of the molecules than is melting (the average distance between the molecules is greatly increased), and hence the heat of vaporization is generally much greater than the heat of fusion for a given substance.

19—6 The First Law of ThermodynamicsUp to now in this Chapter we have discussed internal energy and heat. But work too is often involved in thermodynamic processes.

In Chapter 8 we saw that work is done when energy is transferred from one object to another by mechanical means. In Section 19-1 we saw that heat is a transfer of energy from one object to a second one at a lower temperature. Thus, heat is much like work. To distinguish them, heat is defined as a transfer o f energy due to a difference in temperature, whereas work is a transfer of energy that is not due to a temperature difference.

In Section 19-2, we defined the internal energy of a system as the sum total of all the energy of the molecules within the system. We would expect that the internal energy of a system would be increased if work was done on the system, or if heat were added to it. Similarly the internal energy would be decreased if heat flowed out of the system or if work were done by the system on something in the surroundings.

tAccording to kinetic theory, evaporation is a cooling process because it is the fastest-moving moleculestV ia t AO panA f r n m f l i p c n r fa p p LTprir ’A t l iA a irA ra rrp on A A rl n f f l ip r p m a in in a m n lp p n lp c ie Ip cg c n In n 1 8 _A

0 P H Y S I C S A P P L I E DB ody temperature

FIRST LAW OF THERMODYNAMICS

H eat added is + Heat lost is -

Work on system is — W ork by system is +

Thus it is reasonable to extend conservation of energy and propose an impor­tant law: the change in internal energy of a closed system, AEint, will be equal to the energy added to the system by heating minus the work done by the system on the surroundings. In equation form we write

A-Eint = Q - W (19-4)

where Q is the net heat added to the system and W is the net work done by the system.* We must be careful and consistent in following the sign conventions for Q and W. Because W in Eq. 19-4 is the work done by the system, then if work is done on the system, W will be negative and Emt will increase. Similarly, Q is positive for heat added to the system, so if heat leaves the system, Q is negative.

Equation 19-4 is known as the first law of thermodynamics. It is one of the great laws of physics, and its validity rests on experiments (such as Joule’s) to which no exceptions have been seen. Since Q and W represent energy transferred into or out of the system, the internal energy changes accordingly. Thus, the first law of thermodynamics is a great and broad statement of the law o f conservation o f energy.

It is worth noting that the conservation of energy law was not formulated until the nineteenth century, for it depended on the interpretation of heat as a transfer of energy.

Equation 19-4 applies to a closed system. It also applies to an open system (Section 19-4) if we take into account the change in internal energy due to the increase or decrease in the amount of matter. For an isolated system (p. 500), no work is done and no heat enters or leaves the system, so W = Q = 0, and hence A £ in , = 0 .

A given system at any moment is in a particular state and can be said to have a certain amount of internal energy, -Eint • But a system does not “have” a certain amount of heat or work. Rather, when work is done on a system (such as compressing a gas), or when heat is added or removed from a system, the state of the system changes. Thus, work and heat are involved in thermodynamic processes that can change the system from one state to another; they are not characteristic of the state itself. Quantities which describe the state of a system, such as internal energy E-mt, pressure P, volume V, temperature T, and mass m or number of moles n, are called state variables. Q and W are not state variables.

Because Eint is a state variable, which depends only on the state of the system and not on how the system arrived in that state, we can write

AEint = -Eint, 2 — Eint = Q — W

where Eint, 1 and Eint, 2 represent the internal energy of the system in states 1 and 2, and Q and W are the heat added to the system and work done by the system in going from state 1 to state 2.

It is sometimes useful to write the first law of thermodynamics in differential form:

dEmt = dQ - dW.

Here, dE-mX represents an infinitesimal change in internal energy when an infinitesimal amount of heat dQ is added to the system, and the system does an infinitesimal amount of work dW}

trrhis convention relates historically to steam engines: the interest was in the heat input and the work output, both regarded as positive. In other books you may see the first law of thermodynamics written as A£int = Q + W, in which case W refers to the work done on the system.*The differential form of the first law is often written

dEint = r n - aw,where the bars on the differential sign (d) are used to remind us that W and Q are not functions of the state variables (such as P, V, T, n). Internal energy, E-mt, is a function of the state variables, and so dEmi represents the differential (called an exact differential) of some function E-m\ • The differentials dW and

EXAMPLE 19-7 Using the first law. 2500 J of heat is added to a system, and 1800 J of work is done on the system. What is the change in internal energy of the system?APPROACH We apply the first law of thermodynamics, Eq. 19-4, to our system. SOLUTION The heat added to the system is Q = 2500 J. The work W done by the system is -1800 J. Why the minus sign? Because 1800 J done on the system (as given) equals -1800 J done by the system, and it is the latter we need for the sign conventions we used in Eq. 19-4. Hence

AEint = 2500 J - (-1800 J) = 2500 J + 1800 J = 4300 J.You may have intuitively thought that the 2500 J and the 1800 J would need to be added together, since both refer to energy added to the system. You would have been right.

EXERCISE D What would be the internal energy change in Example 19 -7 if 2500 J of heat is added to the system and 1800 J of work is done by the system (i.e., as output)?

*The First Law of Thermodynamics ExtendedTo write the first law of thermodynamics in its complete form, consider a system that has kinetic energy K (there is motion) as well as potential energy U. Then the first law of thermodynamics would have to include these terms and would be written as

AK + A U + AEint = Q - W. (19-5)

EXAMPLE 19-8 Kinetic energy transformed to thermal energy. A 3.0-g bullet traveling at a speed of 400 m/s enters a tree and exits the other side with a speed of 200 m/s. Where did the bullet’s lost kinetic energy go, and what was the energy transferred?APPROACH Take the bullet and tree as our system. No potential energy is involved. No work is done on (or by) the system by outside forces, nor is any heat added because no energy was transferred to or from the system due to a temperature difference. Thus the kinetic energy gets transformed into internal energy of the bullet and tree. SOLUTION From the first law of thermodynamics as given in Eq. 19-5, we are given Q = W = A U = 0, so we have

A K + A£int = 0 or, using subscripts i and f for initial and final velocities

A-Eint = -A X = -(X f - Ki) = \m (v\ - vj)= |(3.0 X l(T3kg)[(400m/s)2 - (200m/s)2] = 180J.

NOTE The internal energy of the bullet and tree both increase, as both experience a rise in temperature. If we had chosen the bullet alone as our system, work would be done on it and heat transfer would occur.

1 9 -7 The First Law of Thermodynamics Applied; Calculating the Work

Let us analyze some simple processes in the light of the first law of thermodynamics.

Isothermal Processes (AT = 0)First we consider an idealized process that is carried out at constant temperature. Such a process is called an isothermal process (from the Greek meaning “same temperature”). If the system is an ideal gas, then PV = nRT (Eq. 17-3), so for a fixed amount of gas kept at constant temperature, PV = constant. Thus the process follows a curve like AB on the PV diagram shown in Fig. 19-6, which is a curve for PV = constant. Each point on the curve, such as point A, represents the state of the system at a given moment—that is, its pressure P and volume V. At a lower temperature, another isothermal process would be represented by a curve

FIGURE 19-6 PV diagram for an ideal gas undergoing isothermal processes at two different temperatures.

FIGURE 19-7 A n ideal gas in a cylinder fitted with a movable piston.

^ I s o t h e r m a l

Adiabatic''"* ̂

FIGURE 19-8 P V diagram for adiabatic (AC) and isothermal (A B ) processes on an ideal gas.

FIGURE 19-9 (a) Isobaric (“same pressure”) process, (b) Isovolumetric (“same volume”) process.

A B

V(a) Isobaric

0 V(b) Isovolumetric

FIGURE 19-10 The work done by a gas when its volume increases by d V = A d i is dW = P dV.

di

Let us assume that the gas is enclosed in a container fitted with a movable piston, Fig. 19-7, and that the gas is in contact with a heat reservoir (a body whose mass is so large that, ideally, its temperature does not change significantly when heat is exchanged with our system). We also assume that the process of compression (volume decreases) or expansion (volume increases) is done quasistatically (“almost statically”), by which we mean extremely slowly, so that all of the gas moves between a series of equilibrium states each of which are at the same constant temperature. If an amount of heat Q is added to the system and temperature is to remain constant, the gas will expand and do an amount of work W on the environment (it exerts a force on the piston and moves it through a distance). The temperature and mass are kept constant so, from Eq. 19-1, the internal energy does not change: A£jnt = \nR AT = 0. Hence, by the first law of thermodynamics, Eq. 19-4, AE-mt = Q — W = 0, so W = Q: the work done by the gas in an isothermal process equals the heat added to the gas.

Adiabatic Processes (Q = 0)An adiabatic process is one in which no heat is allowed to flow into or out of the system: Q = 0. This situation can occur if the system is extremely well insulated, or the process happens so quickly that heat—which flows slowly—has no time to flow in or out. The very rapid expansion of gases in an internal combustion engine is one example of a process that is very nearly adiabatic. A slow adiabatic expansion of an ideal gas follows a curve like that labeled AC in Fig. 19-8. Since Q = 0, we have from Eq. 19-4 that AEint = —W. That is, the internal energy decreases if the gas expands; hence the temperature decreases as well (because AEint = \nR AT).This is evident in Fig. 19-8 where the product PV (= nRT) is less at point C than at point B (curve AB is for an isothermal process, for which AEint = 0 and AT = 0). In the reverse operation, an adiabatic compression (going from C to A, for example), work is done on the gas, and hence the internal energy increases and the temperature rises. In a diesel engine, the fuel-air mixture is rapidly compressed adiabatically by a factor of 15 or more; the temperature rise is so great that the mixture ignites spontaneously.

Isobaric and Isovolumetric ProcessesIsothermal and adiabatic processes are just two possible processes that can occur. Two other simple thermodynamic processes are illustrated on the PV diagrams of Fig. 19-9: (a) an isobaric process is one in which the pressure is kept constant, so the process is represented by a horizontal straight line on the PV diagram, Fig. 19-9a; (b) an isovolumetric (or isochoric) process is one in which the volume does not change (Fig. 19-9b). In these, and in all other processes, the first law of thermodynamics holds.

Work Done in Volume ChangesWe often want to calculate the work done in a process. Suppose we have a gas confined to a cylindrical container fitted with a movable piston (Fig. 19-10). We must always be careful to define exactly what our system is. In this case we choose our system to be the gas; so the container’s walls and the piston are parts of the environment. Now let us calculate the work done by the gas when it expands quasistatically, so that P and T are defined for the system at all instants.f The gas expands against the piston, whose area is A. The gas exerts a force F = PA on the piston, where P is the pressure in the gas. The work done by the gas to move the piston an infinitesimal displacement dl is

dW = F-dl = P A di = P dV (19-6)

since the infinitesimal increase in volume is dV = A dl. If the gas was compressed so that di pointed into the gas, the volume would decrease and dV < 0. The work done by the gas in this case would then be negative, which is equivalent to saying that positive work was done on the gas, not by it. For a finite change in volume

from VA to VB, the work W done by the gas will be

W = jrfW = | "pdV. (19-7)

Equations 19-6 and 19-7 are valid for the work done in any volume change—by a gas, a liquid, or a solid—as long as it is done quasistatically.

In order to integrate Eq. 19-7, we need to know how the pressure varies during the process, and this depends on the type of process. Let us first consider a quasistatic isothermal expansion of an ideal gas. This process is represented by the curve between points A and B on the PV diagram of Fig. 19-11. The work done by the gas in this process, according to Eq. 19-7, is just the area between the PV curve and theV axis, and is shown shaded in Fig. 19-11. We can do the integral in Eq. 19-7 for an ideal gas by using the ideal gas law, P = nRT/V. The work done at constant T is

w = \ V"pdV = nRT I*' — = nRTln^- [isothermalprocess;] (W_8) JvA JvA y Va LiaealSas J

Let us next consider a different way of taking an ideal gas between the same states A and B. This time, let us lower the pressure in the gas from PA to PB, as indicated by the line AD in Fig. 19-12. (In this isovolumetric process, heat must be allowed to flow out of the gas so its temperature drops.) Then let the gas expand from VA to at constant pressure (= PB), which is indicated by the line DB in Fig. 19-12. (In this isobaric process, heat is added to the gas to raise its temperature.) No work is done in the isovolumetric process AD, since dV = 0:

W = 0. [isovolumetric process]In the isobaric process DB the pressure remains constant, so

fVBW = P dV = PB{VB - >A.) = P AF. [isobaric process] (19-9a)

JVaThe work done is again represented on the PV diagram by the area between the curve (ADB) and the V axis, as indicated by the shading in Fig. 19-12. Using the ideal gas law, we can also write

w = P jv n - VA) = n R lJy l - J spr0CeSS;] (19-9b)

As can be seen from the shaded areas in Figs. 19-11 and 19-12, or by putting in numbers in Eqs. 19-8 and 19-9 (try it for VB = 2VA ), the work done in these two processes is different. This is a general result. The work done in taking a system from one state to another depends not only on the initial and final states but also on the type o f process (or “path”).

This result reemphasizes the fact that work cannot be considered a property of a system. The same is true of heat. The heat input required to change the gas from state A to state B depends on the process; for the isothermal process of Fig. 19-11, the heat input turns out to be greater than for the process ADB of Fig. 19-12. In general, the amount o f heat added or removed in taking a system from one state to another depends not only on the initial and final states but also on the path or process.

CONCEPTUAL EXAMPLE 19-9~| Work in isothermal and adiabatic processes.In Fig. 19-8 we saw the PV diagrams for a gas expanding in two ways, isothermally and adiabatically. The initial volume VA was the same in each case, and the final volumes were the same (VB = Vc). In which process was more work done by the gas? RESPONSE Our system is the gas. More work was done by the gas in the isothermal process, which we can see in two simple ways by looking at Fig. 19-8. First, the “average” pressure was higher during the isothermal process AB, so W = P AV was greater (AV is the same for both processes). Second, we can look at the area under each curve: the area under curve AB, which represents the work done, was greater (since curve AB is higher) than that under AC.

P

FIGURE 19-11 Work done by an ideal gas in an isothermal process equals the area under the P V curve. Shaded area equals the work done by the gas when it expands from VA to V B.

FIGURE 1 9 -1 2 Process A D B consists of an isovolumetric (A D ) and an isobaric (D B ) process.

P

\ A

O I

Isov

olum

etric

I I ----- 1-----1-----1-----1-----1—0 2 4 6 8 10 V (L )

FIGURE 19-13 Example 19-10.

First law in isobaric and isovolumetric processes. An ideal gas is slowly compressed at a constant pressure of 2.0 atm from 10.0 L to 2.0 L. This process is represented in Fig. 19-13 as the path B to D. (In this process, some heat flows out of the gas and the temperature drops.) Heat is then added to the gas, holding the volume constant, and the pressure and temperature are allowed to rise (line DA) until the temperature reaches its original value (TA = TB). Calculate (a) the total work done by the gas in the process BDA, and (b) the total heat flow into the gas.APPROACH (a) Work is done only in the compression process BD. In process DA, the volume is constant so AV = 0 and no work is done. (b) We use the first law of thermodynamics, Eq. 19-4.SOLUTION (a) During the compression BD, the pressure is 2.0 atm = 2(1.01 X 105 N/m2) and the work done is (since 1 L = 103 cm3 = 10-3 m3)

W = P AV = (2.02 X 105N/m2)(2.0 X 10-3m3 - 10.0 X 10-3m3) = -1.6 X 103 J.

The total work done by the gas is -1.6 X 103 J, where the minus sign means that +1.6 X 103 J of work is done on the gas.(b) Because the temperature at the beginning and at the end of process BDA is the same, there is no change in internal energy: AE mt = 0. From the first law of thermodynamics we have

0 = A Eint = Q - Wso Q = W = -1.6 X 103 J. Because Q is negative, 1600 J of heat flows out of the gas for the whole process, BDA.

EXERCISE F In Example 19-10, if the heat lost from the gas in the process B D is8.4 X 103 J, what is the change in internal energy of the gas during process BD?

Work done in an engine. In an engine, 0.25 mol of an ideal monatomic gas in the cylinder expands rapidly and adiabatically against the piston. In the process, the temperature of the gas drops from 1150 K to 400 K. How much work does the gas do?APPROACH We take the gas as our system (the piston is part of the surroundings). The pressure is not constant, and its varying value is not given. Instead, we can use the first law of thermodynamics because we can determine A£int given Q = 0 (the process is adiabatic).SOLUTION We determine A£int from Eq. 19-1 for the internal energy of an ideal monatomic gas, using subscripts f and i for final and initial states:

AEint = £int,f - -Bin,,! = \n R {T t - 71)= I (0.25 mol) (8.314 J/mol-K) (400 K - 1150 K) = -2300 J.

Then, from the first law of thermodynamics, Eq. 19-4,W = Q - AEint = 0 - (-2300J) = 2300J.

Table 19-3 gives a brief summary of the processes we have discussed.

Free ExpansionOne type of adiabatic process is a so-called free expansion in which a gas is allowed to expand in volume adiabatically without doing any work. The apparatus to

TABLE 19-3 Simple Thermodynamic Processes and the First LawProcess What is constant: The first law predicts:

Isothermal T = constant AT = 0 makes A E ^ = 0, so Q = WT> — + /O — A TP. _L W — A 77. _1_ P A T /

EXAMPLE 19-11

EXAMPLE 19-10

accomplish a free expansion is shown in Fig. 19-14. It consists of two well-insulated compartments (to ensure no heat flow in or out) connected by a valve or stopcock.One compartment is filled with gas, the other is empty. When the valve is opened, the gas expands to fill both containers. No heat flows in or out (Q = 0), and no work is done because the gas does not move any other object. Thus Q = W = 0 and by the first law of thermodynamics, A Eint = 0. The internal energy o f a gas does not change in a free expansion. For an ideal gas, AT = 0 also, since Eint depends FIGURE 19-14 Free expansion, only on T (Section 19-2). Experimentally, the free expansion has been used to determine if the internal energy of real gases depends only on T. The experiments are very difficult to do accurately, but it has been found that the temperature of a real gas drops very slightly in a free expansion. Thus the internal energy of real gases does depend, a little, on pressure or volume as well as on temperature.

A free expansion can not be plotted on a PV diagram, because the process is rapid, not quasistatic. The intermediate states are not equilibrium states, and hence the pressure (and even the volume at some instants) is not clearly defined.

19—8 Molar Specific Heats for Gases, and the Equipartition of Energy

In Section 19-3 we discussed the concept of specific heat and applied it to solids and liquids. Much more than for solids and liquids, the values of the specific heat for gases depends on how the process is carried out. Two important processes are those in which either the volume or the pressure is kept constant. Although for solids and liquids it matters little, Table 19-4 shows that the specific heats of gases at constant volume (cy) and at constant pressure (cP) are quite different.

Molar Specific Heats for GasesThe difference in specific heats for gases is nicely explained in terms of the first law of thermodynamics and kinetic theory. Our discussion is simplified if we use molar specific heats, Cv and CP, which are defined as the heat required to raise 1 mol of the gas by 1 C° at constant volume and at constant pressure, respectively. That is, in analogy to Eq. 19-2, the heat Q needed to raise the temperature of n moles of gas by AT is

Q = nCyAT [volume constant] (19-10a)Q = nCpAT. [pressure constant] (19-10b)

It is clear from the definition of molar specific heat (or by comparing Eqs. 19-2 and 19-10) that

Cy = A/ Cy

Cp = Mcp,where M is the molecular mass of the gas (M = m /n in grams/mol). The values for molar specific heats are included in Table 19-4, and we see that the values are nearly the same for different gases that have the same number of atoms per molecule.

TABLE 19-4 Specific Heats of Gases at 15°CSpecific heats Molar specific heats(kcal/kg • K) (cal/mol • K) C„ - C „ CP

Gas C y Cp C y Cp (cal/mol • K) C y

M onatomicH e 0.75 1.15 2.98 4.97 1.99 1.67N e

Diatomic0.148 0.246 2.98 4.97 1.99 1.67

n 2 0.177 0.248 4.96 6.95 1.99 1.40

o 2 0.155 0.218 5.03 7.03 2.00 1.40

FIGURE 1 9 -1 5 A diatomic m olecule can rotate about two different axes.

Axis

Now we use kinetic theory and imagine that an ideal gas is slowly heated via two different processes—first at constant volume, and then at constant pressure. In both of these processes, we let the temperature increase by the same amount, AT. In the process done at constant volume, no work is done since AV = 0. Thus, according to the first law of thermodynamics, the heat added (which we denote by Qv) all goes into increasing the internal energy of the gas:

Qv = A E-mi.In the process carried out at constant pressure, work is done, and hence the heat added, QP, must not only increase the internal energy but also is used to do the work W = PAV. Thus, more heat must be added in this process at constant pressure than in the first process at constant volume. For the process at constant pressure, we have from the first law of thermodynamics

Qp = A Eint + P AV.Since AE^ is the same in the two processes (AT was chosen to be the same), we can combine the two above equations:

QP - Q V = P AV.From the ideal gas law, V = nRT/P, so for a process at constant pressure we have AV = nR AT/P. Putting this into the above equation and using Eqs. 19-10, we find

nCPAT - nCv AT = p ( ^ y ^ J

or, after cancellations,CP - Cv = R. (19-11)

Since the gas constant R = 8.314 J/mol-K = 1.99 cal/mol-K, our prediction is that CP will be larger than Cv by about 1.99 cal/mol • K. Indeed, this is very close to what is obtained experimentally, as shown in the next to last column of Table 19-4.

Now we calculate the molar specific heat of a monatomic gas using kinetic theory. In a process carried out at constant volume, no work is done; so the first law of thermodynamics tells us that if heat Q is added to the gas, the internal energy of the gas changes by

A £ i n t = f i .

For an ideal monatomic gas, the internal energy -Eint is the total kinetic energy of all the molecules,

-Eint = N (jm v2) = I nRT as we saw in Section 19-2. Then, using Eq. 19-10a, we can write AE'int = Q in the form

A£int = \n R A T = nCv AT (19-12)or

Cv = \R . (19-13)Since R = 8.314J/mol-K = 1.99cal/mol-K, kinetic theory predicts that Cv = 2.98cal/mol-K for an ideal monatomic gas. This is very close to the experimental values for monatomic gases such as helium and neon (Table 19-4). From Eq. 19-11, CP is predicted to be about 4.97 cal/mol-K, also in agreement with experiment.

Equipartition of EnergyThe measured molar specific heats for more complex gases (Table 19-4), such as diatomic (two atoms) and triatomic (three atoms) gases, increase with the increased number of atoms per molecule. We can explain this by assuming that the internal energy includes not only translational kinetic energy but other forms of energy as well. Take, for example, a diatomic gas. As shown in Fig. 19-15 the two atoms can rotate about two different axes (but rotation about a third axis passing through the two atoms would give rise to very little energy since the moment of inertia is so small). The molecules can have rotational as well as translational kinetic energy. It is useful to introduce the idea of degrees of freedom, by which we mean the number of independent ways molecules can possess energy. For example, a monatomic gas is said to have three degrees of freedom, since an atom can have velocity alone the x axis, the

energy plus two more degrees of freedom associated with rotational kinetic energy, for a total of five degrees of freedom. A quick look at Table 19-4 indicates that the Cy for diatomic gases is about f times as great as for a monatomic gas—that is, in the same ratio as their degrees of freedom. This result led nineteenth-century physicists to an important idea, the principle of equipartition of energy. This principle states that energy is shared equally among the active degrees of freedom, and in particular each active degree of freedom of a molecule has on the average an energy equal to \k T . Thus, the average energy for a molecule of a monatomic gas would be \ k T (which we already knew) and of a diatomic gas f kT. Hence the internal energy of a diatomic gas would be EMt = N (^kT) = f nRT, where n is the number of moles. Using the same argument we did for monatomic gases, we see that for diatomic gases the molar specific heat at constant volume would be ^R = 4.97 cal/mol-K, in accordance with measured values. More complex molecules have even more degrees of freedom and thus greater molar specific heats.

The situation was complicated, however, by measurements that showed that for diatomic gases at very low temperatures, Cv has a value of only § R, as if it had only three degrees of freedom. And at very high temperatures, Cv was about \ R, as if there were seven degrees of freedom. The explanation is that at low temperatures, nearly all molecules have only translational kinetic energy. That is, no energy goes into rotational energy, so only three degrees of freedom are “active.” At very high temperatures, on the other hand, all five degrees of freedom are active plus two additional ones. We can interpret the two new degrees of freedom as being associated with the two atoms vibrating as if they were connected by a spring, as shown in Fig. 19-16. One degree of freedom comes from the kinetic energy of the vibrational motion, and the second comes from the potential energy of vibrational motion Q kx 2). At room temperature, these two degrees of freedom are apparently not active. See Fig. 19-17.

Q m Q

FIGURE 19-16 A diatom ic m olecule can vibrate, as if the two atoms w ere connected by a spring. O f course they are not connected by a spring; rather they exert forces on each other that are electrical in nature, but o f a form that resem bles a spring force.

4R +C

3 R -

2R --

R -

l RVibrational kinetic energy

Rotational kinetic energy

Translational kinetic energy

20 50 100 200 500 1000 2000 5000 10,000h— r(K)

FIGURE 19-17 M olar specific heat C v as a function o f temperature for hydrogen m olecules (H 2). A s the temperature is increased, som e o f the translational kinetic energy can be transferred in collisions into rotational kinetic energy and, at still higher temperature, into vibrational kinetic energy. [Note: H 2 dissociates into two atoms at about 3200 K, so the last part o f the curve is shown dashed .1

Just why fewer degrees of freedom are “active” at lower temperatures was eventually explained by Einstein using the quantum theory. [According to quantum theory, energy does not take on continuous values but is quantized—it can have only certain values, and there is a certain minimum energy. The minimum rotational and vibrational energies are higher than for simple translational kinetic energy, so at lower temperatures and lower translational kinetic energy, there is not enough energy to excite the rotational or vibrational kinetic energy.] Calculations based on kinetic theory and the principle of equipartition of energy (as modified by the quantum theory) give numerical results in accord with experiment.

* SolidsThe principle of equipartition of energy can be applied to solids as well. The molar specific heat of any solid, at high temperature, is close to 3R (6.0cal/mol-K), Fig. 19-18. This is called the Dulong and Petit value after the scientists who first measured it in 1819. (Note that Table 19-1 gives the specific heats per kilogram, not per mole.) At high temperatures, each atom apparently has six degrees of freedom, although some are not active at low temperatures. Each atom in a crystalline solid can vibrate about its eauilibrium position as if it were connected

FIGURE 19-18 M olar specific heats o f solids as a function o f temperature.

Copper

0 100 300 500 700 900

19—9 Adiabatic Expansion of a GasThe PV curve for the quasistatic (slow) adiabatic expansion (Q = 0) of an ideal gas was shown in Fig. 19-8 (curve AC). It is somewhat steeper than for an isothermal process (AT =0) , which indicates that for the same change in volume the change in pressure will be greater. Hence the temperature of the gas must drop during an adia­batic expansion. Conversely, the temperature rises during an adiabatic compression.

We can derive the relation between the pressure P and the volume V of an ideal gas that is allowed to slowly expand adiabatically. We begin with the first law of thermodynamics, written in differential form:

dEmX = dQ - dW = -d W = -P d V , since dQ = 0 for an adiabatic process. Equation 19-12 gives us a relation between A E^ and Cv , which is valid for any ideal gas process since -̂ int is a function only of T for an ideal gas. We write this in differential form:

dEmi = nCv dT.When we combine these last two equations, we obtain

nCv dT + P dV = 0.We next take the differential of the ideal gas law, PV = nRT, allowing P, V, and T to vary:

P dV + VdP = nRdT.We solve for dT in this relation and substitute it into the previous relation and get

„ , P dV + VdP . nCv \ -------- —-------I + P dV = 0

or, multiplying through by R and rearranging,(CV + R)PdV + Cv V dP = 0.

We note from Eq. 19-11 that Cv + R = CP, so we have CpPdV + Cv VdP = 0,

or

We define

CP-z^P dV + VdP = 0.Cy

Cpy = (w-14)Uy

so that our last equation becomes dP dV p + 7 V ~ '

This is integrated to become\nP + 7 \n V = constant.

This simplifies (using the rules for addition and multiplication of logarithms) to

PVy = constant. [q uasistaticadiabatic] / i q _ i c \ process; ideal gas J ̂ '

This is the relation between P and V for a quasistatic adiabatic expansion or contraction. We will find it very useful when we discuss heat engines in the next Chapter. Table 19-4 (p. 511) gives values of 7 for some real gases. Figure 19-8 compares an adiabatic expansion (Eq. 19-15) in curve AC to an isothermal expansion (PV = constant) in curve AB. It is important to remember that the ideal gas law, PV = nRT, continues to hold even for an adiabatic expansion (PVy = constant); clearly PV is not constant, meaning T is not constant.

Compressing an ideal gas. An ideal monatomic gas is compressed starting at point A in the PV diagram of Fig. 19-19, where PA = 100 kPa, VA = 1.00 m3, and TA = 300 K. The gas is first compressed adia- baticallv to state B (Pa = 200kPa>). The gas is then further compressed from

APPROACH Volume VB is obtained using Eq. 19-15. The work done by a gas is given by Eq. 19-7, W = J PdV. The work on the gas is the negative of this: Won = - J PdV. SOLUTION In the adiabatic process, Eq. 19-15 tells us PVy = constant. Therefore, PV '1 = PAV yA = PBV l where for a monatomic gas 7 = CP/CV = (5/2)/(3/2) = f.(a) Eq. 19-15 gives VB = VA(PJPBfr = (l.00m3)(l00kPa/200kPa)! = 0.66 m3.(b) The pressure P at any instant during the adiabatic process is given by P = Pa V aV~7. The work done on the gas in going from VA to VB is

WAB == - [ P d v = - p Av \ f BJA JVA

V y dV = -P AV l1

■7 + 1 ( n - y

Since 7 = §.

=

then - 7 + 1 = 1 — 7 = —| , so

PAv j | iiva 1| = + 2 P* V* ~~ I — 1

= + -(1 0 0 kPa) (1.00 m3) | (0.66)[(0.66)"*- l ] = +48 kJ.

For the isothermal process from B to C, the work is done at constant temperature, so the pressure at any instant during the process is P = nRTB/V and

c dV Vc Vc— = -nRT„\ n-r- = - P « K ln - # = +37 kJ.IPdV = -nRTr f JVB

The total work done on the gas is 48 kJ + 37 kJ = 85 kJ.

19—10 Heat Transfer: Conduction, Convection, Radiation

Heat is transferred from one place or body to another in three different ways: by conduction, convection, and radiation. We now discuss each of these in turn; but in practical situations, any two or all three may be operating at the same time. We start with conduction.

ConductionWhen a metal poker is put in a hot fire, or a silver spoon is placed in a hot bowl of soup, the exposed end of the poker or spoon soon becomes hot as well, even though it is not directly in contact with the source of heat. We say that heat has been conducted from the hot end to the cold end.

Heat conduction in many materials can be visualized as being carried out via molecular collisions. As one end of an object is heated, the molecules there move faster and faster. As they collide with slower-moving neighbors, they transfer some of their kinetic energy to these other molecules, which in turn transfer energy by collision with molecules still farther along the object. In metals, collisions of free electrons are mainly responsible for conduction.

Heat conduction from one point to another takes place only if there is a difference in temperature between the two points. Indeed, it is found experimentally that the rate of heat flow through a substance is proportional to the difference in temperature between its ends. The rate of heat flow also depends on the size and shape of the object. To investigate this quantitatively, let us consider the heat flow through a uniform cylinder, as illustrated in Fig. 19-20. It is found experimentally that the heat flow A Q over a time interval At is given by the relation

AQ At

= kA (19-16a)

where A is the cross-sectional area of the object, i is the distance between the two ends, which are at temperatures Tx and T2, and A: is a proportionality constantp q I I p H f l i p f h u r m Q l p n n r l i i p f l v i t i ; Yx/Viir*Vi i c p V i Q r Q p f p r i c t i p r \ f f l i p m a t e r i a l T 7 r r \m

Isothermal

0 0.50 1.00

FIGURE 19-19 Example 19-12

V (m3)

Hotter Cooler

Ti

A

t2

h r /Heat J flowI

1/ v

J-------1 --------H

FIGURE 19-20 Heat conduction between areas at temperatures Ti and T2 . If Ti is greater than T2, the heat flows to the right; the rate is given by Eq. 19-16a.

TABLE 19-5 Thermal Conductivities

Substance

Thermal conductivity, k kcal J

Silver Copper Aluminum Steel Ice Glass Brick Concrete Water Human tissue 0.5 Wood 0.3Fiberglass 0.12 Cork 0.1Wool 0.1G oose down 0.06 Polyurethane 0.06 Air 0.055

( sm-C°) ( sm-C°)

420 380 200

40 20.84 0.84 0.84 0.56 0.2 0.1 0.048 0.042 0.040 0.025 0.024 0.023

X 10“2 X 10“2 X 10“2 X 10“2 X 10“4 X 10“4 X 10“4 X 10“4 X 10“4 X 10“4 X 10“4 X 10“4 X 10“4 X 10“4 X 10“4 X 10“4 X 10“4

FIGURE 19-21 Example 19-13.

(^ P H Y S I C S A P P L I E DThermal w indow s

In some cases (such as when k or A cannot be considered constant) we need to consider the limit of an infinitesimally thin slab of thickness dx. Then Eq. 19-16a becomes

¥■ -dt dx (19-16b)

where dT/dx is the temperature gradient and the negative sign is included since the heat flow is in the direction opposite to the temperature gradient.

The thermal conductivities, k, for a variety of substances are given in Table 19-5. Substances for which k is large conduct heat rapidly and are said to be good thermal conductors. Most metals fall in this category, although there is a wide range even among them, as you may observe by holding the ends of a silver spoon and a stainless-steel spoon immersed in the same hot cup of soup. Substances for which k is small, such as wool, fiberglass, polyurethane, and goose down, are poor conductors of heat and are therefore good thermal insulators.

The relative magnitudes of k can explain simple phenomena such as why a tile floor is much colder on the feet than a rug-covered floor at the same temperature. Tile is a better conductor of heat than the rug. Heat that flows from your foot to the rug is not conducted away rapidly, so the rug’s surface quickly warms up to the temper­ature of your foot and feels good. But the tile conducts the heat away rapidly and thus can take more heat from your foot quickly, so your foot’s surface temperature drops.

EXAMPLE 19-13 Heat loss through windows. A major source of heat loss from a house is through the windows. Calculate the rate of heat flow through a glass window 2.0 m X 1.5 m in area and 3.2 mm thick, if the temperatures at the inner and outer surfaces are 15.0°C and 14.0°C, respectively (Fig. 19-21).APPROACH Heat flows by conduction through the glass from the higher inside temperature to the lower outside temperature. We use the heat conduction equation, Eq. 19-16a.SOLUTION Here A = (2.0m)(1.5m) = 3.0m2 and I = 3.2 X 10_3m. Using Table 19-5 to get k, we have

AQ At

= kA71 - T2 (0.84 J/s • m • C°)(3.0 m2)(15.0°C - 14.0°C)

(3.2 X 10-3 m= 790 J/s.

NOTE This rate of heat flow is equivalent to (790 J/s)/(4.19 X 103J/kcal) = 0.19kcal/s, or (0.19kcal/s) X (3600 s/h) = 680kcal/h.

You might notice in Example 19-13 that 15°C is not very warm for the living room of a house. The room itself may indeed be much warmer, and the outside might be colder than 14°C. But the temperatures of 15°C and 14°C were specified as those at the window surfaces, and there is usually a considerable drop in temperature of the air in the vicinity of the window both on the inside and the outside. That is, the layer of air on either side of the window acts as an insulator, and normally the major part of the temperature drop between the inside and outside of the house takes place across the air layer. If there is a heavy wind, the air outside a window will constantly be replaced with cold air; the temperature gradient across the glass will be greater and there will be a much greater rate of heat loss. Increasing the width of the air layer, such as using two panes of glass separated by an air gap, will reduce the heat loss more than simply increasing the glass thickness, since the thermal conductivity of air is much less than that for glass.

The insulating properties of clothing come from the insulating properties of air. Without clothes, our bodies in still air would heat the air in contact with the skin and would soon become reasonably comfortable because air is a very good insulator.

Equations 19-16 are quite similar to the relations describing diffusion (Section 18-7) and the flow of

But since air moves—there are breezes and drafts, and people move about—the warm air would be replaced by cold air, thus increasing the temperature difference and the heat loss from the body. Clothes keep us warm by trapping air so it cannot move readily. It is not the cloth that insulates us, but the air that the cloth traps. Goose down is a very good insulator because even a small amount of it fluffs up and traps a great amount of air.

[For practical purposes the thermal properties of building materials, particularly when considered as insulation, are usually specified by i?-values (or “thermal resistance”), defined for a given thickness £ of material as:

R = i/k .The i?-value of a given piece of material combines the thickness £ and the thermal conductivity k in one number. In the United States, R -values are given in British units as ft2 • h • F°/Btu (for example, R-19 means R = 19 ft2 • h • F°/Btu ). Table 19-6 gives i?-values for some common building materials. ^-values increase directly with material thickness: for example, 2 inches of fiberglass is R-6, whereas 4 inches is R-12.]

ConvectionAlthough liquids and gases are generally not very good conductors of heat, they can transfer heat quite rapidly by convection. Convection is the process whereby heat flows by the mass movement of molecules from one place to another. Whereas conduction involves molecules (and/or electrons) moving only over small distances and colliding, convection involves the movement of large numbers of molecules over large distances.

A forced-air furnace, in which air is heated and then blown by a fan into a room, is an example of forced convection. Natural convection occurs as well, and one familiar example is that hot air rises. For instance, the air above a radiator (or other type of heater) expands as it is heated (Chapter 17), and hence its density decreases. Because its density is less than that of the surrounding cooler air, it rises, just as a log submerged in water floats upward because its density is less than that of water. Warm or cold ocean currents, such as the balmy Gulf Stream, represent natural convection on a global scale. Wind is another example of convection, and weather in general is strongly influenced by convective air currents.

When a pot of water is heated (Fig. 19-22), convection currents are set up as the heated water at the bottom of the pot rises because of its reduced density. That heated water is replaced by cooler water from above. This principle is used in many heating systems, such as the hot-water radiator system shown in Fig. 19-23. Water is heated in the furnace, and as its temperature increases, it expands and rises as shown. This causes the water to circulate in the heating system. Hot water then enters the radiators, heat is transferred by conduction to the air, and the cooled water returns to the furnace. Thus, the water circulates because of convection; pumps are sometimes used to improve circulation. The air throughout the room also becomes heated as a result of convection. The air heated by the radiators rises and is replaced by cooler air, resulting in convective air currents, as shown by the green arrows in Fig. 19-23.

Other types of furnaces also depend on convection. Hot-air furnaces with registers (openings) near the floor often do not have fans but depend on natural convection, which can be appreciable. In other systems, a fan is used. In either case, it is important that cold air can return to the furnace so that convective currents circulate throughout the room if the room is to be uniformly heated. Convection is not always favorable. Much of the heat from a fireplace, for example, goes up the chimney and not out into the room.

RadiationConvection and conduction require the presence of matter as a medium to carry the heat from the hotter to the colder region. But a third type of heat transfer occurs without any medium at all. All life on Earth depends on the transfer of enerev from the Sun. and this enerev is transferred to the Earth over emotv for

[g P H Y S I C S A P P L I E DClothes insulate by trapping an air layer

0 P H Y S I C S A P P L I E DR-values o f thermal insulation

TABLE 19-6 / -̂valuesMaterial Thickness R -yalue

(ft2 • h • F°/Btu)

Glass |in ch 1Brick 3 \ inches 0.6-1Plywood \ inch 0.6Fiberglass

insulation 4 inches 12

Coolerwater

FIGURE 19-22 Convection currents in a pot of water being heated on a stove.

( ^ P H Y S I C S A P P L I E DConvective home heating

FIGURE 19-23 Convection plays a role in heating a house. The circular arrows show the convective air currents in the rooms.

@ P H Y S I C S A P P L I E DD ark vs. light clothing

^ P H Y S I C S A P P L I E DThe b o d y ’s radiative heat loss

P R O B L E M S O L V I N GMust use the Kelvin temperature

As we shall see in later Chapters, radiation consists essentially of electromagnetic waves. Suffice it to say for now that radiation from the Sun consists of visible light plus many other wavelengths that the eye is not sensitive to, including infrared (IR) radiation.

The rate at which an object radiates energy has been found to be proportional to the fourth power of the Kelvin temperature, T. That is, an object at 2000 K, as compared to one at 1000 K, radiates energy at a rate 24 = 16 times as much. The rate of radiation is also proportional to the area A of the emitting object, so the rate at which energy leaves the object, AQ / At, is

AQ At

= ecrAT4. (19-17)

This is called the Stefan-Boltzmann equation, and a is a universal constant called the Stefan-Boltzmann constant which has the value

o- = 5.67 X 10_8W/m2*K4.The factor e (Greek letter epsilon), called the emissivity, is a number between 0 and 1 that is characteristic of the surface of the radiating material. Very black surfaces, such as charcoal, have emissivity close to 1, whereas shiny metal surfaces have e close to zero and thus emit correspondingly less radiation. The value depends somewhat on the temperature of the material.

Not only do shiny surfaces emit less radiation, but they absorb little of the radiation that falls upon them (most is reflected). Black and very dark objects are good emitters (e ~ 1), and they also absorb nearly all the radiation that falls on them—which is why light-colored clothing is usually preferable to dark clothing on a hot day. Thus, a good absorber is also a good emitter.

Any object not only emits energy by radiation but also absorbs energy radiated by other objects. If an object of emissivity e and area A is at a temperature 7\, it radiates energy at a rate eaA T 4. If the object is surrounded by an environment at temperature T2, the rate at which the surroundings radiate energy is proportional to T \ , and the rate that energy is absorbed by the object is proportional to T\. The net rate of radiant heat flow from the object is given by

^ = «rA (T i ~ T i l (19-18)

where A is the surface area of the object, Tx its temperature and e its emissivity (at temperature 7\), and T2 is the temperature of the surroundings. This equation is consistent with the experimental fact that equilibrium between the object and its surroundings is reached when they come to the same temperature. That is, AQ /At must equal zero when Tx = T2, so e must be the same for emission and absorption. This confirms the idea that a good emitter is a good absorber. Because both the object and its surroundings radiate energy, there is a net transfer of energy from one to the other unless everything is at the same temperature.

EXAMPLE 19-14 ESTIMATE"! Cooling by radiation. An athlete is sitting unclothed in a locker room whose dark walls are at a temperature of 15°C. Estimate his rate of heat loss by radiation, assuming a skin temperature of 34°C and e = 0.70. Take the surface area of the body not in contact with the chair to be 1.5 m2.APPROACH We use Eq. 19-18, with Kelvin temperatures.SOLUTION We have

^ = eaA(T\ - 71)

= (0.70)(5.67 x 10“8W/m2-K4)(l.5m 2)[(307K)4 - (288 K)4] = 120 W.

NOTE The “output” of this resting person is a bit more than what a 100-W

A resting person naturally produces heat internally at a rate of about 100 W, which is less than the heat loss by radiation as calculated in Example 19-14. Hence, the person’s temperature would drop, causing considerable discomfort. The body responds to excessive heat loss by increasing its metabolic rate, and shivering is one method by which the body increases its metabolism. Naturally, clothes help a lot. Example 19-14 illustrates that a person may be uncomfortable even if the temperature of the air is, say, 25°C, which is quite a warm room. If the walls or floor are cold, radiation to them occurs no matter how warm the air is. Indeed, it is estimated that radiation accounts for about 50% of the heat loss from a sedentary person in a normal room. Rooms are most comfortable when the walls and floor are warm and the air is not so warm. Floors and walls can be heated by means of hot-water conduits or electric heating elements. Such first-rate heating systems are becoming more common today, and it is interesting to note that 2000 years ago the Romans, even in houses in the remote province of Great Britain, made use of hot-water and steam conduits in the floor to heat their houses.

Heating of an object by radiation from the Sun cannot be calculated using Eq. 19-18 since this equation assumes a uniform temperature, T2, of the environment surrounding the object, whereas the Sun is essentially a point source. Hence the Sun must be treated as a separate source of energy. Heating by the Sun is calculated using the fact that about 1350 J of energy strikes the atmosphere of the Earth from the Sun per second per square meter of area at right angles to the Sun’s rays. This number, 1350 W/m2, is called the solar constant. The atmosphere may absorb as much as 70% of this energy before it reaches the ground, depending on the cloud cover. On a clear day, about 1000 W/m2 reaches the Earth’s surface. An object of emissivity e with area A facing the Sun absorbs energy from the Sun at a rate, in watts, of about

= (1000 W/m2) € A cos 6, (19-19)

where 0 is the angle between the Sun’s rays and a line perpendicular to the area A (Fig. 19-24). That is, A cos 0 is the “effective” area, at right angles to the Sun’s rays.

The explanation for the seasons and the polar ice caps (see Fig. 19-25) depends on this cos 0 factor in Eq. 19-19. The seasons are not a result of how close the Earth is to the Sun—in fact, in the Northern Hemisphere, summer occurs when the Earth is farthest from the Sun. It is the angle (i.e., cos0) that really matters. Furthermore, the reason the Sun heats the Earth more at midday than at sunrise or sunset is also related to this cos 0 factor.

An interesting application of thermal radiation to diagnostic medicine is thermography. A special instrument, the thermograph, scans the body, measuring the intensity of radiation from many points and forming a picture that resembles an X-ray (Fig. 19-26). Areas where metabolic activity is high, such as in tumors, can often be detected on a thermogram as a result of their higher temperature and consequent increased radiation.

FIGURE 19-24 Radiant energy striking a body at an angle 6.

FIGURE 19-25 (a) Earth’s seasons arise from the 231° angle Earth’s axis makes with its orbit around the Sun.(b) June sunlight makes an angle of about 23° with the equator. Thus 6 in the southern United States (A) is near 0° (direct summer sunlight), whereas in the Southern Hemisphere (B), 6 is 50° or 60°, and less heat can be absorbed— hence it is winter. Near the poles (C), there is never strong direct sunlight; cos 6 varies from about \ in summer to 0 in winter; so with little heating, ice can form.

Axis

>Equator

Earth I JuuuI

Sun EarilitlX-cL’iiihcri

(a)

June Axis(A>0- 0= (Summer*

( N(CI<H9<JC (R tfl = 50°

(Cold) (Winter)

(h)

FIGURE 19-26 Thermograms of a healthy person’s arms and hands (a) before and (b) after smoking a cigarette, showing a temperature decrease due to impaired blood circulation associated with smoking. The thermograms have been color-coded according to temperature; the scale on the right goes from blue (cold) to white (hot).

®LAstronomy— size

o f a star

P H Y S I C S A P P L I E D EXAMPLE 19-15 ESTIMATE I Star radius. The giant star Betelgeuse emits radiant energy at a rate 104 times greater than our Sun, whereas its surface temperature is only half (2900 K) that of our Sun. Estimate the radius of Betelgeuse, assuming e = 1 for both. The Sun’s radius is rs = 7 X 108 m.

APPROACH We assume both Betelgeuse and the Sun are spherical, with surface area Airr2.SOLUTION We solve Eq. 19-17 for A:

( A Q / A t )

Then

Airr A =t o - r

( A Q / A ( ) b T\( A G / A O s T

= (104)(24) = 16 X 104.

H ence rB = V 16 x 1q4 rs = (400)(7 X 108m) « 3 X 10u m.NOTE If Betelgeuse were our Sun, it would envelop us (Earth is 1.5 X 1011 m from the Sun).

EXERCISE G Fanning yourself on a hot day cools you by (a) increasing the radiation rate of the skin; (b) increasing conductivity; (c) decreasing the mean free path of air;(d) increasing the evaporation of perspiration; (e) none of these.

SummaryInternal energy, E{nt, refers to the total energy of all the molecules in an object. For an ideal monatomic gas,

Eint = \N k T = I nRT (19-1)

where N is the number of molecules or n is the number of moles.

Heat refers to the transfer of energy from one object to another because of a difference of temperature. Heat is thus measured in energy units, such as joules.

Heat and internal energy are also sometimes specified in calories or kilocalories (kcal), where

1 kcal = 4.186 kJ

is the amount of heat needed to raise the temperature of 1 kg of water by 1 C°.

The specific heat, c, of a substance is defined as the energy (or heat) required to change the temperature of unit mass of substance by 1 degree; as an equation,

Q = mcAT, (19-2)

where Q is the heat absorbed or given off, AT is the temperature increase or decrease, and m is the mass of the substance.

When heat flows between parts of an isolated system, conservation of energy tells us that the heat gained by one part of the system is equal to the heat lost by the other part of the system. This is the basis of calorimetry, which is the quantitative measurement of heat exchange.

Exchange of energy occurs, without a change in temperature, whenever a substance changes phase. The heat of fusion is the heat required to melt 1 kg of a solid into the liquid phase; it is also equal to the heat given off when the substance changes from liquid to solid. The heat of vaporization is the energy required to change 1 kg of a substance from the liquid to the vanor nhase: it is also the enerov piven off when the substance

The first law of thermodynamics states that the change in internal energy AE[nt of a system is equal to the heat added to the system, Q, minus the work, W, done by the system:

A E i n t = Q - w. (19-4)This important law is a broad restatement of the conservation of energy and is found to hold for all processes.

Two simple thermodynamic processes are isothermal, which is a process carried out at constant temperature, and adiabatic, a process in which no heat is exchanged. Two more are isobaric (a process carried out at constant pressure) and isovolumetric (a process at constant volume).

The work done by (or on) a gas to change its volume by dV is dW = P dV, where P is the pressure.

Work and heat are not functions of the state of a system (as are P, V, T,n, and £int) but depend on the type of process that takes a system from one state to another.

The molar specific heat of an ideal gas at constant volume, C y , and at constant pressure, CP, are related by

CP - Cv = R, (19-11)where R is the gas constant. For a monatomic ideal gas, Cy = \R .

For ideal gases made up of diatomic or more complex molecules, Cy is equal to \R times the number of degrees of freedom of the molecule. Unless the temperature is very high, some of the degrees of freedom may not be active and so do not contribute. According to the principle of equipartition of energy, energy is shared equally among the active degrees of freedom in an amount \k T per molecule on average.

When an ideal gas expands (or contracts) adiabatically (Q = 0), the relation PVy = constant holds, where

H eat is transferred from one nlar.e (or ohieet'i to another in

In conduction, energy is transferred by collisions between molecules or electrons with higher kinetic energy to slower- moving neighbors.

Convection is the transfer of energy by the mass movement of molecules over considerable distances.

Radiation, which does not require the presence of matter, is energy transfer by electromagnetic waves, such as from the Sun.

All objects radiate energy in an amount that is proportional to the fourth power of their Kelvin temperature (T4) and to their surface area. The energy radiated (or absorbed) also depends on the nature of the surface, which is characterized by the emissivity, e (dark surfaces absorb and radiate more than do bright shiny ones).

Radiation from the Sun arrives at the surface of the Earth on a clear day at a rate of about 1000 W/m2.

Questions1. What happens to the work done on a jar of orange juice

when it is vigorously shaken?2. When a hot object warms a cooler object, does temperature

flow between them? Are the temperature changes of the two objects equal? Explain.

3. (a) If two objects of different temperature are placed in contact, will heat naturally flow from the object with higher internal energy to the object with lower internal energy?(b) Is it possible for heat to flow even if the internal energies of the two objects are the same? Explain.

4. In warm regions where tropical plants grow but the temperature may drop below freezing a few times in the winter, the destruction of sensitive plants due to freezing can be reduced by watering them in the evening. Explain.

5. The specific heat of water is quite large. Explain why this fact makes water particularly good for heating systems (that is, hot-water radiators).

6. Why does water in a canteen stay cooler if the cloth jacket surrounding the canteen is kept moist?

7. Explain why burns caused by steam at 100°C on the skin are often more severe than burns caused by water at 100°C.

8. Explain why water cools (its temperature drops) when it evaporates, using the concepts of latent heat and internal energy.

9. Will potatoes cook faster if the water is boiling more vigorously?

10. Very high in the Earth’s atmosphere the temperature can be 700°C. Yet an animal there would freeze to death rather than roast. Explain.

11. What happens to the internal energy of water vapor in the air that condenses on the outside of a cold glass of water? Is work done or heat exchanged? Explain.

12. Use the conservation of energy to explain why the temperature of a well-insulated gas increases when it is compressed—say, by pushing down on a piston—whereas the temperature decreases when the gas expands.

13. In an isothermal process, 3700 J of work is done by an ideal gas. Is this enough information to tell how much heat has been added to the system? If so, how much?

14. Explorers on failed Arctic expeditions have survived by covering themselves with snow. Why would they do that?

15. Why is wet sand at the beach cooler to walk on than dry sand?

16. When hot-air furnaces are used to heat a house, why is it important that there be a vent for air to return to the furnace? What happens if this vent is blocked by a bookcase?

17. Is it possible for the temperature of a system to remain constant even though heat flows into or out of it? Tf so. oive

18. Discuss how the first law of thermodynamics can apply to metabolism in humans. In particular, note that a person does work W, but very little heat Q is added to the body (rather, it tends to flow out). Why then doesn’t the internal energy drop drastically in time?

19. Explain in words why Cp is greater than Cy.20. Explain why the temperature of a gas increases when it is

adiabatically compressed.21. An ideal monatomic gas is allowed to expand slowly to twice

its volume (1) isothermally; (2) adiabatically; (3) isobarically. Plot each on a PV diagram. In which process is A£int the greatest, and in which is A£jnt the least? In which is W the greatest and the least? In which is Q the greatest and the least?

22. Ceiling fans are sometimes reversible, so that they drive the air down in one season and pull it up in another season. Which way should you set the fan for summer? For winter?

23. Goose down sleeping bags and parkas are often specified as so many inches or centimeters of loft, the actual thickness of the garment when it is fluffed up. Explain.

24. Microprocessor chips nowadays have a “heat sink” glued on top that looks like a series of fins. Why is it shaped like that?

25. Sea breezes are often encountered on sunny days at the shore of a large body of water. Explain, assuming the temperature of the land rises more rapidly than that of the nearby water.

26. The Earth cools off at night much more quickly when the weather is clear than when cloudy. Why?

27. Explain why air-temperature readings are always taken with the thermometer in the shade.

28. A premature baby in an incubator can be dangerously cooled even when the air temperature in the incubator is warm. Explain.

29. The floor of a house on a foundation under which the air can flow is often cooler than a floor that rests directly on the ground (such as a concrete slab foundation). Explain.

30. Why is the liner of a thermos bottle silvered (Fig. 19-27), and why does it have avacuum between its two walls?

FIGURE 19-27

Stopper

Outer casing

Air

Liner----- -

Liquid.hot ur cold - - - _

Vacuum---------

Insulating support —

CSV

31. A 22°C day is warm, while a swimming pool at 22°C feels cool. Why?

32. In the Northern Hemisphere the amount of heat required to heat a room where the windows face north is much higher than that required where the windows face south. Explain. 35.

33. Heat loss occurs through windows by the following processes:(1) ventilation around edges; (2) through the frame, particu­larly if it is metal; (3) through the glass panes; and (4) radia- 36. tion. (a) For the first three, what is (are) the mechanism(s): conduction, convection, or radiation? (b) Heavy curtains 37. reduce which of these heat losses? Explain in detail.

34. Early in the day, after the Sun has reached the slope of a mountain, there tends to be a gentle upward movement of air. Later, after a slope goes into shadow, there is a gentle downdraft. Explain.A piece of wood lying in the Sun absorbs more heat than a piece of shiny metal. Yet the metal feels hotter than the wood when you pick it up. Explain.An “emergency blanket” is a thin shiny (metal-coated) plastic foil. Explain how it can help to keep an immobile person warm. Explain why cities situated by the ocean tend to have less extreme temperatures than inland cities at the same latitude.

| Problems19-1 Heat as Energy Transfer1. (I) To what temperature will 8700 J of heat raise 3.0 kg of

water that is initially at 10.0°C?2. (II) When a diver jumps into the ocean, water leaks into the

gap region between the diver’s skin and her wetsuit, forming a water layer about 0.5 mm thick. Assuming the total surface area of the wetsuit covering the diver is about1.0 m2, and that ocean water enters the suit at 10°C and is warmed by the diver to skin temperature of 35 °C, estimate how much energy (in units of candy bars = 300kcal) is required by this heating process.

3. (II) An average active person consumes about 2500 Cal a day. (a) What is this in joules? (b) What is this in kilowatt- hours? (c) If your power company charges about 10 0 per kilowatt-hour, how much would your energy cost per day if you bought it from the power company? Could you feed yourself on this much money per day?

4. (II) A British thermal unit (Btu) is a unit of heat in the British system of units. One Btu is defined as the heat needed to raise 1 lb of water by 1 F°. Show that

1 Btu = 0.252 kcal = 1056 J.

5. (II) How many joules and kilocalories are generated when the brakes are used to bring a 1200-kg car to rest from a speed of 95 km /h?

6. (II) A small immersion heater is rated at 350 W. Estimate how long it will take to heat a cup of soup (assume this is 250 mL of water) from 15°C to 75°C.

19-3 and 19-4 Specific Heat; Calorimetry7. (I) An automobile cooling system holds 18 L of water. How

much heat does it absorb if its temperature rises from 15°C to 95°C?

8. (I) What is the specific heat of a metal substance if 135 kJ of heat is needed to raise 5.1 kg of the metal from 18.0°C to 37.2°C?

9. (II) (a) How much energy is required to bring a 1.0-L pot of water at 20°C to 100°C? (b) For how long could this amount of energy run a 100-W lightbulb?

10. (II) Samples of copper, aluminum, and water experience the same temperature rise when they absorb the same amount of heat. What is the ratio of their masses?

11. (II) How long does it take a 750-W coffeepot to bring to a boil 0.75 L of water initially at 8.0°C? Assume that the partn f the. n n t w h ir .h is h e a te d w ith the. w a te r is m a d e n f 9 8 0 o n f

12. (II) A hot iron horseshoe (mass = 0.40 kg), just forged (Fig. 19-28), is dropped into 1.05 L of water in a 0.30-kg iron pot initially at 20.0°C. If the final equilibrium temper­ature is 25.0°C, estimate the initial temperature of the hot horseshoe.

FIGURE 19-28Problem 12.

13. (II) A 31.5-g glass thermometer reads 23.6°C before it is placed in 135 mL of water. When the water and thermometer come to equilibrium, the thermometer reads 39.2°C. What was the original temperature of the water? [Hint: Ignore the mass of fluid inside the glass thermometer.]

14. (II) Estimate the Calorie content of 65 g of candy from the following measurements. A 15-g sample of the candy is placed in a small aluminum container of mass 0.325 kg filled with oxygen. This container is placed in 2.00 kg of water in an aluminum calorimeter cup of mass 0.624 kg at an initial temperature of 15.0°C. The oxygen-candy mixture in the small container is ignited, and the final temperature of the whole system is 53.5°C.

15. (II) When a 290-g piece of iron at 180°C is placed in a 95-g aluminum calorimeter cup containing 250 g of glycerin at 10°C, the final temperature is observed to be 38°C. Estimate the specific heat of glycerin.

16. (II) The heat capacity, C, of an object is defined as the amount of heat needed to raise its temperature by 1 C°. Thus, to raise the temperature by AT requires heat Q given by

Q = CAT.(a) Write the heat capacity C in terms of the specific heat, c, of the material, (b) What is the heat capacity of 1.0 kg of water? (c) Of 35 kg of water?

17. (II) The 1.20-kg head of a hammer has a speed of 7.5 m/s just before it strikes a nail (Fig. 19-29) and is brought to rest. Estimate the temperature rise of a 14-g iron nail generated by 10 such hammer blows done in quick succession. Assume the nail absorbs all theenergy.

FIGURE 19-29

19-5 Latent Heat18. (I) How much heat is needed to melt 26.50 kg of silver that

is initially at 25 °C?19. (I) During exercise, a person may give off 180 kcal of heat in

25 min by evaporation of water from the skin. How much water has been lost?

20. (II) A 35-g ice cube at its melting point is dropped into an insulated container of liquid nitrogen. How much nitrogen evaporates if it is at its boiling point of 77 K and has a latent heat of vaporization of 200 kJ/kg? Assume for simplicity that the specific heat of ice is a constant and is equal to its value near its melting point.

21. (II) High-altitude mountain climbers do not eat snow, but always melt it first with a stove. To see why, calculate the energy absorbed from your body if you (a) eat 1.0 kg of —10°C snow which your body warms to body temperature of 37°C. (ib) You melt 1.0 kg of — 10°C snow using a stove and drink the resulting 1.0 kg of water at 2°C, which your body has to warm to 37°C.

22. (II) An iron boiler of mass 180 kg contains 730 kg of water at 18°C. A heater supplies energy at the rate of 52,000 kJ/h. How long does it take for the water (a) to reach the boiling point, and (b) to all have changed to steam?

23. (II) In a hot day’s race, a bicyclist consumes 8.0 L of water over the span of 3.5 hours. Making the approximation that all of the cyclist’s energy goes into evaporating this water as sweat, how much energy in kcal did the rider use during the ride? (Since the efficiency of the rider is only about 20%, most of the energy consumed does go to heat, so our approximation is not far off.)

24. (II) The specific heat of mercury is 138 J/kg - C°. Determine the latent heat of fusion of mercury using the following calorimeter data: 1.00 kg of solid Hg at its melting point of —39.0°C is placed in a 0.620-kg aluminum calorimeter with 0.400 kg of water at 12.80°C; the resulting equilibrium temperature is 5.06°C.

25. (II) At a crime scene, the forensic investigator notes that the 7.2-g lead bullet that was stopped in a doorframe apparently melted completely on impact. Assuming the bullet was shot at room temperature (20° C), what does the investigator calculate as the minimum muzzle velocity of the gun?

26. (II) A 58-kg ice-skater moving at 7.5 m/s glides to a stop. Assuming the ice is at 0°C and that 50% of the heat generated by friction is absorbed by the ice, how much ice melts?

19-6 and 19-7 First Law of Thermodynamics27. (I) Sketch a PV diagram of the following process: 2.0 L of

ideal gas at atmospheric pressure are cooled at constant pressure to a volume of 1.0 L, and then expanded isothermally back to 2.0 L, whereupon the pressure is increased at constant volume until the original pressure is reached.

28. (I) A gas is enclosed in a cylinder fitted with a light frictionless piston and maintained at atmospheric pressure. When 1250 kcal of heat is added to the gas, the volume is observed to increase slowly from 12.0 m3 to 18.2 m3. Calculate (a) the work done by the gas and (b) the change in internal energy of the gas.

29. (II) The pressure in an ideal gas is cut in half slowly, while being kept in a container with rigid walls. In the process, 365 kJ of heat left the gas. (a) How much work was done fh ir in p th is n ro ce ss? (hi\ W h a t w as th e ehanpe in in te rn a l

30. (II) A 1.0-L volume of air initially at 3.5 atm of (absolute) pressure is allowed to expand isothermally until the pressure is 1.0 atm. It is then compressed at constant pressure to its initial volume, and lastly is brought back to its original pressure by heating at constant volume. Draw the process on a PV diagram, including numbers and labels for the axes.

31. (II) Consider the following two-step process. Heat is allowed to flow out of an ideal gas at constant volume so that its pressure drops from 2.2 atm to 1.4 atm. Then the gas expands at constant pressure, from a volume of 5.9 L to9.3 L, where the temperature reaches its original value. See Fig. 19-30. Calculate (a) the total work done by the gas in the process,(b) the change in internal energy of the gas in the process, and (c) the total heat flow into or out of the gas.

FIGURE 19-30Problem 31.

32. (II) The PV diagram in Fig. 19-31 shows two possible states of a system containing 1.55 moles of a monatomic ideal gas. (Px = P2 = 455 N/m2, ^ = 2.00 m3, V2 = 8.00 m3.)(a) Draw the process which depicts an isobaric expansion from state 1 to state 2, and label this process A. (b) Find the work done by the gas and the change in internal energy of the gas in process A. (c) Draw the two-step process which depicts an isothermal expansion from state 1 to the volume V2, followed by an isovolumetric increase in temperature to state 2, and label this process B. (d) Find the change ininternal energy of the gas for the two-step process B.

FIGURE 19-31Problem 32.

P(N/m2)

500-4 00-300-200 -

100-H------1------1------h

0 2 4 6 8 10 V (m3)

33. (II) Suppose 2.60 mol of an ideal gas of volume Vi = 3.50 m3 at T\ = 290 K is allowed to expand isothermally to V2 = 7.00 m3 at T2 = 290 K. Determine(a) the work done by the gas, (b) the heat added to the gas, and (c) the change in internal energy of the gas.

34. (II) In an engine, an almost ideal gas is compressed adiabatically to half its volume. In doing so, 2850 J of work is done on the gas. (a) How much heat flows into or out of the gas? (b) What is the change in internal energy of the gas? (c) Does its temperature rise or fall?

35. (II) One and one-half moles of an ideal monatomic gas expand adiabatically, performing 7500 J of work in the process. What is the change in temperature of the gas during this expansion?

36. (II) Determine (a) the work done and (b) the change in in te rn a l e n e rp v n f 1 .00 kp n f w a te r w h en it is a ll b o iled to

37. (II) How much work is done by a pump to slowly compress, isothermally, 3.50 L of nitrogen at 0°C and 1.00 atm to 1.80 L at 0°C?

38. (II) When a gas is taken from a to c along the curved path in Fig. 19-32, the work done by the gas is W = -35 J and the heat added to the gas is Q = -63 J. Along path abc, the work done is W = — 54 J. (a) What is Q for path abc? (b) If PC = \P\) , what is W for path cda? (c) What is Q for path cda? (d) What is Eint>a ~~ Eint, c? (e) If Eint, d — Eint, c = 12 J, what is Q for path da?

P

FIGURE 19-32Problems 38,39, _________________________and 40. 0 y

39. (Ill) In the process of taking a gas from state a to state c along the curved path shown in Fig. 19-32, 85 J of heat leaves the system and 55 J of work is done on the system.(a) Determine the change in internal energy, Ejnt a — Ejnt c.(b) When the gas is taken along the path cda, the work done by the gas is W = 38 J. How much heat Q is added to the gas in the process cda? (c) If Pa = 22P&, how much work is done by the gas in the process abc? (d) What is Q for path abc? (e) If Eintj a - Eint; b = 15 J, what is Q for the process be? Here is a summary of what is given:

Qa-»c = — 85 JWa_>c = -55 JWcda = 38 J

Eint, a — E^t, b — 15 JPa = 2.2 Pd.

40. (Ill) Suppose a gas is taken clockwise around the rectangular cycle shown in Fig. 19-32, starting at b, then toa, to d, to c, and returning to b. Using the values given in Problem 39, (a) describe each leg of the process, and then calculate (b) the net work done during the cycle, (c) the total internal energy change during the cycle, and (d) the net heat flow during the cycle. (e) What percentage of the intake heat was turned into usable work: i.e., how efficient is this “rectangular” cycle (give as a percentage)?

*41. (Ill) Determine the work done by 1.00 mol of a van der Waals gas (Section 18-5) when it expands from volume V\ to V2 isothermally.

19-8 Molecular Specific Heat for Gases; Equipartition of Energy42. (I) What is the internal energy of 4.50 mol of an ideal

diatomic gas at 645 K, assuming all degrees of freedom are active?

43. (I) If a heater supplies 1.8 X 106J/h to a room3.5 m X 4.6 m X 3.0 m containing air at 20°C and 1.0 atm, by how much will the temperature rise in one hour, assu m in g n n lo sses n f h e a t n r a ir m ass tn th e n n ts id e ?

44. (I) Show that if the molecules of a gas have n degrees of freedom, then theory predicts C y = \nR and Cp = \ {n + 2 )i?.

45. (II) A certain monatomic gas has specific heat cv = 0.0356 kcal/kg *C°, which changes little over a wide temperature range. What is the atomic mass of this gas? What gas is it?

46. (II) Show that the work done by n moles of an ideal gas when it expands adiabatically is W = nCv (T\ - T2), where T\ and T2 are the initial and final temperatures, and C y is the molar specific heat at constant volume.

47. (II) An audience of 1800 fills a concert hall of volume22.000 m3. If there were no ventilation, by how much would the temperature of the air rise over a period of 2.0 h due to the metabolism of the people (70 W/person)?

48. (II) The specific heat at constant volume of a particular gas is 0.182 kcal/kg • K at room temperature, and its molecular mass is 34. (a) What is its specific heat at constant pressure? (b) What do you think is the molecular structure of this gas?

49. (II) A 2.00 mole sample of N2 gas at 0°C is heated to 150°C at constant pressure (1.00 atm). Determine (a) the change in internal energy, (b) the work the gas does, and (c) the heat added to it.

50. (Ill) A 1.00-mol sample of an ideal diatomic gas at a pressure of 1.00 atm and temperature of 420 K undergoes a process in which its pressure increases linearly with temperature. The final temperature and pressure are 720 K and 1.60 atm. Determine (a) the change in internal energy,(b) the work done by the gas, and (c) the heat added to the gas. (Assume five active degrees of freedom.)

19-9 Adiabatic Expansion of a Gas51. (I) A 1.00-mol sample of an ideal diatomic gas, originally at

1.00 atm and 20°C, expands adiabatically to 1.75 times its initial volume. What are the final pressure and temperature for the gas? (Assume no molecular vibration.)

52. (II) Show, using Eqs. 19-6 and 19-15, that the work done by a gas that slowly expands adiabatically from pressure Pi and volume Vl5 to P2 and V2, is given byw = (PtVi - p2v2)/(y - l).

53. (II) A 3.65-mol sample of an ideal diatomic gas expands adiabatically from a volume of 0.1210 m3 to 0.750 m3. Initially the pressure was 1.00 atm. Determine: (a) the initial and final temperatures; (b) the change in internal energy;(c) the heat lost by the gas; (d) the work done on the gas. (Assume no molecular vibration.)

54. (II) An ideal monatomic gas, consisting of 2.8 mol of volume 0.086 m3, expands adiabatically. The initial and final temperatures are 25°C and — 68°C. What is the final volume of the gas?

55. (Ill) A 1.00-mol sample of an ideal monatomic gas, originally at a pressure of 1.00 atm, undergoes a three-step process: (1) it is expanded adiabatically from 7i = 588 K to T2 = 389 K; (2) it is compressed at constant pressure until its temperature reaches T3; (3) it then returns to its original pressure and temperature by a constant-volume process,(a) Plot these processes on a PV diagram, (b) Determine T3.(c) Calculate the change in internal energy, the work done hv th e pas. and th e heat added tn th e pas fnr each nrncess.

56. (Ill) Consider a parcel of air moving to a different altitude y in the Earth’s atmosphere (Fig. 19-33). As the parcel changes altitude it acquires the pressure P of the surrounding air. From Eq. 13-4 we have

dPdy = ~ P g

where p is the parcel’s altitude-dependent mass density.

y = 0 J .

i

/ “Parcel” of N air molecules

FIGURE 19-33Problem 56.

During this motion, the parcel’s volume will change and, because air is a poor heat conductor, we assume this expansion or contraction will take place adiabatically. (a) Starting with Eq. 19-15, P V 7 = constant, show that for an ideal gas undergoing an adiabatic process, p l yT y = constant. Then show that the parcel’s pressure and temperature are related by

,d P P d T^ - 7^ + y T ^ = ° and thus

x/ x P d T(i - y)(-pg) + y = °-

(b) Use the ideal gas law with the result from part (a) to show that the change in the parcel’s temperature with change in altitude is given by

dTdy

1 - 7 mg 7 k

where m is the average mass of an air molecule and k is the Boltzmann constant, (c) Given that air is a diatomic gas with an average molecular mass of 29, show that dT/dy = — 9.8C°/km. This value is called the adiabatic lapse rate for dry air. (d) In California, the prevailing west­erly winds descend from one of the highest elevations (the 4000-m Sierra Nevada mountains) to one of the lowest elevations (Death Valley, -100 m) in the continental United States. If a dry wind has a temperature of — 5°C at the top of the Sierra Nevada, what is the wind’s temperature after it has descended to Death Valley?

19-10 Conduction, Convection, Radiation57. (I) (a) How much power is radiated by a tungsten sphere

(emissivity e = 0.35 ) of radius 16 cm at a temperature of 25 °C? (b) If the sphere is enclosed in a room whose wallsare k e n t at — 5°C w h a t is th e n e t f lo w ra te o f en e r o v n u t n f

58. (I) One end of a 45-cm-long copper rod with a diameter of 2.0 cm is kept at 460° C, and the other is immersed in water at 22°C. Calculate the heat conduction rate along the rod.

59. (II) How long does it take the Sun to melt a block of ice at 0°C with a flat horizontal area 1.0 m2 and thickness 1.0 cm? Assume that the Sun’s rays make an angle of 35° with the vertical and that the emissivity of ice is 0.050.

60. (II) Heat conduction to skin. Suppose 150 W of heat flows by conduction from the blood capillaries beneath the skin to the body’s surface area of 1.5 m2. If the temperature difference is 0.50 C°, estimate the average distance of capillaries below the skin surface.

61. (II) A ceramic teapot (e = 0.70) and a shiny one (e = 0.10) each hold 0.55 L of tea at 95°C. (a) Estimate the rate of heat loss from each, and (b) estimate the temperature drop after 30 min for each. Consider only radiation, and assume the surroundings are at 20° C.

62. (II) A copper rod and an aluminum rod of the same length and cross-sectional area are attached end to end (Fig. 19-34). The copper end is placed in a furnace maintained at a constant temperature of 225°C. The aluminum end is placed in an ice bath held at constant temperature of 0.0°C. Calculate the temperature at the point where the two rods are joined.

Cu A1

225°C r = ? 0.0°C

FIGURE 19-34 Problem 62.

63. (II) (a) Using the solar constant, estimate the rate at which the whole Earth receives energy from the Sun. (b) Assume the Earth radiates an equal amount back into space (that is, the Earth is in equilibrium). Then, assuming the Earth is a perfect emitter (e = 1.0), estimate its average surface temperature. [Hint: Use area A = 4 t7 t| , and state why.]

64. (II) A 100-W lightbulb generates 95 W of heat, which is dissipated through a glass bulb that has a radius of 3.0 cm and is 0.50 mm thick. What is the difference in temperature between the inner and outer surfaces of the glass?

65. (Ill) A house thermostat is normally set to 22°C, but at night it is turned down to 12°C for 9.0 h. Estimate how much more heat would be needed (state as a percentage of daily usage) if the thermostat were not turned down at night. Assume that the outside temperature averages 0°C for the 9.0 h at night and 8°C for the remainder of the day, and that the heat loss from the house is proportional to the difference in temperature inside and out. To obtain an estimate from the data, you will have to make other simplifying assumptions; state what these are.

66. (Ill) Approximately how long should it take 9.5 kg of ice at 0°C to melt when it is placed in a carefully sealed Styrofoam ice chest of dimensions 25 cm X 35 cm X 55 cm whose walls are 1.5 cm thick? Assume that the conductivityn f S tvrnfn am is d n n h le th a t n f air and th a t th e n iits id e

67. (Ill) A cylindrical pipe has inner radius R 1 and outer radius R2. The interior of the pipe carries hot water at temperature 7 \ . The temperature outside is T2 (< Tj).(a) Show that the rate of heat loss for a length L of pipe is

dQ _ 2ir k fc - T2)L dt ~ \n(R2/R i)

where k is the thermal conductivity of the pipe, (b) Suppose the pipe is steel with R\ = 3.3 cm, R2 = 4.0 cm, and T2 = 18°C. If the pipe holds still water at Tx = 71 °C, what will be the initial rate of change of its temperature?(c) Suppose water at 71 °C enters the pipe and moves at a speed of 8.0 cm/s. What will be its temperature drop per centimeter of travel?

Brick Insulation itt |> iR2)

68. (Ill) Suppose the insulating qualities of the wall of a come mainly from a 4.0-in. layer of brick and an R -19 layer of insulation, as shown in Fig. 19-35. What is the total rate of heat loss through such a wall, if its total area is 195 ft2 and the temperature difference across it is 12 F°?

house

FIGURE 19-35 Problem 68. Two layers insulating a wall.

| General Problems__________69. A soft-drink can contains about 0.20 kg of liquid at 5°C.

Drinking this liquid can actually consume some of the fat in the body, since energy is needed to warm the liquid to body temperature (37°C). How many food Calories should the drink have so that it is in perfect balance with the heat needed to warm the liquid (essentially water)?

70. (a) Find the total power radiated into space by the Sun, assuming it to be a perfect emitter at T = 5500 K. The Sun’s radius is 7.0 X 108m. ( b ) From this, determine the power per unit area arriving at the Earth, 1.5 X 1011m away.

71. To get an idea of how much thermal energy is contained in the world’s oceans, estimate the heat liberated when a cube of ocean water, 1 km on each side, is cooled by 1 K. (Approximate the ocean water as pure water for this estimate.)

72. A mountain climber wears a goose-down jacket 3.5 cm thick with total surface area 0.95 m2. The temperature at the surface of the clothing is — 18°C and at the skin is 34°C. Determine the rate of heat flow by conduction through the jacket(a) assuming it is dry and the thermal conductivity k is that of goose down, and (b) assuming the jacket is wet, so k is that of water and the jacket has matted to 0.50 cm thickness.

73. During light activity, a 70-kg person may generate 200kcal/h. Assuming that 20% of this goes into useful work and the other 80% is converted to heat, estimate the temperature rise of the body after 30 min if none of this heat is transferred to the environment.

74. Estimate the rate at which heat can be conducted from the interior of the body to the surface. Assume that the thickness of tissue is 4.0 cm, that the skin is at 34°C and the interior at 37°C, and that the surface area is 1.5 m2. Compare this to the measured value of about 230 W that must be dissipated by a person working lightly. This clearly shows the necessity of convective cooling by the blood.

75. A marathon runner has an average metabolism rate of about 950 kcal/h during a race. If the runner has a mass of 55 k o e s t im a te h n w rrmr.h w a te r sh e w o u ld lo se to e va n o ra t io n

76. A house has well-insulated walls 19.5 cm thick (assume conductivity of air) and area 410 m2, a roof of wood5.5 cm thick and area 280 m2, and uncovered windows0.65 cm thick and total area 33 m2. (a) Assuming that heat is lost only by conduction, calculate the rate at which heat must be supplied to this house to maintain its inside temperature at 23°C if the outside temperature is -15°C. (b) If the house is initially at 12°C, estimate how much heat must be supplied to raise the temperature to 23°C within 30 min. Assume that only the air needs to be heated and that its volume is 750 m3. (c) If natural gas costs $0,080 per kilogram and its heat of combustion is 5.4 X 107J/kg, how much is the monthly cost to maintain the house as in part (a) for 24 h each day, assuming 90% of the heat produced is used to heat the house? Take the specific heat of air to be0.24 kcal/kg-C°.

77. In a typical game of squash (Fig. 19-36), two people hit a soft rubber ball at a wall until they are about to drop due to dehydration and exhaustion. Assume that the ball hits the wall at a velocity of 22 m /s and bounces back with a velocity of 12 m/s, and that the kinetic energy lost in the process heats the ball. What will be the temperature increase of the ball after one bounce? (The specific heat of rubber is about 1200 J/kg-C°.)

78. A bicycle pump is a cylinder 22 cm long and 3.0 cm in diameter. The pump contains air at 20.0° C and 1.0 atm. If the outlet at the base of the pump is blocked and the handle is pushed in very quickly, compressing the air to half its original volume, how hot does the air in the pump become?

79. A microwave oven is used to heat 250 g of water. On its maximum setting, the oven can raise the temperature of the liquid water from 20°C to 100°C in 1 min 45 s (= 105 s). (a) At what rate does the oven input energy to the liquid water?(b) If the power input from the oven to the water remains constant, determine how many grams of water will boil away if the oven is operated for 2 min (rather than just 1 min 45 s).

80. The temperature within the Earth’s crust increases about1.0 C° for each 30 m of depth. The thermal conductivity of the crust is 0.80W/C°-m. (a) Determine the heat transferred from the interior to the surface for the entire Earth in 1.0 h. (b) Compare this heat to the amount of energy incident on the Earth in 1.0 h due to radiation from the Sun.

81. An ice sheet forms on a lake. The air above the sheet is at — 18°C, whereas the water is at 0°C. Assume that the heat of fusion of the water freezing on the lower surface is conducted through the sheet to the air above. How much time will it take to form a sheet of ice 15 cm thick?

82. An iron meteorite melts when it enters the Earth’s atmosphere. If its initial temperature was -105°C outside of Earth’s atmosphere, calculate the minimum velocity the meteorite must have had before it entered Earth’s atmosphere.

83. A scuba diver releases a 3.60-cm-diameter (spherical) bubble of air from a depth of 14.0 m. Assume the temperature is constant at 298 K, and that the air behaves as an ideal gas.(a) How large is the bubble when it reaches the surface?(b) Sketch a PV diagram for the process, (c) Apply the first law of thermodynamics to the bubble, and find the work done by the air in rising to the surface, the change in its internal energy, and the heat added or removed from the air in the bubble as it rises. Take the density of water to be 1000 kg/m3.

84. A reciprocating compressor is a device that compresses air by a back-and-forth straight-line motion, like a piston in a cylinder. Consider a reciprocating compressor running at 150 rpm. During a compression stroke, 1.00 mol of air is compressed. The initial temperature of the air is 390 K, the engine of the compressor is supplying 7.5 kW of power to compress the air, and heat is being removed at the rate of 1.5 kW. Calculate the temperature change per compression stroke.

85. The temperature of the glass surface of a 75-W lightbulb is 75°C when the room temperature is 18°C. Estimate the temperature of a 150-W lightbulb with a glass bulb the same size. Consider only radiation, and assume that 90% of the energy is emitted as heat.

86. Suppose 3.0 mol of neon (an ideal monatomic gas) at STP are compressed slowly and isothermally to 0.22 the original volume. The gas is then allowed to expand quickly and adiabatically back to its original volume. Find the highest and lowest temperatures and pressures attained by the gas, and show on a PV diagram where these values occur.

87. At very low temperatures, the molar specific heat of many substances varies as the cube of the absolute temperature:

which is sometimes called Debye’s law. For rock salt, T0 = 281 K and k = 1940J/mol*K. Determine the heat needed to raise 2.75 mol of salt from 22.0 K to 48.0 K.

88. A diesel engine accomplishes ignition without a spark plug by an adiabatic compression of air to a temperature above the ignition temperature of the diesel fuel, which is injected into the cylinder at the peak of the compression. Suppose air is taken into the cylinder at 2 8 0 K and volume V\ and is compressed adiabatically to 5 6 0 ° C ( ~ 1 0 0 0 ° F ) and volume V2. Assuming that the air behaves as an ideal gas whose ratio of C p to C y is 1 .4 , calculate the compression ratio V\/V2 of the engine.

89. When 6.30 X 105 J of heat is added to a gas enclosed in a cylinder fitted with a light frictionless piston maintained at atmospheric pressure, the volume is observed to increase from 2.2 m3 to 4.1 m3. Calculate (a) the work done by the gas, and (b) the change in internal energy of the gas.(c) Graph this process on a PV diagram.

90. In a cold environment, a person can lose heat by conduction and radiation at a rate of about 200 W. Estimate how long it would take for the body temperature to drop from 36.6°C to 35.6°C if metabolism were nearly to stop. Assume a mass of 70 kg. (See Table 19-1.)

* Numerical/Computer*91. (II) Suppose 1.0 mol of steam at 100°C of volume

0.50 m3 is expanded isothermally to volume 1.00 m3. Assume steam obeys the van der Waals equation (P + n2a /V 2)(V /n - b) = RT, Eq. 18-9, with a =0.55 N • m4/mol2 and b = 3.0 X 10_5m3/mol. Using the expression dW = P dV, determine numerically the total work done W. Your result should agree within 2% of the result obtained by integrating the expression for dW.

Answers to Exercises

A: (b). E: Less.B: (c). F: -6 .8 X 103J.C: 0.21 kg. G: (d).r » . t