sm chapter20

20Heat and the First Law of Thermodynamics

CHAPTER OUTLINE

20.1 Heat and Internal Energy20.2 Specifi c Heat and Calorimetry20.3 Latent Heat20.4 Work and Heat in Thermodynamic

Processes20.5 The First Law of Thermodynamics20.6 Some Applications of the First Law

of Thermodynamics20.7 Energy Transfer Mechanisms

ANSWERS TO QUESTIONS

Q20.1 Temperature is a measure of molecular motion. Heat is energy in the process of being transferred between objects by random molecular collisions. Internal energy is an object’s energy of random molecular motion and molecular interaction.

*Q20.2 With a specifi c heat half as large, the ∆T is twice as great in the ethyl alcohol. Answer (c).

Q20.3 Heat is energy being transferred, not energy contained in an object. Further, a low-temperature object with large mass, or an object made of a material with high specifi c heat, can contain more internal energy than a higher-temperature object.

*Q20.4 We think of the product mc∆T in each case, with c = 1 for water and about 0.5 for beryllium. For (a) we have 1 ⋅ 1 ⋅ 6 = 6. For (b), 2 ⋅ 1 ⋅ 3 = 6. For (c), 2 ⋅ 1 ⋅ 3 = 6. For (d), 2(0.5)3 = 3. For (e), a large quantity of energy input is required to melt the ice. Then we have e > a = b = c > d.

Q20.5 There are three properties to consider here: thermal conductivity, specifi c heat, and mass. With dry aluminum, the thermal conductivity of aluminum is much greater than that of (dry) skin. This means that the internal energy in the aluminum can more readily be transferred to the atmosphere than to your fi ngers. In essence, your skin acts as a thermal insulator to some degree (pun intended). If the aluminum is wet, it can wet the outer layer of your skin to make it into a good conductor of heat; then more internal energy from the aluminum can get into you. Further, the water itself, with additional mass and with a relatively large specifi c heat compared to aluminum, can be a signifi cant source of extra energy to burn you. In practical terms, when you let go of a hot, dry piece of aluminum foil, the heat transfer immediately ends. When you let go of a hot and wet piece of aluminum foil, the hot water sticks to your skin, continuing the heat transfer, and resulting in more energy transfer to you!

Q20.6 Write 1 000 1 1 3 1 000kg 4 186 J kg C C kg m J3⋅( )( ) = ( )° ° V . kkg C C⋅( )( )° °1 to fi ndV = ×3 2 103 3. m .

*Q20.7 Answer (a). Do a few trials with water at different original temperatures and choose the one where room temperature is halfway between the original and the fi nal temperature of the water. Then you can reasonably assume that the contents of the calorimeter gained and lost equal quantities of heat to the surroundings, for net transfer zero. James Joule did it like this in his basement in London.

Q20.8 If the system is isolated, no energy enters or leaves the system by heat, work, or other transfer proces-ses. Within the system energy can change from one form to another, but since energy is conserved these transformations cannot affect the total amount of energy. The total energy is constant.

*Q20.9 (i) Answer (d). (ii) Answer (d). Internal energy and temperature both increase by minuscule amounts due to the work input.

519

13794_20_ch20_p519-542.indd 51913794_20_ch20_p519-542.indd 519 12/22/06 5:08:40 PM12/22/06 5:08:40 PM

Q20.10 The steam locomotive engine is one perfect example of turning internal energy into mechanical energy. Liquid water is heated past the point of vaporization. Through a controlled mechanical process, the expanding water vapor is allowed to push a piston. The translational kinetic energy of the piston is usually turned into rotational kinetic energy of the drive wheel.

Q20.11 The tile is a better thermal conductor than carpet. Thus, energy is conducted away from your feet more rapidly by the tile than by the carpeted fl oor.

*Q20.12 Yes, wrap the blanket around the ice chest. The insulation will slow the transfer of heat from the exterior to the interior. Explain to your little sister that her winter coat helps to keep the outdoors cold to the same extent that it helps to keep her warm. If that is too advanced, promise her a really cold can of Dr. Pepper at the picnic.

Q20.13 The sunlight hitting the peaks warms the air immediately around them. This air, which is slightly warmer and less dense than the surrounding air, rises, as it is buoyed up by cooler air from the val-ley below. The air from the valley fl ows up toward the sunny peaks, creating the morning breeze.

*Q20.14 Answer (d). The high specifi c heat will keep the end in the fi re from warming up very fast. The low conductivity will make your end warm up only very slowly.

*Q20.15 Twice the radius means four times the surface area. Twice the absolute temperature makes T 4 sixteen times larger in Stefan’s law. We multiply 4 times 16 to get answer (e).

Q20.16 The bit of water immediately over the fl ame warms up and expands. It is buoyed up and rises through the rest of the water. Colder, more dense water fl ows in to take its place. Convection currents are set up. They effectively warm the bulk of the water all at once, much more rapidly than it would be warmed by heat being conducted through the water from the fl ame.

Q20.17 Keep them dry. The air pockets in the pad conduct energy by heat, but only slowly. Wet pads would absorb some energy in warming up themselves, but the pot would still be hot and the water would quickly conduct and convect a lot of energy right into you.

Q20.18 The person should add the cream immediately when the coffee is poured. Then the smaller temperature difference between coffee and environment will reduce the rate of energy loss during the several minutes.

*Q20.19 Convection: answer (b). The bridge deck loses energy rapidly to the air both above it and below it.

Q20.20 The marshmallow has very small mass compared to the saliva in the teacher’s mouth and the surrounding tissues. Mostly air and sugar, the marshmallow also has a low specifi c heat compared to living matter. Then the marshmallow can zoom up through a large temperature change while causing only a small temperature drop of the teacher’s mouth. The marshmallow is a foam with closed cells and it carries very little liquid nitrogen into the mouth. (Note that microwaving the marshmallow beforehand might change it into an open-cell sponge, with disasterous effects.) The liquid nitrogen still on the undamaged marshmallow comes in contact with the much hotter saliva and immediately boils into cold gaseous nitrogen. This nitrogen gas has very low thermal conduc-tivity. It creates an insulating thermal barrier between the marshmallow and the teacher’s mouth (the Leydenfrost effect). A similar effect can be seen when water droplets are put on a hot skillet. Each one dances around as it slowly shrinks, because it is levitated on a thin fi lm of steam. Upon application to the author of this manual, a teacher who does this demonstration for a class using the Serway-Jewett textbook may have a button reading “I am a professional. Do not try this at home.”

The most extreme demonstration of this effect is pouring liquid nitrogen into one’s mouth and blowing out a plume of nitrogen gas. We strongly recommended that you read of Jearl Walker’s adventures with this demonstration rather than trying it.

520 Chapter 20


Heat and the First Law of Thermodynamics 521

Q20.21 (a) Warm a pot of coffee on a hot stove.

(b) Place an ice cube at 0°C in warm water—the ice will absorb energy while melting, but not increase in temperature.

(c) Let a high-pressure gas at room temperature slowly expand by pushing on a piston. Work comes out of the gas in a constant-temperature expansion as the same quantity of heat fl ows in from the surroundings.

(d) Warm your hands by rubbing them together. Heat your tepid coffee in a microwave oven. Energy input by work, by electromagnetic radiation, or by other means, can all alike produce a temperature increase.

(e) Davy’s experiment is an example of this process.

(f) This is not necessarily true. Consider some supercooled liquid water, unstable but with temperature below 0°C. Drop in a snowfl ake or a grain of dust to trigger its freezing into ice, and the loss of internal energy measured by its latent heat of fusion can actually push its temperature up.

Q20.22 Heat is conducted from the warm oil to the pipe that carries it. That heat is then conducted to the cooling fi ns and up through the solid material of the fi ns. The energy then radiates off in all direc-tions and is effi ciently carried away by convection into the air. The ground below is left frozen.

SOLUTIONS TO PROBLEMS

Section 20.1 Heat and Internal Energy

P20.1 Taking m = 1 00. kg, we have

∆U mghg = = ( )( )( ) =1 00 9 80 50 0 490. . .kg m s m J2

But ∆ ∆ ∆U Q mc T Tg = = = ( ) ⋅( ) =1 00 4 186 490. kg J kg C J° so ∆T = 0 117. °C

T T Tf i= + = +( )∆ 10 0 0 117. . °C

P20.2 The container is thermally insulated, so no energy fl ows by heat:

Q = 0 and ∆E Q W W mghint = + = + =input input0 2

The work on the falling weights is equal to the work done on the water in the container by the rotating blades. This work results in an increase in internal energy of the water:

2mgh E m c T= =∆ ∆int water

∆Tmgh

m c= =

× ( )( )2 2 1 50 9 80 3 00

0water

2kg m s m. . .

..

.

.

200 4 186

88 2

0 105

kg J kg C

J

837 J C

C

⋅( ) =

=

° °

°

FIG. P20.2


522 Chapter 20

Section 20.2 Specifi c Heat and Calorimetry

P20.3 ∆ ∆Q mc T= silver

1 23 0 525 10 0

0

. . .

.

kJ kg Csilver

silver

= ( ) ( )=

c

c

°

2234 kJ kg C⋅°

P20.4 The laser energy output:

P∆t = ×( ) × = ×−1 60 10 4 00 1013 9 4. .J s 2.50 10 s J

The teakettle input:

Q mc T= = ⋅( ) = ×∆ 0 800 2 68 105. .kg 4 186 J kg C 80 C° ° JJ

The energy input to the water is 6.70 times larger than the laser output of 40 kJ.

P20.5 We imagine the stone energy reservoir has a large area in contact with air and is always at nearly the same temperature as the air. Its overnight loss of energy is described by

P

P

= =

= =−( )( )

Q

t

mc T

t

mt

c T

∆∆

∆

∆∆

6 000 14 3 600J s h s hh

J kg C C C

J kg( )⋅( ) −( )

=× ⋅ ⋅

850 18 38

3 02 108

° ° °

°. CC

850 J 20 Ckg

°( )= ×1 78 104.

P20.6 Let us fi nd the energy transferred in one minute.

Q m c m c T

Q

= +⎡⎣ ⎤⎦= ( )

cup cup water water

kg

∆

0 200 90. 00 0 800 4 186 1 5J kg C kg J kg C⋅( ) + ( ) ⋅( )⎡⎣ ⎤⎦ −° °. . 00 5 290°C J( ) = − If this much energy is removed from the system each minute, the rate of removal is

P = = = =Q

t∆5 290

88 2 88 2J

60.0 sJ s W. .

P20.7 Q Qcold hot= −

mc T mc T∆ ∆( ) = − ( )⋅( )

water iron

kg J kg C20 0 4 186. ° TT Tf f−( ) = − ( ) ⋅( ) −25 0 1 50 448 600. .° ° °C kg J kg C C(( )=Tf 29 6. °C



*P20.8 (a) Work that the bit does in deforming the block, breaking chips off, and giving them kinetic energy is not a fi nal destination for energy. All of this work turns entirely into internal energy as soon as the chips stop their macroscopic motion. The amount of energy input to the steel is the work done by the bit:

W = ⋅ = ( )( )( ) =� �F r∆ 3 2 40 15 0 1 920. cosN m s s J°

To evaluate the temperature change produced by this energy we imagine injecting the same quantity of energy as heat from a stove. The bit, chips, and block all undergo the same tem-perature change. Any difference in temperature between one bit of steel and another would erase itself by causing a heat transfer from the temporarily hotter to the colder region.

Q mc T

TQ

mc

=

= = ⋅ ⋅( )( ) =

∆

∆ 1 920

448

J kg C

0.267 kg J

°116 1. °C

(b) See part (a). 16.1 C°

(c) It makes no difference whether the drill bit is dull or sharp, how far into the block it cuts, or what its diameter is. The answers to (a) and (b) are the same because work (or ‘work to produce deformation’) cannot be a fi nal form of energy: all of the work done by the bit constitutes energy being transferred into the internal energy of the steel.

*P20.9 (a) Q Qcold hot= −

m c m c T T m c T T m cw w c c f c f unk unk+( ) −( ) = − −( ) −Cu Cu Cu TT Tf unk−( ) where w is for water, c the calorimeter, Cu the copper sample, and unk the unknown.

250 100g 1.00 cal g C g 0.215 cal g C⋅( ) + ⋅( )⎡⎣ ⎤° ° ⎦⎦ −( )20 0 10 0. . °C

= − ( ) ⋅( )50 0 0 092 4 20. . °g cal g C .. . .0 80 0 20 0 100−( ) − ( ) −( )° °C 70.0 g Ccunk

.2 44 1× 00 5 60 103 3cal g C= × ⋅( ). ° cunk

or

cunk = ⋅0 435. cal g C°

(b) We cannot make a defi nite identifi cation. The material might be beryllium. It might be some alloy or a material not listed in the table.

P20.10 (a) f mgh mc T( )( ) = ∆

0 600 3 00 10 9 80 50 0

4 18

3. . . .

.

( ) ×( )( )( )− kg m s m2

663 00 0 092 4

0 76

J calg cal g C= ( ) ⋅( )( )

=

. .

.

° ∆

∆

T

T 00 25 8° °C C; .T =

(b) The fi nal temperature does not depend on the mass. Both the change in potential energy, and the heat that would be required from a stove to produce the temperature change, are proportional to the mass; hence, the mass divides out in the energy relation.


524 Chapter 20

P20.11 We do not know whether the aluminum will rise or drop in temperature. The energy the water can

absorb in rising to 26°C is mc T∆ = =0 25 4 186 6 6 279. kgJ

kg CC J

°° . The energy the copper can

put out in dropping to 26°C is mc T∆ = =0 1 387 74 2 864. kgJ

kg CC J

°° . Since 6 279 2 864J J> ,

the fi nal temperature is less than 26°C. We can write Q Qh c= − as

Q Q Q

Tf

water Al Cu

kgJ

kg CC

+ + =

−( )0

0 25 4 186 20.°

° ++ −( )0 4 900 26. kgJ

kg CC

°°Tf

0 1 387. kg

J

kg+

°°°

CCTf −( ) =100 0

° °CT Tf f− + −1 046 5 20 930 360 9 360. CC C

C

C

+ − =

=

=

38 7 3 870 0

1 445 2 34 160

23 6

.

.

.

T

T

T

f

f

f

°

°

°

P20.12 Vessel one contains oxygen described by PV nRT= :

nPV

RTc = =×( ) × −1 75 1 013 10 16 8 10

8 314

5 3. . .

.

Pa m3

Nm mol K 300 Kmol

⋅= 1 194.

Vessel two contains this much oxygen:

nh =×( ) ×

( ) =−2 25 1 013 10 22 4 10

8 314 4501

5 3. . .

.mol ..365 mol

(a) The gas comes to an equilibrium temperature according to

mc T mc T

n Mc T n Mc Tc f h f

∆ ∆( ) = − ( )−( ) + −

cold hot

K300 4450 0K( ) =

The molar mass M and specifi c heat divide out:

1 194 358 2 1 365 614 1 0

972 3

. . . .

.

T T

T

f f

f

− + − =

=

K K

K

22.559K= 380

(b) The pressure of the whole sample in its fi nal state is

PnRT

V= =

+2 559

16

.

.

mol 8.314 J 380 K

mol K 22.4 88 102 06 10 2 043

5

( ) ×= × =− m

Pa atm3 . .



Section 20.3 Latent Heat

P20.13 The energy input needed is the sum of the following terms:

Qneeded heat to reach melting point heat= ( ) + tto melt

heat to reach boiling point he

( )+( ) + aat to vaporize heat to reach 110 C( ) + ( )°

Thus, we have

Qneeded kg J kg C C= ⋅( )( ) +0 040 0 2 090 10 0 3 33. . .° ° ××( )⎡⎣

+ ⋅( )( ) + ×

10

4 186 100 2 26 10

5

6

J kg

J kg C C° ° . J kg J kg C C

needed

( ) + ⋅( )( )⎤⎦=

2 010 10 0

1 22

° °.

.Q ×× 105 J

P20.14 Q Qcold hot= −

m c m c T T m L c Tw w c c f i s w f+( ) −( ) = − − + −( )⎡⎣ ⎤⎦v 100

0 2. 550 0 050 0kg 4 186 J kg C kg 387 J kg C⋅( ) + ⋅( )⎡ ° °.⎣⎣ ⎤⎦ −( )50 0 20 0. .° °C C

= − − × +2 26 10 4 1866. J kg Jms kkg C C C⋅( ) −( )⎡⎣ ⎤⎦° ° °50 0 100.

J

2.47=

×3 20 104.ms ××

= =10 J kg

kg g steam6 0 012 9 12 9. .

P20.15 The bullet will not melt all the ice, so its fi nal temperature is 0°C.

Then 1

22m mc T m Lw fv +⎛

⎝⎜⎞⎠⎟ =∆

bullet

where mw is the melt water mass

mw =×( )( ) + ×− −0 500 3 00 10 240 3 00 103 2 3. . .kg m s kgg 128 J kg C C

J kg

J

⋅( )( )×

=

° °30 0

3 33 10

86 4

5

.

.

.mw

++ =11 50 294

..

J

333 000 J kgg

P20.16 (a) Q1 = heat to melt all the ice

= ×( ) ×( ) = ×−50 0 10 3 33 10 1 67 103 5 4. . .kg J kg J

Q2

50

= ( )=

heat to raise temp of ice to 100 C°

.00 10 4 186 100 2 09 103 4×( ) ⋅( )( ) = ×− kg J kg C C J° ° .

Thus, the total heat to melt ice and raise temp to 100°C = ×3 76 104. J

Q3 10 0 10= = ×heat availableas steam condenses

. −−( ) ×( ) = ×3 6 42 26 10 2 26 10kg J kg J. .

Thus, we see that Q Q3 1> , but Q Q Q3 1 2< + .

Therefore, all the ice melts but Tf < 100°C. Let us now fi nd Tf

Q Qcold hot

kg J kg

= −

×( ) ×( ) +−50 0 10 3 33 10 503 5. . .00 10 4 186 0

10 0 10

3×( ) ⋅( ) −( )= − ×

−

−

kg J kg C C° °Tf

. 33 6 32 26 10 10 0 10 4 186kg J kg kg J( ) − ×( ) − ×( )−. . kkg C C⋅( ) −( )° °Tf 100

From which,

Tf = 40 4. °C

continued on next page


526 Chapter 20

(b) Q1 = heat to melt all ice = 1.67 � 10 4 J [See part (a)]

Q2310= = ( −heat given up

as steam condenseskg)) ×( ) = ×

=

2 26 10 2 26 106 3

3

. .J kg J

heat given upQ

as condensedsteam cools to 0 C

kg°

= ( )−10 43 1186 100 419J kg C C J⋅( )( ) =° °

Note that Q Q Q2 3 1+ < . Therefore, the final temperature will be 0°C with some ice remaining. Let us fi nd the mass of ice which must melt to condense the steam andcool the condensate to 0°C.

mL Q Qf = + = ×2 332 68 10. J

Thus,

m = ××

= × =−2 68 108 04 10 8 04

33.

. .J

3.33 10 J kgkg5 gg of ice melts

Therefore, there is 42.0 g of ice left over, also at 0°C.

P20.17 Q m c T m L= = ( )Cu Cu N vap N22

∆

1 00 293 77 3 48 0. . .kg 0.092 0 cal g C C c⋅( ) −( ) =° ° m aal g

kg

( )=m 0 414.

*P20.18 (a) Let n represent the number of stops. Follow the energy:

n 12 1500 6( kg)(25 m/s) kg(900 J/kg C)(6602 = ⋅° −−

= ××

=

20

3 46 10

4 69 107 37

6

5

)

.

..

°C

nJ

J

Thus seven stops can happen before melting begins.

(b) As the car is moving or stopping it transfers part of its kinetic energy into the air and into its rubber tires. As soon as the brakes rise above the air temperature they lose energy by heat, and lost it very fast if they attain a high temperature.

P20.19 (a) Since the heat required to melt 250 g of ice at 0°C exceeds the heat required to cool 600 g of

water from 18°C to 0°C, the fi nal temperature of the system (water + ice) must be 0°C .

(b) Let m represent the mass of ice that melts before the system reaches equilibrium at 0°C.

Q Q

mL m c T

m

f w w i

cold hot

C

J kg

= −

= − −( )×( )

0

3 33 105

°

. == − ( ) ⋅( ) −( )

=

0 600 4 186 0 18 0

136

. .kg J kg C C C° ° °

m g, so the ice remaining g g g= − =250 136 114



*P20.20 The left-hand side of the equation is the kinetic energy of a 12-g object moving at 300 m �s together with an 8-g object moving at 400 m �s. If they are moving in opposite directions, collide head-on, and stick together, momentum conservation implies that we have a 20-g object moving with speed given by 8(400) − 12(300) = 20v |v| = 20 m�s, and the kinetic energy of a 20-g object moving at 20 m �s appears on the right-hand side. Thus we state

(a) Two speeding lead bullets, one of mass 12.0 g moving to the right at 300 m/s and one of mass 8.00 g moving to the left at 400 m/s, collide head-on and all of the material sticks together. Both bullets are originally at temperature 30.0°C. Describe the state of the system immediately thereafter.

(b) We fi nd 540 J + 640 J = 4 J + 761 J + m�

(24500 J�kg)

So the mass of lead melted is m� = 415 J�(24500 J�kg) = 0.0169 kg.

After the completely inelastic collision, a glob comprising 3.10 g of solid lead and 16.9 g of liquid lead is moving to the right at 20.0 m/s. Its temperature is 327.3°C.

Section 20.4 Work and Heat in Thermodynamic Processes

P20.21 W PdVif

i

f

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

P V= α 2 between Vi and Vf .

W V dV V V

V V

if

i

f

f i

f i

= − = − −( )= = (

∫α α2 3 31

3

2 2 1 00. m3 )) = 2 00. m3

Wif = − ( ) ×( )⎡⎣ ⎤⎦1

35 00 1 013 10 2 05. . .atm m Pa atm6 00 1 00 1 18

3 3m m MJ3 3( ) − ( )⎡

⎣⎤⎦ = −. .

P20.22 (a) W PdV= −∫ W = − ×( ) −( ) +

− ×

6 00 10 2 00 1 00

4 00 10

6

6

. . .

.

Pa m

Pa

3

(( ) −( ) +

− ×( ) −( )3 00 2 00

2 00 10 4 00 3 006

. .

. . .

m

Pa

3

m

MJ

3

Wi f→ = −12 0.

(b) Wf i→ = +12 0. MJ

P20.23 W P V PnR

PT T nR Tf i= − = − ⎛

⎝⎞⎠ −( ) = − = − ( )∆ ∆ 0 200 8 314. .(( )( ) = −280 466 J

P

OV

1.00 m3 2.00 m3

P = aV2

f

i

FIG. P20.21

FIG. P20.22


528 Chapter 20

P20.24 W PdV P dV P V nR T nR T Ti

f

i

f

= − = − = − = − = − −( )∫ ∫ ∆ ∆ 2 1 The negative sign for work on the

sample indicates that the expanding gas does positive work. The quantity of work is directly proportional to the quantity of gas and to the temperature change.

P20.25 During the heating process PP

VVi

i

=⎛⎝⎜

⎞⎠⎟

(a) W PdVP

VVdV

i

f

i

iV

V

i

i

= − = −⎛⎝⎜

⎞⎠⎟∫ ∫

3

WP

V

V P

VV V Pi

i V

V

i

ii i i

i

i

= −⎛⎝⎜

⎞⎠⎟

= − −( ) = −2 3

2 2

2 29 4 VVi

(b) PV nRT=

P

VV V nRT

TP

nRVV

i

i

i

i

⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥ =

=⎛⎝⎜

⎞⎠⎟

2

Temperature must be proportional to the square of volume, rising to nine times its original value.

Section 20.5 The First Law of Thermodynamics

P20.26 (a) Q = −W = Area of triangle

Q = ( )( ) =1

24 00 6 00 12 0. . .m kPa kJ3

(b) Q W= − = −12 0. kJ

P20.27 ∆E Q Wint = +

Q E W= − = − − = −∆ int J J J500 220 720

The negative sign indicates that positive energy is transferred from the system by heat.

FIG. P20.26



P20.28 W P V VBC B C B

3atm 0.400 m

= − −( )= − −( )= −

3 00 0 090 0. .

994 2. kJ

∆E Q W

E E

E

int

int, C int, B kJ

= +− = −( )100 94 2.

int, CC int, B kJ− =E 5 79.

Since T is constant,

E Eint, D int, C− = 0

W P V VDA D A D

3atm m= − −( ) = − −( )= +

1 00 0 200 1 20

10

. . .

11 kJ

E Eint, A int, D kJ kJ kJ− = − + +( ) = −150 101 48 7.

Now, E E E E Eint, B int, A int, C int, B int, D− = − −( ) + − EE E Eint, C int, A int, D( ) + −( )⎡⎣ ⎤⎦

E Eint, B int, A kJ kJ k− = − + −[ ] =5 79 0 48 7 42 9. . . JJ

P20.29 Q W ∆Eint

BC − 0 −CA − + –

AB + − +

Q E WBC= =( )∆ int since 0

∆E W Qint and so< > <( )0 0 0,

( ,

; )

W E E

B C A Q

< > <→ → >

0 0 0

0

∆ ∆int intsince

for so

Section 20.6 Some Applications of the First Law of Thermodynamics

P20.30 (a) W nRTV

VP V

V

Vf

if f

f

i

= −⎛⎝⎜

⎞⎠⎟

= −⎛⎝⎜

⎞⎠⎟

ln ln

so

`V VW

P Vi ff f

= +⎛

⎝⎜⎞

⎠⎟= ( ) −

exp . exp.

0 025 03 000

0 025 0 11 013 100 007 655..

×( )⎡

⎣⎢⎢

⎤

⎦⎥⎥

= m3

(b) TP V

nRff f= =

× ( )1 013 10 0 025 0

1 00

5. .

.

Pa m

mol 8.

3

3314 J K molK

⋅( ) = 305

P20.31 (a) ∆ ∆E Q P Vint3kJ kPa 3.00 m= − = − −( ) =12 5 2 50 1 00. . . 77 50. kJ

(b) V

T

V

T1

1

2

2

=

TV

VT2

2

11

3 00

1 00300 900= = ( ) =.

.K K

1.0

3.0

P(atm)

0.090 0.20 0.40 1.2

A

CB

D

V(m3)

FIG. P20.28


530 Chapter 20

P20.32 (a) W P V P V T= − = − [ ]

= − ×( ) × −

∆ ∆3

1 013 10 3 24 0 105 6

α

. .N m2 °° °Ckg

2.70 10 kg mC3 3

−( )×

⎛⎝⎜

⎞⎠⎟

( )⎡

⎣1 1 00

18 0.

.⎢⎢⎤

⎦⎥

= −W 48 6. mJ

(b) Q cm T= = ⋅( )( )( ) =∆ 900 1 00 18 0 16 2J kg C kg C k° °. . . JJ

(c) ∆E Q Wint kJ mJ kJ= + = − =16 2 48 6 16 2. . .

P20.33 W P V P V VP nRT

PPs w= − = − −( ) = − ( )

+ (∆ 18 0. g

1.00 g cm3 ))( )⎡

⎣⎢⎢

⎤

⎦⎥⎥106 cm m3 3

W = −( ) ⋅( )( ) + ×1 00 8 314 373 1 013 10. . .mol J K mol K 55 18 0

3 10

0

N mg

10 g mkJ2

6 3( )⎛⎝⎜

⎞⎠⎟

= −

= =

..

Q mLv .. .018 0 40 7

37

kg 2.26 10 J kg kJ6

int

×( ) =

= + =∆E Q W ..6 kJ

P20.34 (a) The work done during each step of the cycle equals the negative of the area under that segment of the PV curve.

W W W W W

W P V V P V V

DA AB BC CD

i i i i i i

= + + +

= − −( ) + − −(3 0 3 3 )) + = −0 4PVi i

(b) The initial and fi nal values of T for the system are equal.

Therefore,

∆Eint = 0 and Q W PVi i= − = 4

(c) W PV nRTi i i= − = − = − ( )( )( ) = −4 4 4 1 00 8 314 273 9 08. . . kkJ

P20.35 (a) PV P V nRTi i f f= = = ⋅( )( )2 00 300. mol 8.314 J K mol K == ×4 99 103. J

VnRT

P

VnRT

P

ii

ff

= = ×

= = ×

4 99 10

4 99 1

3.

.

J

0.400 atm

00 1

30 041 0

3 J

1.20 atmm3= =Vi .

(b) W PdV nRTV

Vf

i

= − = −⎛⎝⎜

⎞⎠⎟

= − ×( ) ⎛⎝

⎞∫ ln . ln4 99 101

33

⎠⎠ = +5 48. kJ

(c) ∆E Q Wint = = +0

Q = −5 48. kJ

FIG. P20.34



P20.36 ∆ ∆E EABC ACint, int,= (conservation of energy)

(a) ∆E Q WABC ABC ABCint, = + (First Law)

QABC = + =800 500 1 300J J J

(b) W P VCD C CD= − ∆ , ∆ ∆V VAB CD= − , and P PA C= 5

Then, W P V WCD A AB AB= = − =1

5

1

5100∆ J

(+ means that work is done on the system)

(c) W WCDA CD= so that Q E WCA CA CDA= − = − − = −∆ int, J J J800 100 900

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

(d) ∆ ∆ ∆E E ECD CDA DAint, int, int, J J= − = − −800 500 == −1 300 J

and Q E WCD CD CD= − = − − = −∆ int, J J J1 300 100 1 400

Section 20.7 Energy Transfer Mechanisms

P20.37 P = =( )( )( )

×kA T

L

∆ 0 800 3 00 25 0

6 00

. . .

.

W m C m C2⋅° °

1101 00 10 10 03

4− = × =

mW kW. .

P20.38 P = =( )( )

×(∑ −

A T

L ki ii

∆�

6 00 50 0

2 4 00 10 3

. .

.

m C

m

2 °

)) ⋅[ ] + ×⎡⎣ ⎤⎦ ⋅−0 800 5 00 10 0 023 43. . .W m C m W m C° °[[ ]

= 1 34. kW

P20.39 In the steady state condition, P PAu Ag=

so that k AT

xk A

T

xAu AuAu

Ag AgAg

∆∆

∆∆

⎛⎝

⎞⎠ = ⎛

⎝⎞⎠

In this case A AAu Ag=

∆ ∆

∆

x x

T T

Au Ag

Au

=

= −( )80 0.

and ∆T TAg = −( )30 0.

where T is the temperature of the junction.

Therefore, k T k TAu Ag80 0 30 0. .−( ) = −( )

And T = 51 2. °C

FIG. P20.36

FIG. P20.39


532 Chapter 20

P20.40 From the table of thermal resistances in the chapter text,

(a) R = ⋅ ⋅0 890. ft F h Btu2 °

(b) The insulating glass in the table must have sheets of glass less than 1

8 inch thick. So we

estimate the R-value of a 0.250-inch air space as 0.250

3.50 times that of the thicker air space.

Then for the double glazing

Rb = + ⎛⎝

⎞⎠ +⎡

⎣⎢⎤⎦⎥

⋅0 890

0 250

3 501 01 0 890.

.

.. .

ft2 °FF h

Btu

ft F h

Btu

2⋅ = ⋅ ⋅1 85.

°

(c) Since A and T T2 1−( ) are constants, heat fl ow is reduced by a factor of 1 85

0 8902 08

.

..= .

*P20.41 The net rate of energy loss from his skin is

Pnet2 4 2W m K m= −( ) = × ⋅( )(−σ Ae T T4

04 85 67 10 1 50. . ))( ) ( ) − ( )⎡⎣ ⎤⎦

=

0 900 308 293

125

4 4. K K

W.

Note that the temperatures must be in kelvins. The energy loss in ten minutes is

Q t= = ( )( ) = ×Pnet J s 600 s J∆ 125 7 48 104.

In the infrared, the person shines brighter than a hundred-watt light bulb.

P20.42 P = = × ⋅( ) ×( )−σ πAeT 4 8 8 25 669 6 10 4 6 96 10. .W m K m2 4 ⎡⎡

⎣⎤⎦( )( )0 986 5 800

4. K

P = ×3 85 1026. W

*P20.43 (a) The heat leaving the box during the day is given by P =−( )

=kAT T

L

Q

tH c

∆

Q = −0 012 0 49

37 23

0 04512

3 600. .

.

W

m Cm

C C

mh2

°

° ° s

hJ

17 90 104⎛

⎝⎞⎠ = ×.

The heat lost at night is

Q = −0 012 0 49

37 16

0 04512

3 600. .

.

W

m Cm

C C

mh2

°

° ° s

hJ

11 19 105⎛

⎝⎞⎠ = ×.

The total heat is 1 19 10 7 90 10 1 98 105 4 5. . .× + × = ×J J J. It must be supplied by the

solidifying wax: Q mL=

mQ

L= = ×

×=1 98 10

205 100 964

5

3

..

J

J kgkg or more

(b) The test samples and the inner surface of the insulation can be prewarmed to 37°C as the box is assembled. If this is done, nothing changes in temperature during the test period. Then the masses of the test samples and the insulation make no difference.



P20.44 Suppose the pizza is 70 cm in diameter and � = 2.0 cm thick, sizzling at 100°C. It cannot lose heat

by conduction or convection. It radiates according to P = σ AeT 4. Here, A is its surface area,

A r r= + = ( ) + ( )( ) =2 2 2 0 35 2 0 35 0 022 2π π π π� . . .m m m 00 81. m2

Suppose it is dark in the infrared, with emissivity about 0.8. Then

P = × ⋅( )( )( )( )−5 67 10 0 81 0 80 3738 4. . .W m K m K2 4 2 == 710 103W W~

If the density of the pizza is half that of water, its mass is

m V r= = = ( ) ( ) ( ) =ρ ρπ π2 2500 0 35 0 02 4� kg m m m k3 . . gg

Suppose its specifi c heat is c = ⋅0 6. cal g C° . The drop in temperature of the pizza is described by

Q mc T T

dQ

dtmc

dT

dt

dT

dt mc

f i

f

f

= −( )= = −

= =

P 0

710P J s

44 0 6 4 1860 07 10 1

kg J kg CC s K s( ) ⋅ ⋅( ) = −

.. ~

°°

P20.45 P = σ AeT 4

2 00 5 67 10 0 250 10 0 98 6. . . .W W m K m2 4 2= × ⋅( ) ×( )− − 550

1 49 10 3 49 10

4

14 1 4 3

( )

= ×( ) = ×

T

T . .K K4

P20.46 We suppose the earth below is an insulator. The square meter must radiate in the infrared as much energy as it absorbs, P = σ AeT 4. Assuming that e = 1 00. for blackbody blacktop:

1 000 5 67 10 1 00 1 008 4W W m K m2 4 2= × ⋅( )( )( )−. . . T

T = ×( ) =1 76 10 36410 1 4. K K4 (You can cook an egg on it.)

*P20.47 Intensity is defi ned as power per area perpendicular to the direction of energy fl ow. The direction of sunlight is along the line from the sun to the object. The perpendicular area is the projected fl at circular area enclosed by the terminator—the line that separates day from night on the object. The object radiates infrared light outward in all directions. The area perpendicular to this energy fl ow is its spherical surface area.

The sphere of radius R absorbs sunlight over area π R2. It radiates over area 4 2π R . Then, in steady state,

P P

in out

2W m

=

( ) = ( )e R e R T1 370 42 2 4π σ π

The emissivity e, the radius R, and π all cancel.

Therefore,

T =× ⋅( )

⎡

⎣⎢⎢

⎤

⎦⎥⎥

=−

1 370

4 5 67 102798

1 4

W m

W m K

2

2 4.K C= 6°

It is chilly, well below room temperatures wwe find comfortable.


534 Chapter 20

*P20.48 (a) Because the bulb is evacuated, the fi lament loses energy by radiation but not by convection; we ignore energy loss by conduction. From Stefan’s law, the power ratio is

eσA T h 4 �eσA T c

4 = (2373�2273)4 = 1.19

(b) The radiating area is the lateral surface area of the cylindrical fi lament, 2πr�. Now we want

eσ 2π rh � T h

4 = eσ 2π rc� T c

4 so rc�r

h = 1.19

Additional Problems

P20.49 The increase in internal energy required to melt 1.00 kg of snow is

∆Eint kg J kg J= ( ) ×( ) = ×1 00 3 33 10 3 33 105 5. . .

The force of friction is

f n mg= = = ( )( ) =µ µ 0 200 75 0 9 80 147. . .kg m s N2

According to the problem statement, the loss of mechanical energy of the skier is assumed to be equal to the increase in internal energy of the snow. This increase in internal energy is

∆ ∆ ∆E f r rint N J= = ( ) = ×147 3 33 105.

and

∆r = ×2 27 103. m

P20.50 (a) The energy thus far gained by the copper equals the energy loss by the silver. Your down parka is an excellent insulator.

Q Qcold hot= −

or m c T T m c T Tf i f iCu Cu Cu Ag Ag Ag−( ) = − −( )

9 00 387 16 0 14 0 234. . .g J kg C C g J( ) ⋅( )( ) = − ( )° ° kkg C C

C C

Ag

Ag

⋅( ) −( )−( ) = −

° °

° °

T

T

f

f

30 0

30 0 17 0

.

. .

so Tf , .Ag C= 13 0°

(b) Differentiating the energy gain-and-loss equation gives: m cdT

dtm c

dT

dtAg AgAg

Cu CuCu

⎛⎝

⎞⎠ = − ⎛

⎝⎞⎠

dT

dt

m c

m c

dT

dt⎛⎝

⎞⎠ = − ⎛

⎝⎞⎠ = −

Ag

Cu Cu

Ag Ag Cu

g 39 00. 887 J kg C

g 234 J kg CC s

⋅( )⋅( ) +( )°

°°

14 00 500

..

dT

ddt⎛⎝

⎞⎠ = − ⇒

Ag

C s negative sign decreas0 532. ° iing temperature( )

534 Chapter 20



P20.51 (a) Before conduction has time to become important, the energy lost by the rod equals the energy gained by the helium. Therefore,

mL mc Tv( ) = ( )He Al

∆

or ρ ρVL Vc Tv( ) = ( )He Al

∆

so VVc T

LHeAl

He

=( )

( )ρ

ρ∆

v

VHe

3 3g cm cm cal g C=

( )( ) ⋅( )2 70 62 5 0 210 295. . . ° ..

. . .

8

0 125 2 09 10 1 004

°C

g cm J kg cal 4.3

( )( ) ×( ) 1186 J kg 1 000 g

cmHe3

( )( )= × =

1 00

1 68 10 164

.

. .V 88 liters

(b) The rate at which energy is supplied to the rod in order to maintain constant temperatures is given by

P = ⎛⎝

⎞⎠ = ⋅ ⋅( )( )kA

dT

dx31 0 2 50

295 8. .

.J s cm K cm2 KK

25.0 cmW⎛

⎝⎞⎠ = 917

This power supplied to the helium will produce a “boil-off ” rate of

PρLv

=( )( )

( ) ×917 10

0 125 2 09 10

3

4

W g kg

g cm J3. . kkgcm s L s3

( ) = =351 0 351.

*P20.52 (a) Work done by the gas is the negative of the area under

the PV curve W PV

VPV

ii

ii i= − −⎛

⎝⎞⎠ = +

2 2

Put the cylinder into a refrigerator at absolute tempera-ture T

i �2. Let the piston move freely as the gas cools.

(b) In this case the area under the curve is W PdV= −∫ . Since the process is isothermal,

PV PV PV

nRTi i ii

i= = ⎛⎝

⎞⎠ =4

4

and

WdV

VPV PV

V

VPi i

V

V

i ii

ii

i

= − ⎛⎝

⎞⎠ ( ) = −

⎛⎝⎜

⎞⎠⎟

=∫4

4ln ii iV ln 4

= +1 39. PVi i

With the gas in a constant-temperature bath at , slowly push the piston in.Ti

(c) The area under the curve is 0 and W = 0 .

Lock the piston in place and hold the cylindder over a hotplate at 3 .Ti

The student may be confused that the integral in part (c) is not explicitly covered in calculus class. Mathematicians ordinarily study integrals of functions, but the pressure is not a single-valued function of volume in a isovolumetric process. Our physics idea of an integral is more general. It still corresponds to the idea of area under the graph line.

a

b c

P

V

FIG. P20.52


536 Chapter 20

P20.53 Q mc T V c T= = ( )∆ ∆ρ so that when a constant temperature difference ∆T is maintained, the rate

of adding energy to the liquid is P = = ⎛⎝

⎞⎠ =dQ

dt

dV

dtc T Rc Tρ ρ∆ ∆ and the specifi c heat of the

liquid is cR T

= Pρ ∆

P20.54 The initial moment of inertia of the disk is

1

2

1

2

1

2

1

28 920 282 2 2 2MR VR R tR= = = ( ) ( )ρ ρπ πkg m m3 44 101 2 1 033 10. .m kg m2= × ⋅

The rotation speeds up as the disk cools off, according to

I I

MR MR MR T

i i f f

i i f f i f

ω ω

ω ω α ω

=

= = −( )1

2

1

2

1

212 2 2 2∆

ωω ωαf i

T=

−( ) =− ×( )−

1

125

1

1 17 10 8302 6∆

rad s1 C° °CC

rad s⎡⎣ ⎤⎦

=2 25 720 7.

(a) The kinetic energy increases by

1

2

1

2

1

2

1

2

1

22 2 2I I I I If f i i i i f i i i i fω ω ω ω ω ω ω ω− = − = − ii( )

= × ⋅ ( )1

21 033 10 25 0 720 710. .kg m rad s rad s2 == ×9 31 1010. J

(b) ∆ ∆E mc Tint kg 387 J kg C C= = × ⋅( ) −2 64 10 20 8507. ° ° °CC J( ) = − ×8 47 1012.

(c) As 8 47 1012. × J leaves the fund of internal energy, 9 31 1010. × J changes into extra kinetic

energy, and the rest, 8 38 1012. × J is radiated.

P20.55 The loss of mechanical energy is

1

2

1

2670

6 67 102 2m

GM m

RiE

E

v + = ×( ) + ×kg 1.4 10 m s4 . −− ×

×

=

11 Nm 5.98 10 kg 670 kg

kg 6.37 10 m

2 24

2 6

66 57 10 4 20 10 1 08 1010 10 11. . .× + × = ×J J J

One half becomes extra internal energy in the aluminum: ∆Eint .= ×5 38 1010 J. To raise its temperature to the melting point requires energy

mc T∆ = − −( )( ) = ×670 900 660 15 4 07 108kgJ

kg CC

°° . JJ

To melt it, mL = × = ×670 3 97 10 2 66 105 8kg J kg J. .

To raise it to the boiling point, mc T∆ = ( ) −( ) = ×670 1170 2 450 660 1 40 109J J.

To boil it, mL = × = ×670 1 14 10 7 64 107 9kg J kg J. .

Then

5 38 10 9 71 10 670 1170 2 45010 9. .× = × + ( ) −( )J J CTf ° JJ C

C

°

°Tf = ×5 87 104.



P20.56 From Q mLV= the rate of boiling is described by

P = =Q

t

L m

tV

∆ ∆ ∴ =m

t LV∆P

Model the water vapor as an ideal gas

P V nRTm

MRT

P V

t

m

t

RT

M

P A

0 0

0

0

= = ⎛⎝⎜

⎞⎠⎟

= ⎛⎝⎜

⎞⎠⎟

=

∆ ∆

vPLL

RT

M

RT

ML P A

V

V

⎛⎝⎜

⎞⎠⎟

= =⋅(

vP

0

1 000 W 8.314 J mol K))( )( ) ×( ) ×

373

0 018 0 2 26 10 1 0136

K

kg mol J kg. . . 110 2 00 10

3 76

5 4N m m

m s

2 2( ) ×( )=

−.

.v

P20.57 The power incident on the solar collector is

Pi IA= = ( ) ( )⎡⎣ ⎤⎦ =600 0 300 1702W m m W2 π .

For a 40.0% refl ector, the collected power is Pc = 67 9. W. The total energy required to increase the temperature of the water to the boiling point and to evaporate it is Q cm T mLV= +∆ :

Q = ⋅( )( ) + ×0 500 80 0 2 26 106. . .kg 4 186 J kg C C J k° ° gg

J.

⎡⎣ ⎤⎦= ×1 30 106.

The time interval required is

∆tQ

c

= = × =P

1 30 105 31

6..

J

67.9 Wh

P20.58 (a) The block starts with K mi i= = ( )( ) =1

2

1

21 60 2 50 5 002 2v . . .kg m s J

All this becomes extra internal energy in ice, melting some according to “ ”Q m Lf= ice

Thus, the mass of ice that melts is

mQ

L

K

Lf

i

fice 5

J

3.33 10 J kg= = =

×= × −“ ” 5 00

1 50 10.

. 55 15 0kg mg= .

For the block: Q = 0 (no energy fl ows by heat since there is no temperature difference)

W = −5 00. J

∆Eint = 0 (no temperature change)

and ∆K = −5 00. J

For the ice, Q = 0

W = +5 00. J

∆Eint J= +5 00.

and ∆K = 0

continued on next page

FIG. P20.57


538 Chapter 20

(b) Again, Ki = 5 00. J and mice mg= 15 0.

For the block of ice: Q = 0; ∆Eint J= +5 00. ; ∆K = −5 00. J

so W = 0

For the copper, nothing happens: Q E K W= = = =∆ ∆int 0

(c) Again, Ki = 5 00. J. Both blocks must rise equally in temperature.

“ ”Q mc T= ∆ : ∆TQ

mc= = ( ) ⋅( ) = ×

“ ”

°

5 00

3874 04

..

J

2 1.60 kg J kg C110 3− °C

At any instant, the two blocks are at the same temperature, so for both Q = 0.

For the moving block: ∆K = −5 00. J

and ∆Eint J= +2 50.

so W = −2 50. J

For the stationary block: ∆K = 0

and ∆Eint J= +2 50.

so W = +2 50. J

For each object in each situation, the general continuity equation for energy, in the form∆ ∆K E W Q+ = +int , correctly describes the relationship between energy transfers and changes in the object’s energy content.

P20.59 Energy goes in at a constant rate P. For the period from

50 0

10 0 10

.

.

min to 60.0 min,

min

Q mc T=

( ) =

∆

P kkg J kg C

C C

+( ) ⋅( )−( )mi 4 186

2 00 0

°

° °.

P 10 0 83 7 8 37. . .min kJ kJ kg( ) = + ( )mi (1)

For the period from 0 to 50.0 min, Q m Li f=

P 50 0 3 33 105. .min J kg( ) = ×( )mi

Substitute P =×( )mi 3 33 10

50 0

5.

.

J kg

min into Equation (1)

to fi nd

mm

m

ii

3 33 10

5 0083 7 8 37

5.

.. .

×( )= + ( )J kg

kJ kJ kg

ii =−( )

=83 7

8 371 44

.

..

kJ

66.6 kJ kgkg

0.00

1.00

2.00

3.00

20.0 40.0 60.0

T (° C)

t (min)

FIG. P20.59



P20.60 L Adx

dtk A

T

x

ρ= ⎛

⎝⎜⎞⎠⎟

∆

L xdx k T dt

Lx

k T t

t

ρ

ρ

4 00

8 00

0

2

4 00

8 00

2

.

.

.

.

∫ ∫=

=

∆

∆ ∆

∆

33 33 10 9170 080 0 0 0405

2

.. .

×( )( ) ( ) −J kg kg m

m 03 mW m C C

( )⎛

⎝⎜⎞

⎠⎟= ⋅( )( )

=

2

22 00 10 0

3 66

. .

.

° ° ∆

∆

t

t ×× =10 10 24 s h.

P20.61 A A A A A= + + +end walls ends of attic side walls rooof

A = ×( ) + × × × ( )2 8 00 2 21

24 00. . tanm 5.00 m m 4.00 m 337 0

2 10 0 2 10 04 00

.

. ..

°⎡⎣⎢

⎤⎦⎥

+ ×( ) + ( )m 5.00 m mm

cos37.0

m2

°

⎛⎝⎜

⎞⎠⎟

=

= =× −

A

k A T

L

304

4 80 10 4

P ∆ . kkW m C m C

mkW

2⋅( )( )( )= =

° °304 25 0

0 21017 4 4

.

.. .115 kcal s

Thus, the energy lost per day by heat is 4 15 86 400 3 59 105. .kcal s s kcal day( )( ) = × .

The gas needed to replace this loss is 3 59 10

9 30038 6

5..

× =kcal day

kcal mm day3

3 .

P20.62 See the diagram appearing with the next problem. For a cylindrical shell of radius r, height L, and thickness dr, the equation for thermal conduction,

dQ

dtkA

dT

dx= − becomes

dQ

dtk rL

dT

dr= − ( )2π

Under equilibrium conditions, dQ

dt is constant; therefore,

dTdQ

dt kL

dr

r= − ⎛

⎝⎜⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

1

2π and dT

dQ

dt kL

dr

rT

T

a

b

a

b

∫ ∫= − ⎛⎝⎜

⎞⎠⎟

1

2π

T TdQ

dt kL

b

ab a− = − ⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

1

2πln

But T Ta b> , so dQ

dt

kL T T

b aa b=

−( )( )

2πln


540 Chapter 20

P20.63 From the previous problem, the rate of energy fl ow through the wall is

dQ

dt

kL T T

b a

dQ

dt

a b=−( )

( )

=× −

2

2 4 00 10 5

π

π

ln

. cal s ⋅⋅ ⋅( )( )( )( )

cm C cm C

cm 250 cm

° °3 500 60 0

256

.

ln

dQQ

dt= × =2 23 10 9 323. .cal s kW

This is the rate of energy loss from the plane by heat, and consequently is the rate at which energy must be supplied in order to maintain a constant temperature.

*P20.64 Q Qcold hot= −

or Q Q QAl water calo= − +( )

m c T T m c m c T Tf i w w c c f i wAl Al Al

k

−( ) = − +( ) −( )0 200. gg C kg 4 186 J kg CAl( ) +( ) = − ⋅( )c 39 3 0 400. .° °⎡⎣

+ 0 0. 440 0 kg 630 J kg C⋅( )⎤⎦° 3 70 C−( )°.

6 29 10Al =

×.c

33

800J

7.86 kg CJ kg C

⋅= ⋅

°°

This differs from the tabulated value by (9000–800)/900 =11%, so the values agree withiin 15%.

*P20.65 (a) If the energy fl owing by heat through one spherical surface within the shell were different from the energy fl owing through another sphere, the temperature would be changing at a radius between the layers, so the steady state would not yet be established.

For a spherical shell of radius r and thickness dr, the equation for thermal conduction,

dQ

dtkA

dT

dx= − , becomes dQ

dtk r

dT

dr

dT

dr k= ( ) =P P

= 4 2π so44 2πr

(b) We separate the variables T and r and integrate from the interior to the exterior of the shell:

dTk

dr

r= ∫∫

P4 23

7

5

40

π cm

cm

C

C

°

°

(c) Tk

r540

1

3

7

4 140 5°

° ° °CC

cm

cm

C C=−

−−P

π==

( )− +⎛

⎝⎜⎞⎠⎟

P4 0 8

1

7

1

3π . W/m C cm cm°

P = 35°C(4π)(0.8 W�100 cm⋅°C)�(0.190�cm) = 18.5 W

(d) With P now known, we take the equation from part (a), separate the variables again and integrate between a point on the interior surface and any point within the shell.

dTk

dr

r

rT= ∫∫

P4 235 π cmC°

(e) Tk

rTT

r

5

1

34 1

518 5

° °C

cm

CW

=−

− =−P

π.

44 0 8

1 1

3π . W/m C cm°( )− +⎛

⎝⎜⎞⎠⎟r

T r= 5°C + 184 cm °C (1/3 cm 1/ )⋅ −

(f ) T = 5°C + 184 cm ⋅ °C (1�3 cm − 1�5 cm) = 29.5°C

FIG. P20.63



*P20.66 (a) P = = ×( ) × ( )−σ AeT 4 8 145 67 10 5 1 10 0 965. . .W m K m2 4 2 55 800 3 16 104 22K W( ) = ×.

(b) Tavg K K K= ( ) + ( ) = ×0 1 4 800 0 9 5 890 5 78 103. . .

This is cooler than 5 800 K by 5 800 – 5 781 ____________ 5 800

= 0.328%

(c) P = ×( ) ×( )−5 67 10 0 1 5 1 10 0 965 4 808 14. . . .W m K m2 4 2 004

K( ) + × ×( ) ( ) =− W 0.9 5.1 10145 67 10 0 965 5 890 3 18 4

. . . 77 1022× W

This is larger than 3.158 × 1022 W by 1.29 × 1020 W ____________ 3.16 × 1022 W

= 0.408%

ANSWERS TO EVEN PROBLEMS

P20.2 0 105. °C

P20.4 The energy input to the water is 6.70 times larger than the laser output of 40.0 kJ.

P20.6 88 2. W

P20.8 (a) 16.1°C (b) 16.1°C (c) It makes no difference whether the drill bit is dull or sharp, or how far into the block it cuts. The answers to (a) and (b) are the same because work cannot be a fi nal form of energy: all of the work done by the bit constitutes energy being transferred into the inter-nal energy of the steel.

P20.10 (a) 25 8. °C (b) The fi nal temperature does not depend on the mass. Both the change in potential energy and the heat that would be required (from a stove to produce the change in temperature) are proportional to the mass; hence, the mass cancels in the energy relation.

P20.12 (a) 380 K (b) 2.04 atm

P20.14 12.9 g

P20.16 (a) all the ice melts; 40 4. °C (b) 8.04 g melts; 0°C

P20.18 (a) 7 (b) As the car stops it transfers part of its kinetic energy into the air. As soon as the brakes rise above the air temperature they lose energy by heat, and lose it very fast if they attain a high temperature.

P20.20 (a) Two speeding lead bullets, one of mass 12.0 g moving to the right at 300 m�s and one of mass 8.00 g moving to the left at 400 m �s, collide head-on and all of the material sticks together. Both bullets are originally at temperature 30.0°C. Describe the state of the system immediately thereafter. (b) After the completely inelastic collision, a glob comprising 3.10 g of solid lead and 16.9 g is liquid lead is moving to the right at 20.0 m �s. Its temperature is 327.3°C.

P20.22 (a) −12 0. MJ (b) +12 0. MJ

P20.24 − −( )nR T T2 1

P20.26 (a) 12 0. kJ (b) −12 0. kJ


542 Chapter 20

P20.28 42 9. kJ

P20.30 (a) 7.65 L (b) 305 K

P20.32 (a) −48 6. mJ (b) 16 2. kJ (c) 16 2. kJ

P20.34 (a) −4PVi i (b) +4PVi i (c) −9 08. kJ

P20.36 (a) 1 300 J (b) 100 J (c) −900 J (d) −1 400 J

P20.38 1 34. kW

P20.40 (a) 0 890. ft F h Btu2 ⋅ ⋅° (b) 1 85.ft F h

Btu

2 ⋅ ⋅° (c) 2 08.

P20.42 3.85 × 1026 W

P20.44 (a) ~103 W (b) decreasing at ~10–1 K �s

P20.46 364 K

P20.48 (a) 1.19 (b) 1.19

P20.50 (a) 13.0°C (b) –0.532°C�s

P20.52 (a) PiV

i �2. Put the cylinder into a refrigerator at absolute temperature T

i �2. Let the piston move

freely as the gas cools. (b) 1.39PiV

i. With the gas in a constant-temperature bath at T

i, slowly

push the piston in. (c) 0. Lock the piston in place and hold the cylinder over a hotplate at 3Ti.

See the solution.

P20.54 (a) 9.31 × 1010 J (b) –8.47 × 1012 J (c) 8.38 × 1012 J

P20.56 3 76. m s

P20.58 (a) 15 0. mg. Block: Q = 0, W = −5 00. J, ∆Eint = 0, ∆K = −5 00. J. Ice: Q = 0, W = 5 00. J, ∆Eint J= 5 00. , ∆K = 0. (b) 15 0. mg. Block: Q = 0, W = 0, ∆Eint J= 5 00. , ∆K = −5 00. J.Metal: Q = 0, W = 0, ∆Eint = 0, ∆K = 0. (c) 0 004 04. °C. Moving block: Q = 0, W = −2 50. J, ∆Eint J= 2 50. , ∆K = −5 00. J. Stationary block: Q = 0, W = 2 50. J, ∆Eint J= 2 50. , ∆K = 0.

P20.60 10 2. h

P20.62 see the solution

P20.64 800 J� kg⋅°C. This differs from the tabulated value by 11%, so they agree within 15%.

P20.66 (a) 3.16 × 1022 W (b) 5.78 × 103 K, 0.328% less than 5 800 K (c) 3.17 × 1022 W, 0.408% larger


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