Heat 14 (01 of 32)

Physics Lecture Notes

Heat

Heat

1) Heat As Energy Transfer

2) Internal Energy

3) Specific Heat

4) Calorimetry

5) Latent Heat

6) Heat Transfer: Conduction

Heat 14 (02 of 32)

Heat As Energy Transfer

Unit of heat: calorie (cal)

1 cal is the amount of heat necessary to raise the temperature of 1 g of water by 1 Celsius degree.

1 kcal is the amount of heat necessary to raise the temperature of 1 kg of water by 1 Celsius degree.

Heat is random thermal Energy

Heat 14 (13 of 32)

Heat is a form of energy and can be equated to mechanical energy.

Heat As Energy Transfer

cal 1J 186.4

calk 1J 4186

The apparatus below is used to determine the mechanical equivalent of heat:

Heat 14 (13 of 32)

Heat As Energy Transfer

The sum total of all the energy of all the molecules in a substance is its internal (or thermal) energy.

Temperature: measures molecules’ average kinetic energy

Internal energy: total energy of all molecules

Heat: transfer of energy due to difference in temperature

Definition of heat:

Heat is thermal energy transferred from one object to another because of a difference in temperature.

Heat 14 (13 of 32)

m

TmcQ

(Q) Thermal energy added

(m) Mass of the object

(c) Specific heatof the material

(T) Change intemperature

Heat As Energy Transfer

Heat 14 (13 of 32)

Chapter 14Page 387

Heat 14 (13 of 32)

A 50 g piece of cadmium is at 20 oC. If 400 J of heatis added to the cadmium, what is its final temperature

TmcQ

mcQ

T

oCkg

J 230 kg 050.0

J 400

oC 8.34

TTT of

oof C 8.34C 0.20T C 8.54 o

Heat As Energy Transfer

ocadmium Ckg

J 230c

Problem:

Heat 14 (13 of 32)

A 100 g lead bullet traveling at 300 m/s is stopped by a large tree. Half the kinetic energy of the bullet is transformed into heat energy and remains with the the bullet while the other half is transmitted to the tree. What is the increase in temperature of the bullet?

2

KΔQ bullet

bullet

4mv

Tmc2

c4v

T2

o

2

Ckg

J 031 4

m/s 300

oC 173

Heat As Energy Transfer

olead Ckg

J 130c

Problem:

Heat 14 (13 of 32)

A 3.0 kg block of iron is dropped from rest from the top of a cliff. When the block hits the ground it is observed that its temperature increases by 0.50 oC. Assume that all the potential energy is used to heat the block. How high is the cliff?

blockblock QU

Tmcmgh

gTc

h

2

oo

m/s 8.9

C50.0 Ckg

J 450

m 23

Heat As Energy Transfer

oiron Ckg

J 450c

Problem:

Heat 14 (13 of 32)

A 1.5 kg copper block is given an initial speed of 30 m/s on a rough horizontal surface Because of friction, the block finally comes to rest. If the block absorbs 85% of its initial kinetic energy in the form of heat, Calculate its increase in temperature?

blockblock K85.0Q

2

mv85.0Tmc

2

c2

v85.0T

2

o

2

Ckg

J 390 2

m/s 3085.0

C 0.1 o

Heat As Energy Transfer

ocopper Ckg

J 390c

Problem:

Heat 14 (13 of 32)

A 0.40 kg iron horseshoe that is initially at 500 oC is dropped into a bucket containing 20 kg of water at 22 oC. What is the final equilibrium temperature? Neglect any heat transfer to for from the surroundings.

lossgain QQ

IIIWWW TΔcmTΔcm

ff T5004504.022T418620

ff T180900001841840T83720

C 23T of

CalorimetryProblem:

owater

oiron

Ckg

J 4186c

Ckg

J 450c

Heat 14 (13 of 32)

A 200 g block of copper at a temperature of 90 oC is dropped into 400 g of water at 27 oC. The water is contained in a 300 g glass container. What is the final temperature of the mixture

lossgain QQ

CCCGGGWWW TΔcmTΔcmTΔcm

fff T7870206804T25245208T4.1674

C 5.29T of

Calorimetry

fff T903902.027T8403.027T41864.0

Problem:

oglass

owater

ocopper

Ckg

J 840c

Ckg

J 4186c

Ckg

J 390c

Heat 14 (13 of 32)

Example 1: A 500-g copper coffee mug is filled with 200-g of coffee. How much

heat was required to heat cup and coffee from 20 to 960C?

1. Draw sketch of problem1. Draw sketch of problem.

2. List given 2. List given information.information.Mug massMug mass mmmm = = 0.500 kg0.500 kg

Coffee massCoffee mass mmcc = = 0.200 kg0.200 kg

Initial temperature of coffee and mug:Initial temperature of coffee and mug: tt00 = 20 = 2000CCFinal temperature of coffee and mug:Final temperature of coffee and mug: ttff = 96 = 9600CC

Total heat to raise temperature

of coffee (water) and mug to 960C.

3. List what is to be 3. List what is to be found: found:

Example 1(Cont.): How much heat needed to heat cup and coffee from 20 to 960C?

mm = 0.2 kg; mw = 0.5 kg.

4. Recall applicable formula or law:4. Recall applicable formula or law:Q = mc tHeat Gain or Loss:

5. Decide that TOTAL heat is that 5. Decide that TOTAL heat is that required to raise temperature of required to raise temperature of mug and water (coffee). Write mug and water (coffee). Write equation.equation.

QQTT = = mmmmccmm t + mt + mwwccw w tt

6. Look up specific 6. Look up specific heats in tables:heats in tables:

Copper: cCopper: cmm = 390 J/kg C = 390 J/kg C00 Coffee (water): cCoffee (water): cww = 4186 J/kg = 4186 J/kg

CC00

t = 960C - 200C = 76 C0

t = 960C - 200C = 76 C0

Water: (0.20 kg)(4186 J/kgC0)(76 C0)

Cup: (0.50 kg)(390 J/kgC0)(76 C0)

QT = 63,600 J + 14,800 J QT = 78.4 kJ

QT = 78.4 kJ

7. Substitute info / solve problem:7. Substitute info / solve problem:

QT = mmcm t + mwcw t

Copper: cCopper: cmm = 390 J/kg C = 390 J/kg C00 Coffee (water): cCoffee (water): cww = 4186 J/kg C = 4186 J/kg C00

Example 1(Cont.): How much heat needed to heat cup and coffee from 20 to 960C?

mc = 0.2 kg; mw = 0.5 kg.

m1 m2

TH TL

lossQ gainQ

Conservation of thermal energy:

lossgain QQ

111222 TΔcmTΔcm fH11Lf22 TTcmTTcm

Final Temperature:

2211

L22H11f cmcm

TcmTcmT

Calorimetry

Heat 14 (13 of 32)

mLQ

(Q) Thermal energy added

(L) Latent heatof the fusion or

vaporization

(m) Mass of the object

Latent Heat - Stored / Hidden

Energy is required for a material to change phase,

Even though its temperature is not changing.

Heat 14 (13 of 32)

The water problem

SubstanceLatent Heat

FusionkJ/kg

MeltingPoint

°C

Latent HeatVaporization

kJ/kg

BoilingPoint

°C

Alcohol, ethyl 108 −114 855 78.3

Ammonia 339 −75 1369 −33.34

Carbon dioxide 184 −78 574 −57

Helium 21 −268.93

Hydrogen(2) 58 −259 455 −253

Lead[8] 24.5 327.5 871 1750

Nitrogen 25.7 −210 200 −196

Oxygen 13.9 −219 213 −183

R134a −101 215.9 −26.6

Toluene −93 351 110.6

Turpentine 293

Water 334 0 2260 100

Table of latent heatsThe following table shows the latent heats and change of phase temperatures of some common fluids and gases.

Latent Heat

Heat of fusion, Lf: heat required to change 1.0 kg

of material from solid to liquid

Heat of vaporization, Lv: heat required to change 1.0 kg

of material from liquid to vaporChapter 14 - Page 392

Heat 14 (13 of 32)

50

Q2= mLf

3.33 x 105 J

Q1= mcIt

1.05 x 105 J

Q4= mLv

22.6 x 105 J

Q5= mcSt

1.01 x 105 J

Q3= mcWt

4.19 x 105 J

Latent Heat

1 kgIce

-50 oC

Heat 14 (13 of 32)

Q1 = 1.05 x 105 J

Q2 = 3.33 x 105 J

Q3 = 4.19 x 105 J

Q4 = 22.6 x 105 J

Q5 = 1.01 x 105 J

Heat required to convert 1 kg of ice at -50 oC to steam at 150 oC

3.22 x 106 J

Latent Heat

Heat 14 (13 of 32)

A large block of ice at 0 oC has a hole chipped in it, and 400 g of aluminum pellets at a temperature of 30 oC are poured into the hole. How much of the ice melts?

loss)gain( QQ

AAAfI TcmLm

f

AAAI L

Tcmm

kgJ

10 x 33.3

C300Ckg

J900 kg 4.0

m5

oo

I

kg 032.0

Latent Heat

Problem:

owaterf

oaluminum

Ckg

kJ 333L

Ckg

J 900c

Heat 14 (13 of 32)

Conduction

Heat conduction can be visualized as occurring through molecular collisions.

The heat flow per unit time is given by:

x

x

TTkA

tQ 21

Heat 14 (13 of 32)

The constant k is called the thermal conductivity.

Materials with large k are called conductors; those with small k are called insulators.

Conduction

Chapter 14Page 396

Heat 14 (13 of 32)

A window has a glass surface of 1.6 x 103 cm2 and a thickness of 3.0 mm. Find the rate of heat transfer by conduction through this pane when the temperature of the inside surface of the glass is 20 oC and the outside temperature is 40 oC.

xTkA

tQ

m 10 x 0.3

Co 202

cm 100m 12cm 310 x 6.1

Com

W 84.0

3

s/J 896tΔQΔ

Conduction

Problem:

oglass Cms

J 84.0k

Heat 14 (13 of 32)

Problem:A glass window pane has an area of 3.0 m2 and a thickness of 0.60 cm. If the temperature difference between its faces is 25 oC, how much heat flows through the window per hour?

xTkA

tQ

x

tTkAQ

m .0060

s 3600oC 25 2m 3 oCm

W 84.0

J 710 x 78.3QΔ

Conduction

oglass Cms

J 84.0k

Heat 14 (13 of 32)

Building materials are measured using R−values rather than thermal conductivity:

Where, L is the thickness of the material.

Conduction

kL

R

Chapter 14Page 397

Heat 14 (13 of 32)

Summary

Internal energy U refers to the total energy of all molecules in an object. For an ideal monatomic gas,

nRTNkTU23

23

Heat is the transfer of energy from one object to another due to a temperature difference. Heat can be measured in joules or in calories.

Specific heat of a substance is the energy required to change the temperature of a fixed amount of matter by 1° C.

TΔmcQ

Heat 14 (13 of 32)

In an isolated system, heat gained by one part of the system must be lost by another.

Calorimetry measures heat exchange quantitatively.

Energy in involved in phase changes even though the temperature does not change.

Heat of fusion: amount of energy required to melt 1 kg of material.

Heat of vaporization: amount of energy required to change 1 kg of material from liquid to vapor.

mLQ

Heat 14 (13 of 32)

Summary

Heat transfer takes place by conduction, convection, and radiation.

In conduction, energy is transferred through the collisions of molecules in the substance.

xTkA

tQ

Heat 14 (13 of 32)

Summary

Internal Energy

Internal energy of an ideal (monatomic) gas: 2

21 vm NU

NkTU23

nRTNkT nRTU

23

kTvm232

21

kinetic energy in terms of the temperature

Heat 14 (13 of 32)