Chapter 20

Chapter 20Chapter 20

Heat and the Heat and the

First Law of ThermodynamicsFirst Law of Thermodynamics

Mechanical Equivalent of HeatMechanical Equivalent of Heat

• Joule found that it took approximately Joule found that it took approximately 4.18 J4.18 J of of mechanical energy to raise the temperature of mechanical energy to raise the temperature of 1g1g water at water at 11ooCC

• Later, more precise, measurements determined Later, more precise, measurements determined the amount of mechanical energy needed to the amount of mechanical energy needed to raise the temperature of raise the temperature of 1g1g of water from of water from 14.514.5ooCC to to 15.515.5ooCC

• 1 cal = 4.186 J1 cal = 4.186 JThis is known as the This is known as the mechanical equivalent of mechanical equivalent of

heatheat

Heat CapacityHeat Capacity

• The The heat capacityheat capacity, , CC, of a particular sample is , of a particular sample is defined as the amount of energy needed to raise defined as the amount of energy needed to raise the temperature of that sample by the temperature of that sample by 11ooCC

• If energy If energy QQ produces a change of temperature of produces a change of temperature of TT, then, then

QQ = = CC TT

Specific HeatSpecific Heat

• Specific heatSpecific heat, , cc, is the heat capacity per , is the heat capacity per unit massunit mass

• If energy If energy QQ transfers to a sample of a transfers to a sample of a substance of mass substance of mass mm and the temperature and the temperature changes by changes by TT, then the specific heat is, then the specific heat is

Qc

m T

Specific HeatSpecific Heat• The specific heat is essentially a measure The specific heat is essentially a measure

of how insensitive a substance is to the of how insensitive a substance is to the addition of energyaddition of energy– The greater the substance’s specific heat, the The greater the substance’s specific heat, the

more energy that must be added to cause a more energy that must be added to cause a particular temperature changeparticular temperature change

• The equation is often written in terms of The equation is often written in terms of QQ : :

QQ = = m cm c TT

Some Specific Heat ValuesSome Specific Heat Values

More Specific Heat ValuesMore Specific Heat Values

Sign ConventionsSign Conventions

• If the temperature increases:If the temperature increases:QQ and and TT are positiveare positive

Energy transfers into the systemEnergy transfers into the system

• If the temperature decreases:If the temperature decreases:QQ and and TT are negative are negative

Energy transfers out of the systemEnergy transfers out of the system

Specific Heat Varies With TemperatureSpecific Heat Varies With Temperature

• Technically, the specific heat varies with Technically, the specific heat varies with temperaturetemperature

• The corrected equation isThe corrected equation is

• However, if the temperature intervals are not However, if the temperature intervals are not too large, the variation can be ignored and too large, the variation can be ignored and cc can be treated as a constantcan be treated as a constant

There is only about a There is only about a 1%1% specific heat specific heat variation between variation between 00oo and and 100100ooCC

f

i

T

TQ m c dT

Specific Heat of WaterSpecific Heat of Water

• Water has the highest specific heat of Water has the highest specific heat of common materialscommon materials

• This is responsible for many weather This is responsible for many weather phenomenaphenomena– Moderate temperatures near large bodies of Moderate temperatures near large bodies of

waterwater– Global wind systemsGlobal wind systems– Land and sea breezesLand and sea breezes

CalorimetryCalorimetry

• One technique for measuring specific heat One technique for measuring specific heat involves heating a material, adding it to a involves heating a material, adding it to a sample of water, and recording the final sample of water, and recording the final temperaturetemperature

• This technique is known as This technique is known as calorimetrycalorimetry– A A calorimeter calorimeter is a device in which this energy is a device in which this energy

transfer takes placetransfer takes place


• The system of the sample and the water are The system of the sample and the water are isolatedisolated

• Conservation of energy requires that the amount Conservation of energy requires that the amount of energy that leaves the sample equals the of energy that leaves the sample equals the amount of energy that enters the wateramount of energy that enters the water– Conservation of Energy gives a mathematical Conservation of Energy gives a mathematical

expression of this: expression of this:

QQcoldcold= -= -QQhothot


• The negative sign in the equation is critical for The negative sign in the equation is critical for consistency with the established sign conventionconsistency with the established sign convention

• Since each Since each QQ = = mcmcTT, , ccsamplesample can be found by: can be found by:

Technically, the mass of the container should be Technically, the mass of the container should be included, but if included, but if mmw w >>>>mmcontainercontainer

it can be neglectedit can be neglected

w w f w

s

s s f

m c T Tc

m T T

)()( fssswfww TTcmTTcm

Phase ChangesPhase Changes

• A A phase changephase change is happened when a is happened when a substance changes from one form to anothersubstance changes from one form to another– Two common phase changes areTwo common phase changes are

• Solid to liquid (melting)Solid to liquid (melting)• Liquid to gas (boiling)Liquid to gas (boiling)

• During a phase change, there is During a phase change, there is no change no change in temperaturein temperature of the substance of the substance

Latent HeatLatent Heat

• Different substances react differently to the Different substances react differently to the energy added or removed during a phase energy added or removed during a phase change due to their different molecular change due to their different molecular arrangementsarrangements

• The amount of energy also depends on the The amount of energy also depends on the mass of the samplemass of the sample

• If an amount of energy If an amount of energy QQ is required to change is required to change the phase of a sample of mass the phase of a sample of mass mm, ,

LL = = QQ / /mm


• If an amount of energy If an amount of energy QQ is required to change is required to change the phase of a sample of mass the phase of a sample of mass mm, ,

LL = = QQ / /mm• The quantityThe quantity LL is called the is called the latent heatlatent heat of the of the

materialmaterialLatent means “hidden”Latent means “hidden”

The value of The value of LL depends on the substance as well as depends on the substance as well as the actual phase changethe actual phase change

• The energy required to change the phase is The energy required to change the phase is

Q = Q = mLmL


• The The latent heat of fusionlatent heat of fusion is used when the phase is used when the phase change is from solid to liquidchange is from solid to liquid

• The The latent heat of vaporizationlatent heat of vaporization is used when the is used when the phase change is from liquid to gasphase change is from liquid to gas

• The positive sign is used when the energy is The positive sign is used when the energy is transferred into the systemtransferred into the system– This will result in melting or boilingThis will result in melting or boiling

• The negative sign is used when energy is The negative sign is used when energy is transferred out of the systemtransferred out of the system– This will result in freezing or condensationThis will result in freezing or condensation

Sample Latent Heat ValuesSample Latent Heat Values

Graph of Ice to SteamGraph of Ice to Steam

Warming Ice, Graph Part AWarming Ice, Graph Part A

• Start with Start with one gramone gram of ice at of ice at –30.0–30.0ºCºC

• During phase During phase AA, the , the temperature of the ice temperature of the ice changes from changes from –30.–30.00ºCºC to to 00ºCºC. The specific heat of . The specific heat of ice is ice is 2090 J/kg2090 J/kg··00CC

Q = mQ = mi i ccii ΔT = (1x10 ΔT = (1x10-3-3kg)(2090 kg)(2090

J/kgJ/kg··00C)(30.0C)(30.000C)=62.7JC)=62.7JIn this case, In this case, 62.7 J62.7 J of energy of energy are addedare added

Melting Ice, Graph Part BMelting Ice, Graph Part B

• Once at Once at 00ºCºC, the phase change , the phase change (melting) starts(melting) starts

• The temperature stays the same The temperature stays the same although energy is still being although energy is still being added. The latent heat of fusion added. The latent heat of fusion for water is for water is 3.33x103.33x1055 J/kg J/kg

QQ = =mmi i LLff =(1x10=(1x10-3 -3 kg)(3.33x10kg)(3.33x1055

J/kg)=333JJ/kg)=333J The energy required is The energy required is 333 J333 J

On the graph, the values move from On the graph, the values move from 62.7 J62.7 J to to 396 J396 J

Warming Water, Graph Part CWarming Water, Graph Part C

• Between Between 00ºCºC and and 100ºC100ºC, the , the material is liquid and no material is liquid and no phase changes take placephase changes take place

• Energy added increases the Energy added increases the temperature. Specific heat for temperature. Specific heat for water water 4186 J/kg4186 J/kg··00CC..

QQ = = mmwwccww Δ ΔT = 419J.T = 419J.

419 J419 J are added are added

The total is now The total is now 815 J815 J

Boiling Water, Graph Part DBoiling Water, Graph Part D

• At At 100100ºCºC, a phase , a phase change occurs change occurs (boiling)(boiling)

• Temperature does not Temperature does not change. The latent change. The latent heat of vaporization for heat of vaporization for water is water is 2.26x102.26x1066J/kgJ/kg..

• Use Use QQ = = mmww L Lvv. .

This requires This requires 2260 J2260 JThe total is now The total is now 3075 J3075 J

Heating SteamHeating Steam

• After all the water is converted to After all the water is converted to steam, the steam will heat upsteam, the steam will heat up

• No phase change occursNo phase change occurs

• The added energy goes to increasing The added energy goes to increasing the temperature. The specific heat for the temperature. The specific heat for steam is steam is 2010J/kg2010J/kg∙∙00CC..

• Use Use

QQ = = mmssccss Δ ΔTT

In this case, In this case, 40.2 J40.2 J are needed are needed

The temperature is going to The temperature is going to 120120oo C C

• The total is now The total is now 3115 J3115 J

If water with a mass If water with a mass mmhh at temperature at temperature TThh is is

poured into an aluminum cup of mass poured into an aluminum cup of mass mmAl Al

containing mass containing mass mmcc of water at of water at TTcc, where , where TThh > >

TTcc, what is the equilibrium temperature of the , what is the equilibrium temperature of the

system?system?

If water with a mass If water with a mass mmhh at temperature at temperature TThh is is

poured into an aluminum cup of mass poured into an aluminum cup of mass mmAl Al

containing mass containing mass mmcc of water at of water at TTcc, where , where TThh > >

TTcc, what is the equilibrium temperature of the , what is the equilibrium temperature of the

system?system?

cold hotQ Q

Al Al

Al Al Al Al

Al Al Al Al

Al Al

Al Al

f c c w f c h w f h

c w f c w c h w f h w h

c w h w f c w c h w h

c w c h w hf

c w h w

m c T T mc T T m c T T

m c mc T m c mc T m c T m c T

m c mc m c T m c mc T m c T

m c mc T m c TT

m c mc m c

Molecular View of Phase ChangesMolecular View of Phase Changes

• Phase changes can be described in terms of the Phase changes can be described in terms of the rearrangement of molecules (or atoms in an rearrangement of molecules (or atoms in an elemental substance)elemental substance)

• Liquid to Gas phase changeLiquid to Gas phase changeMolecules in a liquid are close togetherMolecules in a liquid are close togetherThe forces between them are stronger than those in a The forces between them are stronger than those in a gasgasWork must be done to separate the moleculesWork must be done to separate the moleculesThe latent heat of vaporization is the energy per unit The latent heat of vaporization is the energy per unit mass needed to accomplish this separationmass needed to accomplish this separation


Solid to Liquid phase changeSolid to Liquid phase change• The addition of energy will cause the amplitude of The addition of energy will cause the amplitude of

the vibration of the molecules about their equilibrium the vibration of the molecules about their equilibrium position to increaseposition to increase

• At the melting point, the amplitude is great At the melting point, the amplitude is great enough to break apart bonds between the moleculesenough to break apart bonds between the molecules

• The molecules can move to new positionsThe molecules can move to new positions

• The molecules in the liquid are bound together The molecules in the liquid are bound together less strongly than those of the solidless strongly than those of the solid

• The latent heat of fusion is the energy per unit The latent heat of fusion is the energy per unit mass required to go from the solid-type to the liquid-type mass required to go from the solid-type to the liquid-type bondsbonds


• The latent heat of vaporization is greater The latent heat of vaporization is greater than the latent heat of fusionthan the latent heat of fusion

In the liquid-to-gas phase change, the liquid-In the liquid-to-gas phase change, the liquid-type bonds are brokentype bonds are brokenThe gas molecules are essentially not bonded The gas molecules are essentially not bonded to each otherto each other

• It takes more energy to completely break It takes more energy to completely break the bonds than to change the type of the bonds than to change the type of bondsbonds

Calorimetry Problem-Solving StrategyCalorimetry Problem-Solving Strategy

• Units of measure must be consistentUnits of measure must be consistentFor example, if your value of For example, if your value of cc is in is in J/kgJ/kg.o.oCC, then , then your mass must be in your mass must be in kgkg, the temperatures in , the temperatures in ooCC and energies in and energies in JJ

• Transfers of energy are given by Transfers of energy are given by Q Q ==mc mc TT only when no phase change only when no phase change occursoccurs

• If there is a phase change, use If there is a phase change, use QQ = = mLmL

Calorimetry Problem-Solving StrategyCalorimetry Problem-Solving Strategy

• Be sure to select the correct sign for all Be sure to select the correct sign for all energy transfersenergy transfers

• Remember to use Remember to use QQcoldcold = - = - QQhothot

– The The TT is always is always TTff - - TTii

State VariablesState Variables

• State variables describe the state of a systemState variables describe the state of a system

• In the macroscopic approach to In the macroscopic approach to thermodynamics, variables are used to describe thermodynamics, variables are used to describe the state of the systemthe state of the system

Pressure, temperature, volume, internal energyPressure, temperature, volume, internal energyThese are examples of These are examples of state variablesstate variables

• The macroscopic state of an isolated system can The macroscopic state of an isolated system can be specified only if the system is in thermal be specified only if the system is in thermal equilibrium internallyequilibrium internally

Work in ThermodynamicsWork in Thermodynamics

• Work can be done on a deformable Work can be done on a deformable system, such as a gassystem, such as a gas

• Consider a cylinder with a moveable Consider a cylinder with a moveable pistonpiston

• A force is applied to slowly compress A force is applied to slowly compress the gasthe gas

The compression is slow enough The compression is slow enough for all the system to remain for all the system to remain essentially in thermal equilibriumessentially in thermal equilibrium

This is said to occur This is said to occur quasi-quasi-staticallystatically

WorkWork

• The piston is pushed downward by a force The piston is pushed downward by a force FF through a displacement of through a displacement of drdr::

AA··dydy is the change in volume of the gas, is the change in volume of the gas, dV dV Therefore, the work done on the gas is Therefore, the work done on the gas is

dWdW = - = -P dVP dV

ˆ ˆ dW d F dy Fdy PA dy F r j j

WorkWork• Interpreting Interpreting dWdW = - = - P P dVdV

– If the gas is compressed, If the gas is compressed, dVdV is negative and is negative and the work done on the gas is positivethe work done on the gas is positive

– If the gas expands, If the gas expands, dVdV is positive and the is positive and the work done on the gas is negativework done on the gas is negative

– If the volume remains constant, the work done If the volume remains constant, the work done is zerois zero

• The total work done is: The total work done is:

f

i

V

VW P dV

PVPV Diagrams Diagrams

The state of the gas at The state of the gas at each step can be plotted each step can be plotted on a graph called a on a graph called a PVPV diagramdiagram

This allows us to visualize This allows us to visualize the process through which the process through which the gas is progressingthe gas is progressing

• The curve is called the The curve is called the pathpath

• PV diagramsPV diagrams can be can be used when the pressure used when the pressure and volume are known at and volume are known at each step of the processeach step of the process

PVPV Diagrams Diagrams

• The work done on a gas in a quasi-The work done on a gas in a quasi-static process that takes the gas from static process that takes the gas from an initial state to a final state is the an initial state to a final state is the negative of the area under the curve on negative of the area under the curve on the the PVPV diagram, evaluated between the diagram, evaluated between the initial and final statesinitial and final states–This is true whether or not the pressure This is true whether or not the pressure

stays constantstays constant

–The work done The work done doesdoes depend on the path depend on the path takentaken

Work Done By Various PathsWork Done By Various Paths

• Each of these processes has the same initial Each of these processes has the same initial and final statesand final states

• The work done differs in each processThe work done differs in each process• The work done depends on the pathThe work done depends on the path

Work From a Work From a PVPV Diagram Diagram

• The volume of the The volume of the gas is first reduced gas is first reduced from from VVii to to VVff at at

constant pressure constant pressure PPii

• Next, the pressure Next, the pressure increases from increases from PPii to to

PPf f by heating at by heating at

constant volume constant volume VVff

• WW = - = -PPi i ((VVff – – VVii))


• The pressure of the gas The pressure of the gas is increased from is increased from PPii to to

PPff at a constant volumeat a constant volume

• The volume is The volume is decreased from decreased from VVii to to VVff

• WW = - = -PPf f ((VVff – – VVii))


• The pressure and the The pressure and the volume continually changevolume continually change

• The work is some The work is some intermediate value intermediate value between between PPf f ((VVff – – V Vii)) and and PPi i

((VVff – – VVii))

• To evaluate the actual To evaluate the actual amount of work, the amount of work, the function function PP((VV)) must be must be knownknown

V

A sample of ideal gas is expanded to twice its A sample of ideal gas is expanded to twice its original volume of original volume of 1.00 m1.00 m33 in a quasi-static process in a quasi-static process for which for which PP = = V V 22, with , with

= 5.00 atm/m= 5.00 atm/m66, as shown , as shown

in Figure. How much work is in Figure. How much work is

done on the expanding gas?done on the expanding gas?

A sample of ideal gas is expanded to twice its original volume A sample of ideal gas is expanded to twice its original volume of of 1.00 m1.00 m33 in a quasi-static process for which in a quasi-static process for which PP = = V V 22, with , with



done on the expanding gas?done on the expanding gas?f

ifi

W PdV

2P V iV fV

2 3 3

3 3

13

2 2 1.00 m 2.00 m

f

iff ii

fi

W V dV V V

V V

The work done on the gas is the negative of the area under the curve

between

and

.

A sample of ideal gas is expanded to twice its original volume A sample of ideal gas is expanded to twice its original volume of of 1.00 m1.00 m33 in a quasi-static process for which in a quasi-static process for which PP = = V V 22, with , with



done on the expanding gas?done on the expanding gas?

f

ifi

W PdV .

MJ

mmatm

Pa

m

atmWif

18.1

00.100.210013.100.53

1 333356

(a) Determine the work (a) Determine the work done on a fluid that done on a fluid that expands fromexpands from ii to to ff as as indicated in Figure. indicated in Figure. (b) How much work is (b) How much work is performed on the fluid if it performed on the fluid if it is compressed from is compressed from ff to to ii along the same path?along the same path?

(a) Determine the work (a) Determine the work done on a fluid that done on a fluid that expands fromexpands from ii to to ff as as indicated in Figure. (b) indicated in Figure. (b) How much work is How much work is performed on the fluid if it performed on the fluid if it is compressed from is compressed from ff to to ii along the same path?along the same path?

W PdV (a)

6 3

6 3

6 3

6.00 10 Pa 2.00 1.00 m

4.00 10 Pa 3.00 2.00 m

2.00 10 Pa 4.00 3.00 m

12.0 MJi f

W

W

(b)

12.0 MJfiW

Heat TransferHeat Transfer

• The energy transfer, The energy transfer, QQ, into , into or out of a system also or out of a system also depends on the processdepends on the process

• The piston is pushed The piston is pushed upward, the gas is doing upward, the gas is doing work on the pistonwork on the piston

Heat TransferHeat Transfer

• This gas has the same initial This gas has the same initial volume, temperature and volume, temperature and pressure as the previous pressure as the previous exampleexample

• The final states are also The final states are also identicalidentical

• No energy is transferred by No energy is transferred by heat through the insulating heat through the insulating wallwall

• No work is done by the gas No work is done by the gas expanding into the vacuumexpanding into the vacuum

Energy Transfer, SummaryEnergy Transfer, Summary

• Energy transfers by heat, like the work Energy transfers by heat, like the work done, depend on the initial, final, and done, depend on the initial, final, and intermediate states of the systemintermediate states of the system

• Both work and heat depend on the path Both work and heat depend on the path takentaken

• Neither can be determined solely by the Neither can be determined solely by the end points of a thermodynamic processend points of a thermodynamic process

The First Law of ThermodynamicsThe First Law of Thermodynamics

• The The First Law of ThermodynamicsFirst Law of Thermodynamics is a is a special case of the Law of Conservation of special case of the Law of Conservation of EnergyEnergy

It takes into account changes in internal energy and It takes into account changes in internal energy and energy transfers by heat and workenergy transfers by heat and work

• Although Although QQ and and WW each are dependent on the each are dependent on the path, path, QQ + + WW is independent of the path is independent of the path

• The First Law of Thermodynamics states that The First Law of Thermodynamics states that EEintint = = QQ + + W W

All quantities must have the same units of measure All quantities must have the same units of measure of energyof energy

The First Law of ThermodynamicsThe First Law of Thermodynamics

• One consequence of the first law is that there One consequence of the first law is that there must exist some quantity known as internal must exist some quantity known as internal energy which is determined by the state of the energy which is determined by the state of the systemsystem

• For infinitesimal changes in a system For infinitesimal changes in a system dEdEintint = = dQdQ + + dWdW

• The first law is an energy conservation The first law is an energy conservation statement specifying that the only type of energy statement specifying that the only type of energy that changes in a system is internal energythat changes in a system is internal energy

Isolated SystemsIsolated Systems

• An isolated system is one that does not An isolated system is one that does not interact with its surroundingsinteract with its surroundings• No energy transfer by heat takes placeNo energy transfer by heat takes place

• The work done on the system is zeroThe work done on the system is zero

QQ = = WW = 0 = 0, so , so EEintint = 0 = 0

• The internal energy of an isolated system The internal energy of an isolated system remains constantremains constant

Cyclic ProcessesCyclic Processes

• A cyclic process is one that starts and ends in the A cyclic process is one that starts and ends in the same statesame state• This process would not be isolatedThis process would not be isolated

• On a On a PVPV diagram, a cyclic process appears as a diagram, a cyclic process appears as a closed curveclosed curve

• The change in internal energy must be zero since it The change in internal energy must be zero since it is a state variableis a state variable

If If EEintint = 0 = 0, , QQ = - = -WW

• In a cyclic process, the net work done on the system In a cyclic process, the net work done on the system per cycle equals the area enclosed by the path per cycle equals the area enclosed by the path representing the process on a representing the process on a PVPV diagram diagram

Adiabatic ProcessAdiabatic Process

• An adiabatic process is one An adiabatic process is one during which no energy enters during which no energy enters or leaves the system by heator leaves the system by heat

QQ = 0 = 0

This is achieved by:This is achieved by:Thermally insulating the walls Thermally insulating the walls of the systemof the system

Having the process proceed Having the process proceed so quickly that no heat can so quickly that no heat can be exchangedbe exchanged

Adiabatic ProcessAdiabatic Process

• Since Since QQ = 0 = 0, , EEintint = = WW

• If the gas is compressed adiabatically, If the gas is compressed adiabatically, WW is positive so is positive so EEintint is positive and the is positive and the

temperature of the gas increasestemperature of the gas increases

• If the gas expands adiabatically, the If the gas expands adiabatically, the temperature of the gas decreasestemperature of the gas decreases

Adiabatic ProcessesAdiabatic Processes

• Some important examples of adiabatic Some important examples of adiabatic processes related to engineering are:processes related to engineering are:–The expansion of hot gases in an internal The expansion of hot gases in an internal

combustion enginecombustion engine

–The liquefaction of gases in a cooling The liquefaction of gases in a cooling systemsystem

–The compression stroke in a diesel engineThe compression stroke in a diesel engine

Adiabatic Free ExpansionAdiabatic Free Expansion

• This is an example of adiabatic free This is an example of adiabatic free expansionexpansion

• The process is adiabatic because it The process is adiabatic because it takes place in an insulated takes place in an insulated containercontainer

• Because the gas expands into a Because the gas expands into a vacuum, it does not apply a force vacuum, it does not apply a force on a piston and on a piston and WW = 0 = 0

• Since Since QQ = 0 = 0 and and WW = 0 = 0, , EEintint = 0 = 0 and the initial and final states are and the initial and final states are the samethe same– No change in temperature is No change in temperature is

expectedexpected

Isobaric ProcessesIsobaric Processes

• An isobaric process is one that occurs at a An isobaric process is one that occurs at a constant pressureconstant pressure

• The values of the heat and the work are The values of the heat and the work are generally both nonzerogenerally both nonzero

• The work done isThe work done is WW = = PP ( (VVff – – VVii)) wherewhere P

is the constant pressureis the constant pressure

Isovolumetric ProcessesIsovolumetric Processes

• An isovolumetric process is one in which there An isovolumetric process is one in which there is no change in the volumeis no change in the volume

• Since the volume does not change, Since the volume does not change, WW = 0 = 0

• From the first law, From the first law, EEintint = = QQ

• If energy is added by heat to a system kept at If energy is added by heat to a system kept at constant volume, all of the transferred energy constant volume, all of the transferred energy remains in the system as an increase in its remains in the system as an increase in its internal energyinternal energy

Isothermal ProcessIsothermal Process

• An isothermal process is one that occurs An isothermal process is one that occurs at a constant temperatureat a constant temperature

• Since there is no change in temperature, Since there is no change in temperature, EEintint = 0 = 0

• Therefore, Therefore, QQ = - = - WW

• Any energy that enters the system by heat Any energy that enters the system by heat must leave the system by workmust leave the system by work

Isothermal ProcessIsothermal Process

• At right is a At right is a PVPV diagram of an diagram of an isothermal expansionisothermal expansion

• The curve is a The curve is a hyperbolahyperbola

• The curve is called as The curve is called as isothermisotherm

Isothermal ExpansionIsothermal Expansion

• The curve of the The curve of the PVPV diagram indicates diagram indicates PVPV = constant = constant

The equation of a hyperbolaThe equation of a hyperbola

• Because it is an ideal gas and the process Because it is an ideal gas and the process is quasi-static, is quasi-static, PVPV = = nRTnRT and and

ln

f f f

i i i

V V V

V V V

i

f

nRT dVW PdV dV nRT

V V

VW nRT

V

Isothermal ExpansionIsothermal Expansion

• Numerically, the work equals the area Numerically, the work equals the area under theunder the PVPV curvecurve

The shaded area in the diagramThe shaded area in the diagram

• If the gas expands, If the gas expands, VVff > > VVii and the work and the work

done on the gas is negativedone on the gas is negative

• If the gas is compressed, If the gas is compressed, VVff < < VVii and the and the

work done on the gas is positivework done on the gas is positive

Special Processes, SummarySpecial Processes, Summary

• AdiabaticAdiabaticNo heat exchangedNo heat exchanged

QQ = 0 = 0 and and EEintint = = WW

• IsobaricIsobaricConstant pressureConstant pressure

WW = = PP ( (VVff – – VVii)) and and EEintint = = QQ + + WW

• IsothermalIsothermalConstant temperatureConstant temperature

EEintint = 0 = 0 and and QQ = - = -WW

A sample of an ideal gas goes through the process shown A sample of an ideal gas goes through the process shown in Figure. From in Figure. From AA to to BB, the process is adiabatic; from, the process is adiabatic; from BB to to CC, it is isobaric with , it is isobaric with 100 kJ100 kJ of energy entering the of energy entering the system by heat. From system by heat. From CC to to DD, the process is isothermal; , the process is isothermal; from from DD to to AA, it is isobaric with , it is isobaric with 150 kJ150 kJ of energy leaving of energy leaving the system by heat. Determine the difference in internal the system by heat. Determine the difference in internal energy energy EEintint,,BB – E – Eint,int,AA..

A sample of an ideal gas goes through the process shown in Figure. From A sample of an ideal gas goes through the process shown in Figure. From AA to to BB, the process is adiabatic; from, the process is adiabatic; from BB to to CC, it is isobaric with , it is isobaric with 100 kJ100 kJ of of energy entering the system by heat. From energy entering the system by heat. From CC to to DD, the process is , the process is isothermal; from isothermal; from DD to to AA, it is isobaric with , it is isobaric with 150 kJ150 kJ of energy leaving the of energy leaving the system by heat. Determine the difference in internal energy system by heat. Determine the difference in internal energy EEintint,,BB – E – Eint,int,AA..

3BC B C B 3.00 atm 0.400 0.0900 m

94.2 kJ

W P V V

int

int, C int, B

int, C int, B

100 94.2 kJ

5.79 kJ

E Q W

E E

E E


3BC B C B 3.00 atm 0.400 0.0900 m

94.2 kJ

W P V V

int

int, C int, B

int, C int, B

100 94.2 kJ

5.79 kJ

E Q W

E E

E E

T=constantT=constant int, D int, C 0E E

3DA D A D 1.00 atm 0.200 1.20 m

101 kJ

W P V V


int, A int, D 150 kJ 101 kJ 48.7 kJE E

int, B int, A int, C int, B int, D int, C int, A int, DE E E E E E E E

int, B int, A 5.79 kJ 0 48.7 kJ 42.9 kJE E

Mechanisms of Heat TransferMechanisms of Heat Transfer

• We want to know the rate at which energy is transferred

• There are various mechanisms responsible for the transfer:– Conduction– Convection– Radiation

ConductionConduction

• The transfer can be viewed on an atomic The transfer can be viewed on an atomic scalescale– It is an exchange of energy between It is an exchange of energy between

microscopic particles by collisionsmicroscopic particles by collisionsThe microscopic particles can be atoms, The microscopic particles can be atoms, molecules or free electronsmolecules or free electrons

– Less energetic particles gain energy during Less energetic particles gain energy during collisions with more energetic particlescollisions with more energetic particles

• Rate of conduction depends upon the Rate of conduction depends upon the characteristics of the substancecharacteristics of the substance

Conduction exampleConduction example

• The molecules vibrate The molecules vibrate about their equilibrium about their equilibrium positionspositions

• Particles near the heat Particles near the heat source vibrate with larger source vibrate with larger amplitudesamplitudes

• These collide with These collide with adjacent molecules and adjacent molecules and transfer some energytransfer some energy

• Eventually, the energy Eventually, the energy travels entirely through travels entirely through the panthe pan


• In general, metals are good conductorsIn general, metals are good conductorsThey contain large numbers of electrons that are They contain large numbers of electrons that are relatively free to move through the metalrelatively free to move through the metal

They can transport energy from one region to They can transport energy from one region to anotheranother

• Poor conductors include asbestos, paper, and Poor conductors include asbestos, paper, and gasesgases

• Conduction can occur only if there is a Conduction can occur only if there is a difference in temperature between two parts of difference in temperature between two parts of the conducting mediumthe conducting medium


• The slab at right The slab at right allows energy to allows energy to transfer from the transfer from the region of higher region of higher temperature to the temperature to the region of lower region of lower temperaturetemperature

• The rate of transfer is The rate of transfer is given by:given by:

dx

dTkA

t

QP


• AA is the cross-sectional area is the cross-sectional area

• ΔΔxx is the thickness of the slab is the thickness of the slab or the length of a rodor the length of a rod

• PP is in is in WattsWatts when when QQ is in is in JoulesJoules and and tt is in is in secondsseconds

• kk is the is the thermal conductivitythermal conductivity of the of the materialmaterial

Good conductors have highGood conductors have high kk values and values and good insulators have low good insulators have low kk values values

Temperature GradientTemperature Gradient

• The quantity The quantity ||dTdT / / dxdx|| is is called the called the temperature temperature gradientgradient of the material of the material

It measures the rate at It measures the rate at which temperature varies which temperature varies with positionwith position

• For a rod, the For a rod, the temperature gradient temperature gradient can be expressed as:can be expressed as:

h cdT T T

dx L

Rate of Energy Transfer in a RodRate of Energy Transfer in a Rod

• Using the temperature gradient for the rod, the rate of energy transfer becomes:

L

TTkAP ch

Compound SlabCompound Slab

• For a compound slab containing several materials For a compound slab containing several materials of various thicknesses (of various thicknesses (LL11, , LL22, …, …) and various ) and various

thermal conductivities (thermal conductivities (kk11, , kk22, …, …) the rate of ) the rate of

energy transfer depends on the materials and the energy transfer depends on the materials and the temperatures at the outer edges:temperatures at the outer edges:

ii

i

ch

kL

TTAP

Some Thermal ConductivitiesSome Thermal Conductivities

More Thermal ConductivitiesMore Thermal Conductivities

In Figure, the change in internal In Figure, the change in internal energy of a gas that is taken from energy of a gas that is taken from AA to to CC is is +800 J+800 J. The work done on . The work done on the gas along path the gas along path ABCABC is is –500 J–500 J. . (a) How much energy must be (a) How much energy must be added to the system by heat as it added to the system by heat as it goes from goes from AA through through BB to to CC? ? (b) If the pressure at point (b) If the pressure at point AA is five is five times that of point times that of point CC,, what is the what is the work done on the system in going work done on the system in going from from CC to to DD? (c) What is the energy ? (c) What is the energy exchanged with the surroundings exchanged with the surroundings by heat as the cycle goes from by heat as the cycle goes from CC to to AA along the green path? (d) If the along the green path? (d) If the change in internal energy in going change in internal energy in going from point from point DD to point to point AA is is +500 J+500 J, , how much energy must be added how much energy must be added to the system by heat as it goes to the system by heat as it goes from point from point CC to point to point DD??

In Figure, the change in internal energy of In Figure, the change in internal energy of a gas that is taken from a gas that is taken from AA to to CC is is +800 J+800 J. . The work done on the gas along path The work done on the gas along path ABCABC is is –500 J–500 J. . (a) How much energy must be added to (a) How much energy must be added to the system by heat as it goes from the system by heat as it goes from AA through through BB to to CC? ? (b) If the pressure at point (b) If the pressure at point AA is five times is five times that of point that of point CC,, what is the work done on what is the work done on the system in going from the system in going from CC to to DD? (c) What ? (c) What is the energy exchanged with the is the energy exchanged with the surroundings by heat as the cycle goes surroundings by heat as the cycle goes from from CC to to AA along the green path? (d) If along the green path? (d) If the change in internal energy in going the change in internal energy in going from point from point DD to point to point AA is is +500 J+500 J, how , how much energy must be added to the much energy must be added to the system by heat as it goes from point system by heat as it goes from point CC to to point point DD??

int, int, ABC ACE E (conservation of energy)

(a) int, ABC ABC ABCE Q W (First Law)

800 J 500 J 1300 JABCQ


(b)CD C CDW P V

,

AB CDV V , and 5A CP P

Then, 1 1100 J

5 5CD A AB ABW P V W

(+ means that work is done on the system)


(c) CDA CDW W so that

int, 800J 100J 900JCA CA CDAQ E W

(– means that energy must be removed from the system by heat)

In Figure, the change in internal energy of In Figure, the change in internal energy of a gas that is taken from a gas that is taken from AA to to CC is is +800 J+800 J. . The work done on the gas along path The work done on the gas along path ABCABC is is –500 J–500 J. . (a) How much energy must be added to (a) How much energy must be added to the system by heat as it goes from the system by heat as it goes from AA through through BB to to CC? ? (b) If the pressure at point (b) If the pressure at point AA is five times is five times that of point that of point CC,, what is the work done on what is the work done on the system in going from the system in going from CC to to DD? (c) What ? (c) What is the energy exchanged with the is the energy exchanged with the surroundings by heat as the cycle goes surroundings by heat as the cycle goes from from CC to to AA along the green path? (d) If along the green path? (d) If the change in internal energy in going the change in internal energy in going from point from point DD to point to point AA is is +500 J+500 J, how , how much energy must be added to the much energy must be added to the system by heat as it goes from point system by heat as it goes from point CC to to point point DD??(d)

int, int, int, 800 J 500 J 1300 JCD CDA DAE E E

and int, 1300 J 100 J 1400 JCD CD CDQ E W

A bar of gold is in thermal contact with a bar of silver A bar of gold is in thermal contact with a bar of silver of the same length and area. One end of the of the same length and area. One end of the compound bar is maintained at compound bar is maintained at 80.0°C 80.0°C while the while the opposite end is at opposite end is at 30.0°C30.0°C. When the energy transfer . When the energy transfer reaches steady state, what is the temperature at the reaches steady state, what is the temperature at the junction?junction?


In the steady state condition, Au AgP P

so thatAu Au Ag Ag

Au Ag

T Tk A k A

x x

In this case Au AgA A

Au Ag

Au 80.0

x x

T T

Ag 30.0T T


In this case Au AgA A

Au Ag

Au 80.0

x x

T T

Ag 30.0T T

where T is the temperature of the junction.

Au Ag80.0 30.0k T k T

51.2 CT

Therefore,

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Chapter 20