HeatHeatHeatHeat

When two objects at different When two objects at different temperatures are put into temperatures are put into contact, heat spontaneously contact, heat spontaneously flows from the hotter to the flows from the hotter to the cooler one. If kept in contact cooler one. If kept in contact long enough- Their long enough- Their temperatures will become temperatures will become equalequal

We call this We call this thermal thermal equilibriumequilibrium

When two objects at different When two objects at different temperatures are put into temperatures are put into contact, heat spontaneously contact, heat spontaneously flows from the hotter to the flows from the hotter to the cooler one. If kept in contact cooler one. If kept in contact long enough- Their long enough- Their temperatures will become temperatures will become equalequal

We call this We call this thermal thermal equilibriumequilibrium

HEATHEATHeatHeat: the energy that is : the energy that is

transferred from one body to transferred from one body to another because of a another because of a difference in temperature.difference in temperature.

Symbol for HeatSymbol for Heat: Q: QSI unit for heatSI unit for heat: (same as for : (same as for

energy)- Joule (J)energy)- Joule (J)

HeatHeat: the energy that is : the energy that is transferred from one body to transferred from one body to another because of a another because of a difference in temperature.difference in temperature.

Symbol for HeatSymbol for Heat: Q: QSI unit for heatSI unit for heat: (same as for : (same as for

energy)- Joule (J)energy)- Joule (J)

1818thth century model had heat century model had heat flow like a fluid- they called it flow like a fluid- they called it caloric. Due to history we have caloric. Due to history we have various units for heat.various units for heat.

CALORIE:CALORIE:

(cal) – Note the lowercase ‘c’(cal) – Note the lowercase ‘c’ This is the amount of heat This is the amount of heat

needed to raise the needed to raise the temperature of 1g of Htemperature of 1g of H22O by 1O by 1°°CC

1818thth century model had heat century model had heat flow like a fluid- they called it flow like a fluid- they called it caloric. Due to history we have caloric. Due to history we have various units for heat.various units for heat.

CALORIE:CALORIE:

(cal) – Note the lowercase ‘c’(cal) – Note the lowercase ‘c’ This is the amount of heat This is the amount of heat

needed to raise the needed to raise the temperature of 1g of Htemperature of 1g of H22O by 1O by 1°°CC

Kilocalorie:Kilocalorie:

(kcal) this is 1000 calories(kcal) this is 1000 caloriesHeat needed to raise 1kg of Heat needed to raise 1kg of

water by 1water by 1°°C C

CalorieCalorie

Note the capital ‘C’Note the capital ‘C’a Calorie is 1kcal, it is also a Calorie is 1kcal, it is also

called a dietary caloriecalled a dietary calorie

Kilocalorie:Kilocalorie:

(kcal) this is 1000 calories(kcal) this is 1000 caloriesHeat needed to raise 1kg of Heat needed to raise 1kg of

water by 1water by 1°°C C

CalorieCalorie

Note the capital ‘C’Note the capital ‘C’a Calorie is 1kcal, it is also a Calorie is 1kcal, it is also

called a dietary caloriecalled a dietary calorie

British Thermal UnitsBritish Thermal Units

(BTU’s)(BTU’s)1 BTU= .252 kcal = 1055 J1 BTU= .252 kcal = 1055 JHeat needed to raise Heat needed to raise

temperature of 1lb of Htemperature of 1lb of H22O O by 1by 1°°FF

British Thermal UnitsBritish Thermal Units

(BTU’s)(BTU’s)1 BTU= .252 kcal = 1055 J1 BTU= .252 kcal = 1055 JHeat needed to raise Heat needed to raise

temperature of 1lb of Htemperature of 1lb of H22O O by 1by 1°°FF

James Joule ExperimentJames Joule ExperimentJames Joule ExperimentJames Joule Experiment

The break through idea from this The break through idea from this experiment is that work done had experiment is that work done had an equivalent amount of heat an equivalent amount of heat transferred.transferred.““Mechanical Equivalent of Heat”Mechanical Equivalent of Heat”

Doing work (fd, mgh, ½mvDoing work (fd, mgh, ½mv22) on an ) on an object raises its temperature object raises its temperature

1 cal= 4.186 J1 cal= 4.186 J1 kcal = 4186 J1 kcal = 4186 J

The break through idea from this The break through idea from this experiment is that work done had experiment is that work done had an equivalent amount of heat an equivalent amount of heat transferred.transferred.““Mechanical Equivalent of Heat”Mechanical Equivalent of Heat”

Doing work (fd, mgh, ½mvDoing work (fd, mgh, ½mv22) on an ) on an object raises its temperature object raises its temperature

1 cal= 4.186 J1 cal= 4.186 J1 kcal = 4186 J1 kcal = 4186 J

Joule’s Experiment

Cake & Ice Cream= mghCake & Ice Cream= mghCake & Ice Cream= mghCake & Ice Cream= mgh

ExampleExample How many stairs would you need How many stairs would you need

to climb to “work off” 500 to climb to “work off” 500 Calories of cake and ice cream? Calories of cake and ice cream? Assume mass of person = 60kgAssume mass of person = 60kg

How many stairs would you need How many stairs would you need to climb to “work off” 500 to climb to “work off” 500 Calories of cake and ice cream? Calories of cake and ice cream? Assume mass of person = 60kgAssume mass of person = 60kg

500 Calories = 500kcal500 Calories = 500kcal500 Calories = 500kcal500 Calories = 500kcal

500 kcal (4186 J/kcal) = 2.1 x 10500 kcal (4186 J/kcal) = 2.1 x 1066 J J500 kcal (4186 J/kcal) = 2.1 x 10500 kcal (4186 J/kcal) = 2.1 x 1066 J J

Work Done in climbing stairs = mghWork Done in climbing stairs = mghWork Done in climbing stairs = mghWork Done in climbing stairs = mgh

2.1 x 102.1 x 1066 J = 60kg (9.8)h J = 60kg (9.8)h2.1 x 102.1 x 1066 J = 60kg (9.8)h J = 60kg (9.8)h3600m = h3600m = h

About 2 miles About 2 miles notenote- this assumes body is - this assumes body is 100% efficient machine100% efficient machine

About 2 miles About 2 miles notenote- this assumes body is - this assumes body is 100% efficient machine100% efficient machine

ExampleExample A 3g bullet is shot at a tree and A 3g bullet is shot at a tree and

passes through it. The bullet speed passes through it. The bullet speed changes from 400m/s to 200m/s. changes from 400m/s to 200m/s. How much heat (Q) is produces and How much heat (Q) is produces and shared by the bullet and tree.shared by the bullet and tree.

Q = ΔKEQ = ΔKE

Q = KEQ = KEff-KE-KEi i

Q = ½m (VQ = ½m (Vff22 – V – Vii

22))

Q = ½ (.003kg)(400Q = ½ (.003kg)(40022- 200- 20022))

Q= 180 JQ= 180 J

180/4.186 = 43 cal180/4.186 = 43 cal

Q = ΔKEQ = ΔKE

Q = KEQ = KEff-KE-KEi i

Q = ½m (VQ = ½m (Vff22 – V – Vii

22))

Q = ½ (.003kg)(400Q = ½ (.003kg)(40022- 200- 20022))

Q= 180 JQ= 180 J

180/4.186 = 43 cal180/4.186 = 43 cal

Temperature, Heat, Temperature, Heat, Internal EnergyInternal Energy

Temperature, Heat, Temperature, Heat, Internal EnergyInternal Energy

The sum total of all the energy The sum total of all the energy of all the molecules in an of all the molecules in an object is called: object is called:

THERMAL ENERGYTHERMAL ENERGY

or or

INTERNAL ENERGYINTERNAL ENERGY

Distinction between Temperature, Distinction between Temperature, Heat & Internal EnergyHeat & Internal Energy

Distinction between Temperature, Distinction between Temperature, Heat & Internal EnergyHeat & Internal Energy

TemperatureTemperature: (K) is the measure of : (K) is the measure of average KE of individual molecules.average KE of individual molecules.

Thermal EnergyThermal Energy: refers to the total : refers to the total energy of all the molecules in an object. energy of all the molecules in an object. (2 equal- mass chunks of Fe may have the (2 equal- mass chunks of Fe may have the same temperature, but the both of them same temperature, but the both of them have twice as much thermal energy as 1 have twice as much thermal energy as 1 does)does)

HeatHeat: refers to a transfer of energy : refers to a transfer of energy (thermal energy) from one object to (thermal energy) from one object to another because of a difference in another because of a difference in temperature.temperature.

TemperatureTemperature: (K) is the measure of : (K) is the measure of average KE of individual molecules.average KE of individual molecules.

Thermal EnergyThermal Energy: refers to the total : refers to the total energy of all the molecules in an object. energy of all the molecules in an object. (2 equal- mass chunks of Fe may have the (2 equal- mass chunks of Fe may have the same temperature, but the both of them same temperature, but the both of them have twice as much thermal energy as 1 have twice as much thermal energy as 1 does)does)

HeatHeat: refers to a transfer of energy : refers to a transfer of energy (thermal energy) from one object to (thermal energy) from one object to another because of a difference in another because of a difference in temperature.temperature.

Specific HeatSpecific HeatSpecific HeatSpecific Heat The amount of heat required to change The amount of heat required to change

the temperature of a given material is the temperature of a given material is proportional to the mass of the material proportional to the mass of the material and to the temperature change.and to the temperature change.

Q = m c Q = m c ΔTΔTc= specific heat capacityc= specific heat capacity

c = Q/ mΔTc = Q/ mΔTUnits- J/kg°CUnits- J/kg°C

Table 14-1 on page 421 List of common Table 14-1 on page 421 List of common specific heatsspecific heats

The amount of heat required to change The amount of heat required to change the temperature of a given material is the temperature of a given material is proportional to the mass of the material proportional to the mass of the material and to the temperature change.and to the temperature change.

Q = m c Q = m c ΔTΔTc= specific heat capacityc= specific heat capacity

c = Q/ mΔTc = Q/ mΔTUnits- J/kg°CUnits- J/kg°C

Table 14-1 on page 421 List of common Table 14-1 on page 421 List of common specific heatsspecific heats

SubstanceAluminum

Copper

Glass

Iron/Steel

Lead

Marble

Silver

Wood

ExampleExample How much heat is required to raise How much heat is required to raise

the temperature of an empty 20kg the temperature of an empty 20kg vat made of iron from 10vat made of iron from 10°°C to 90C to 90°°C? C? What if the vat is filled with 20kg of What if the vat is filled with 20kg of water?water?

ΔT = 80°CΔT = 80°C

How much heat is required to raise How much heat is required to raise the temperature of an empty 20kg the temperature of an empty 20kg vat made of iron from 10vat made of iron from 10°°C to 90C to 90°°C? C? What if the vat is filled with 20kg of What if the vat is filled with 20kg of water?water?

ΔT = 80°CΔT = 80°CQ = mc Q = mc ΔT 20kg(450)80 = 7.2x10ΔT 20kg(450)80 = 7.2x1055 J JQ = mc Q = mc ΔT 20kg(450)80 = 7.2x10ΔT 20kg(450)80 = 7.2x1055 J J

720 KJ720 KJ

Q = mc ΔT 20(4186)80 = 6.7x 10Q = mc ΔT 20(4186)80 = 6.7x 106 6 J J Q = mc ΔT 20(4186)80 = 6.7x 10Q = mc ΔT 20(4186)80 = 6.7x 106 6 J J

6700 KJ6700 KJ6700 KJ6700 KJ

Total heat= 720 KJ + 6700 KJ = 7400KJTotal heat= 720 KJ + 6700 KJ = 7400KJ

Almost 10x more heat for equal amount of IronAlmost 10x more heat for equal amount of Iron

Total heat= 720 KJ + 6700 KJ = 7400KJTotal heat= 720 KJ + 6700 KJ = 7400KJ

Almost 10x more heat for equal amount of IronAlmost 10x more heat for equal amount of Iron

CalorimetryCalorimetry The quantitative measurement The quantitative measurement

of heat exchangeof heat exchange The main idea is conservation The main idea is conservation

of Energyof Energy Heat Lost = Heat GainedHeat Lost = Heat Gained QQlostlost = Q = Qgainedgained

Heat”flows” from region of Heat”flows” from region of hotter to region of cooler until hotter to region of cooler until thermal equilibrium is reachedthermal equilibrium is reached

The quantitative measurement The quantitative measurement of heat exchangeof heat exchange

The main idea is conservation The main idea is conservation of Energyof Energy Heat Lost = Heat GainedHeat Lost = Heat Gained QQlostlost = Q = Qgainedgained

Heat”flows” from region of Heat”flows” from region of hotter to region of cooler until hotter to region of cooler until thermal equilibrium is reachedthermal equilibrium is reached

ExampleExampleExampleExample If 200cmIf 200cm33 of tea at 95°C is poured into of tea at 95°C is poured into

a 150g glass coffee cup at 25°C, what a 150g glass coffee cup at 25°C, what will be the final temperature (T) of will be the final temperature (T) of both when equilibrium is reached? both when equilibrium is reached?

(assume NO heat is lost to (assume NO heat is lost to surroundings)surroundings)

If 200cmIf 200cm33 of tea at 95°C is poured into of tea at 95°C is poured into a 150g glass coffee cup at 25°C, what a 150g glass coffee cup at 25°C, what

will be the final temperature (T) of will be the final temperature (T) of both when equilibrium is reached? both when equilibrium is reached?

(assume NO heat is lost to (assume NO heat is lost to surroundings)surroundings)

Tea ~ WaterTea ~ Water 1cm1cm33 = 1ml = 1g (for H = 1ml = 1g (for H22O)O)

200cm200cm33 = .2kg = .2kg

QQlostlost teatea = Q = Qgainedgained cupcup

MMTTCCTT (95°C-T) = M (95°C-T) = MccCCcc (T-25°C) (T-25°C)

T is final temp for both tea and T is final temp for both tea and cupcup

continued…continued….2kg (4186)(95-T) = (.15kg)(840)(T-25°C).2kg (4186)(95-T) = (.15kg)(840)(T-25°C)

79,400J – (836T) = 126T – 6300J79,400J – (836T) = 126T – 6300J

T = 89°CT = 89°C

Tea drops 6°CTea drops 6°C

cup rises 64°Ccup rises 64°C

Note Note ΔT is PositiveΔT is Positive Heat lost = THeat lost = Tii – T – Tff

Heat gained = THeat gained = Tff – T – Tii

Calorimeter: tool used in Calorimeter: tool used in calorimetry calorimetry

Note Note ΔT is PositiveΔT is Positive Heat lost = THeat lost = Tii – T – Tff

Heat gained = THeat gained = Tff – T – Tii

Calorimeter: tool used in Calorimeter: tool used in calorimetry calorimetry

It’s a cup within a cup, It It’s a cup within a cup, It insulates and allows for insulates and allows for temperature measuringtemperature measuring

Air Air insulatorinsulator

WaterWater

ExampleExampleExampleExample A mystery alloy has been A mystery alloy has been

discovered in Lacey’s computer discovered in Lacey’s computer room. Find the room. Find the specific heat specific heat

capacitycapacity. The alloy sample is .15kg . The alloy sample is .15kg and is heated to 540°C. It is quickly and is heated to 540°C. It is quickly placed in 400g of water at 10°C. The placed in 400g of water at 10°C. The

calorimeter cup is made of calorimeter cup is made of aluminum and has a mass of 200g. aluminum and has a mass of 200g.

The final temp is 30.5°CThe final temp is 30.5°C

A mystery alloy has been A mystery alloy has been discovered in Lacey’s computer discovered in Lacey’s computer

room. Find the room. Find the specific heat specific heat capacitycapacity. The alloy sample is .15kg . The alloy sample is .15kg and is heated to 540°C. It is quickly and is heated to 540°C. It is quickly placed in 400g of water at 10°C. The placed in 400g of water at 10°C. The

calorimeter cup is made of calorimeter cup is made of aluminum and has a mass of 200g. aluminum and has a mass of 200g.

The final temp is 30.5°CThe final temp is 30.5°C

QQlost alloylost alloy = Q = Qgainedgained by waterby water + Q + Qgainedgained by cupby cupQQlost alloylost alloy = Q = Qgainedgained by waterby water + Q + Qgainedgained by cupby cup

MMAACCAA ΔT ΔTAA = M = MWWCCWW ΔT ΔTWW + M + MCCCCCC ΔT ΔTCCMMAACCAA ΔT ΔTAA = M = MWWCCWW ΔT ΔTWW + M + MCCCCCC ΔT ΔTCC

.15(C.15(CAA)(540-30.5) )(540-30.5)

= .4(4186)(30.5-10) + .2(900)(30.5 – 10)= .4(4186)(30.5-10) + .2(900)(30.5 – 10)

.15(C.15(CAA)(540-30.5) )(540-30.5)

= .4(4186)(30.5-10) + .2(900)(30.5 – 10)= .4(4186)(30.5-10) + .2(900)(30.5 – 10)

76.4C76.4CAA = (34,300 + 3,700) J/kg = (34,300 + 3,700) J/kg00CC76.4C76.4CAA = (34,300 + 3,700) J/kg = (34,300 + 3,700) J/kg00CC

CCAA = 500 J/kg = 500 J/kg00CCCCAA = 500 J/kg = 500 J/kg00CC

Latent HeatLatent Heat Deals with Q during a change Deals with Q during a change

of phase ; solid to liquid, liquid of phase ; solid to liquid, liquid to gasto gas

Latent means hiddenLatent means hidden LLFF: latent heat of fusion: latent heat of fusion

LLVV: latent heat of Vaporization: latent heat of Vaporization

Table 14-3 Table 14-3

Pg. 425Pg. 425

ExampleExampleExampleExample How much Q does a How much Q does a

refrigerator have to remove refrigerator have to remove from 1.5kg of water at 20from 1.5kg of water at 20°°C to C to make ice at –12make ice at –12°°C?C?

How much Q does a How much Q does a refrigerator have to remove refrigerator have to remove from 1.5kg of water at 20from 1.5kg of water at 20°°C to C to make ice at –12make ice at –12°°C?C?

-12-12

00

2020

QQ

11

3322

QQtotaltotal = Q = Q11 + Q + Q22 + Q + Q33

QQtotaltotal = MC = MCWW ΔT + ML ΔT + MLFF + MC + MCiceice ΔT ΔTQQtotaltotal = MC = MCWW ΔT + ML ΔT + MLFF + MC + MCiceice ΔT ΔT

= 1.5(4186)(20) = 1.5(4186)(20)

+ 1.5(3.33 x 10+ 1.5(3.33 x 1055) )

+ 1.5(2100)12+ 1.5(2100)12

= 1.5(4186)(20) = 1.5(4186)(20)

+ 1.5(3.33 x 10+ 1.5(3.33 x 1055) )

+ 1.5(2100)12+ 1.5(2100)12

6.6 x 106.6 x 1055 J J

660kJ660kJ

Read example 14-9 and 14-10 on pg. 426-Read example 14-9 and 14-10 on pg. 426-427427

Conduction 14-7Conduction 14-7Convection 14-8Convection 14-8Radiation 14-9Radiation 14-9

Conduction 14-7Conduction 14-7Convection 14-8Convection 14-8Radiation 14-9Radiation 14-9