energy in thermal processes: the first law of...

461

Energy in ThermalProcesses: The First Law of

ThermodynamicsCHAPTER OUTLINE 17.1 Heat and Internal Energy 17.2 Specific Heat 17.3 Latent Heat and Phase

Changes 17.4 Work in Thermodynamic

Processes 17.5 The First Law of

Thermodynamics 17.6 Some Applications of the

First Law of Thermodynamics

17.7 Molar Specific Heats of Ideal Gases

17.8 Adiabatic Processes for an Ideal Gas

17.9 Molar Specific Heats and the Equipartition of Energy

17.10 Energy Transfer Mechanisms in Thermal Processes

17.11 Context ConnectionEnergy Balance for the Earth

ANSWERS TO QUESTIONS Q17.1 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.

Q17.2 The ∆T is twice as great in the ethyl alcohol. Q17.3 The final equilibrium temperature will show no significant increase

over the initial temperature of the water. Q17.4 Some water may boil away. You would have to very precisely

measure how much, and very quickly measure the temperature of the steam; it is not necessarily 100°C.

Q17.5 Heat is energy being transferred, not energy contained in an object. Further, a large-mass object, or

an object made of a material with high specific heat, can contain more internal energy than a higher-temperature object.

Q17.6 There are three properties to consider here: thermal conductivity, specific 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 fingers. 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 specific heat compared to aluminum, can be a significant 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!

462 Energy in Thermal Processes: The First Law of Thermodynamics

Q17.7 Write 1 000 1 1 3 1 000 1 kg 4 186 J kg C C kg m J kg C C3⋅° ° = ⋅° °b ga f e jb ga fV . to find V = ×3 2 103 3. m .

Q17.8 (a) and (b) both increase by minuscule amounts. Q17.9 If the system is isolated, no energy enters or leaves the system by heat, work, or other transfer

processes. 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.

Q17.10 Yes. If you know the different specific heats of zinc and copper, you can determine the fraction of

each by heating a known mass of pennies to a specific initial temperature, say 100°C, and dumping them into a known quantity of water, at say 20°C . The final temperature T will reveal the metal content:

m xc x c T m c Tpennies Cu Zn H O H OC C

2 2+ − ° − = − °1 100 20a f a f a f.

Since all quantities are known, except x, the fraction of the penny that is copper will be found by

putting in the experimental numbers mpennies , mH O2, T finala f , cZn, and cCu.

Q17.11 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 floor. Q17.12 The question refers to baking in a conventional oven, not to microwaving. The metal has much

higher thermal conductivity than the potato. The metal quickly conducts energy from the hot oven into the center of potato.

Q17.13 In winter the interior of the house is warmer than the air outside. On a summer day we want the

interior to stay cooler than the exterior. Heavy draperies over the windows can slow down energy transfer by conduction, by convection, and by radiation, to make it easier to maintain the desired difference in temperature.

Q17.14 In winter the produce is protected from freezing. The heat capacity of the earth is so high that soil

freezes only to a depth of a few decimeters in temperate regions. Throughout the year the temperature will stay nearly constant all day and night. Factors to be considered are the insulating properties of soil, the absence of a path for energy to be radiated away from or to the vegetables, and the hindrance to the formation of convection currents in the small, enclosed space.

Q17.15 The high mass and specific heat of the barrel of water and its high heat of fusion mean that a large

amount of energy would have to leak out of the cellar before the water and the produce froze solid. Evaporation of the water keeps the relative humidity high to protect foodstuffs from drying out.

Q17.16 The porcelain of the teacup is a thermal insulator. That is, it is a thermal conductor of relatively low

conductivity. When you wrap your hands around a cup of hot tea, you make A large and L small in

the equation P =−

kAT T

Lh c for the rate of energy transfer by heat from tea into you. When you hold

the cup by the handle, you make the rate of energy transfer much smaller by reducing A and increasing L. The air around the cup handle will also reduce the temperature where you are touching it. A paper cup can be fitted into a tubular jacket of corrugated cardboard, with the channels running vertically, for remarkably effective insulation, according to the same principles.

Chapter 17 463

Q17.17 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.

Q17.18 The person should add the cream immediately when the coffee is poured, rather than later when he

is ready to drink it. Then the smaller temperature difference between coffee and environment will reduce the rate of energy loss during the several minutes. Also, the lighter color of the coffee with cream may reduce the emissivity to decrease the radiated power.

Q17.19 Convection. The bridge deck loses energy rapidly to the air both above it and below it. Q17.20 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 specific 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. The liquid nitrogen still on the marshmallow comes in contact with the much hotter saliva and immediately boils into cold gaseous nitrogen. This nitrogen gas has very low thermal conductivity. 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 film of steam. 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.

Q17.21 (a) Warm a pot of coffee on a hot stove.

(b) Consider an ice cube at 0°C in a glass of lemonade on a picnic table. 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 flows 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 snowflake or a grain of dust to trigger its freezing into ice, and the release of internal energy measured by its latent heat of fusion can actually push its temperature up.

Q17.22 Partially evacuating the container is equivalent to letting the remaining gas expand. This means that

the gas does work, making its internal energy and hence its temperature decrease. The liquid in the container will eventually reach thermal equilibrium with the low pressure gas. This effect of an expanding gas decreasing in temperature is a key process in your refrigerator or air conditioner.


SOLUTIONS TO PROBLEMS Section 17.1 Heat and Internal Energy P17.1 Imagine a mass m of water contained in a styrofoam box, which prevents any energy flow into or

out of it by heat. As the water falls the gravitational energy of the water-Earth system turns into kinetic energy of organized (downward) motion of the water, then into kinetic energy of disorganized (splashing around) macroscopic motion, and at last into kinetic energy of random molecular motion, which is internal energy. There is 100% conversion efficiency from gravitational into internal energy. Heat is not involved in the warming process, but imagine a process in which the mass m of water is placed on a stove and energy mgy is put into it by heat. The temperature change is then specified by

mgy mc T

Tgyc

= ∆

∆ = =⋅° ⋅

FHG

IKJ = °

9 80 120

41861

0 281.

. m s m

J kg C J

1 kg m sC

2

2 2

e ja f

P17.2 The container is thermally insulated, so no energy flows 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

0 200 4 18688 2

0 105water

2 kg m s m

kg J kg C J

837 J CC

. . .

..

.e ja fb g

FIG. P17.2

Section 17.2 Specific Heat P17.3 From Q mc T= ∆

we find ∆TQmc

= =⋅°

= °1 200

62 0 J

0.050 0 kg 387 J kg CCb g .

Thus, the final temperature is 87 0. °C .

P17.4 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

= =

= =−

⋅° ° − °=

× ⋅ ⋅°°

= ×

Qt

mc Tt

mt

c T

∆∆

∆

∆∆

6 000 14 3 600

850 18 383 02 10

8501

8 J s h s h

J kg C C C J kg C

J 20 C.78 10 kg4b ga fb g

b ga f a f.

Chapter 17 465

P17.5 Q Qcold hot= −

mc T mc T

T T

T

f f

f

∆ ∆a f a fb gd i b gb gd iwater iron

kg J kg C C kg J kg C C

C

= −

⋅° − ° = − ⋅° − °

= °

20 0 4 186 25 0 1 50 448 600

29 6

. . .

.

*P17.6 Let us find the energy transferred in one minute.

Q m c m c T

Q

= +

= ⋅° + ⋅° − ° = −

cup cup water water

kg J kg C kg J kg C C J

∆

0 200 900 0 800 4 186 1 50 5 290. . .b gb g b gb g a f

If this much energy is removed from the system each minute, the rate of removal is

P = = = =Q

t∆5 290

88 2 88 2 J

60.0 s J s W. . .

*P17.7 (a) Work that the bit does in deforming the block, breaking chips off, and giving them kinetic

energy is not a final 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 = ⋅ = ° =r rF d 3 2 40 15 0 1 920. cos N m s s Ja fb ga f

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 temperature 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

TQmc

=

= =⋅ ⋅°

= °

∆

∆1 920

44816 1

J kg C0.267 kg J

Cb ga f .

(b) See part (a).

P17.8 (a) Q Qcold hot= −

m c m c T T m c T T m c T Tw w c c f c f unk unk f unk+ − = − − − −b gd i d i d iCu Cu Cu

where w is for water, c the calorimeter, Cu the copper sample, and unk the unknown.

250 100 20 0 10 0

50 0 0 092 4 20 0 80 0 20 0 100

2 44 10 5 60 103 3

g 1.00 cal g C g 0.215 cal g C C

g cal g C C 70.0 g C

cal g C

⋅° + ⋅° − °

= − ⋅° − ° − − °

× = × ⋅°

b g b g a fb gb ga f b g a fe j

. .

. . . . .

. .

c

c

unk

unk

or cunk = ⋅°0 435. cal g C .

(b) The material of the sample can be beryllium .


*P17.9 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. kg J

kg CC J . The energy the copper can put

out in dropping to 26°C is mc T∆ =°

° =0 1 387 74 2 864. kg J

kg CC J . Since 6 279 2 864 J J> , the final

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

Q Q Q

T T

T

T T T

T

T

f f

f

f f f

f

f

water Al Cu

kg J

kg CC kg

Jkg C

C

kg J

kg CC

C C C

C

C

+ + =

°− ° +

°− °

+°

− ° =

− ° + − ° + − ° =

= °

= °

0

0 25 4 186 20 0 4 900 26

0 1 387 100 0

1 046 5 20 930 360 9 360 38 7 3 870 0

1 445 2 34 160

23 6

. .

.

. .

.

.

d i d i

d i


m c T T m c T T m c T T

m c m c T m c m c T m c T m c T

m c m c m c T m c m c T m c T

Tm c m c T m c T

m c m c m c

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

fc w c h w h

c w h w

Al Al

Al Al Al Al

Al Al Al Al

Al Al

Al Al

− + − = − −

+ − + = − +

+ + = + +

=+ +

+ +

d i d i d ib g b gb g b gb g

Section 17.3 Latent Heat and Phase Changes P17.11 The heat needed is the sum of the following terms:

Qneeded heat to reach melting point heat to melt

heat to reach boiling point heat to vaporize heat to reach 110 C

= +

+ + + °

b g a fb g b g a f

Thus, we have

Q

Q

needed

needed

kg J kg C C J kg

J kg C C J kg J kg C C

J

= ⋅° ° + ×

+ ⋅° ° + × + ⋅° °

= ×

0 040 0 2 090 10 0 3 33 10

4 186 100 2 26 10 2 010 10 0

1 22 10

5

6

5

. . .

. .

.

b ga f e jb ga f e j b ga f

Chapter 17 467


m c m c T T m L c T

m

m

w w c c f i s v w f

s

s

+ − = − − + −

⋅° + ⋅° ° − °

= − − × + ⋅° ° − °

= ××

= =

b gd i d ib g b g a f

b ga f

100

0 250 0 050 0 50 0 20 0

2 26 10 4 186 50 0 100

3 20 100 012 9 12 9

6

4

. . . .

. .

.. .

kg 4 186 J kg C kg 387 J kg C C C

J kg J kg C C C

J2.47 10 J kg

kg g steam6

P17.13 The bullet will not melt all the ice, so its final temperature is 0°C.

Then 12

2mv mc T m Lw f+FHG

IKJ =∆

bullet

where mw is the melt water mass

m

m

w

w

=× + × ⋅° °

×

= + =

− −0 500 3 00 10 240 3 00 10 30 0

3 33 1086 4 11 5

0 294

3 2 3

5

. . . .

.. .

.

kg m s kg 128 J kg C C

J kg J J

333 000 J kg g

e jb g b ga f

P17.14 Q m c T m L= =Cu Cu N vap N22

∆ e j

1 00 293 77 3 48 0

0 414

. . .

.

kg 0.092 0 cal g C C cal g

kg

⋅° − ° =

=

b ga f b gm

m

P17.15 (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 final 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

m

f w w i

cold hot

C

J kg kg J kg C C C

g, so the ice remaining g g g

= −

= − ° −

× = − ⋅° ° − °

= = − =

0

3 33 10 0 600 4 186 0 18 0

136 250 136 114

5

b ge j b gb ga f. . .

P17.16 The original gravitational energy of the hailstone-Earth system changes entirely into additional

internal energy in the hailstone, to produce its phase change. No temperature change occurs, either in the hailstone, in the air, or in sidewalk. Then

mgy mL

yLg

=

= =× ⋅F

HGIKJ = ×

3 33 109 8

11

3 40 105

4..

. J kg

m s kg m s

J m2

2 2


*P17.17 The original kinetic energy all becomes thermal energy:

12

12

212

5 00 10 500 1 252 2 3 2mv mv+ = FHGIKJ × =−. . kg m s kJe jb g .

Raising the temperature to the melting point requires

Q mc T= = × ⋅° ° − ° =−∆ 10 0 10 327 20 0 3933. . kg 128 J kg C C C Jb ga f . Since 1 250 393 J J> , the lead starts to melt. Melting it all requires

Q mL= = × × =−10 0 10 2 45 10 2453 4. . kg J kg Je je j .

Since 1 250 393 245 J J> + , it all melts. If we assume liquid lead has the same specific heat as solid

lead, the final temperature is given by

1 25 10 393 245 10 0 10 128 327

805

3 3. .× = + + × ⋅° − °

= °

− J J J kg J kg C C

C

b gd iT

T

f

f

P17.18 We find the quantity of water vapor in one exhaled breath.

PV nRT= : nPVRT

= =× ×

⋅ += ×

−−

3 20 10 0 600 10

8 314 273 377 45 10

34

. .

..

3 2 3 N m m

J mol K K K mol

e je jb ga f

The molar mass of water (H O2 ) is M = + =2 1 00 16 0 0 018 0. . .a fc h g mol kg mol. The mass of water

vapor exhaled in one breath is m nMsample mol 0.0180 kg mol kg= = × = ×− −7 45 10 1 34 104 5. .b g . The

energy absorbed from your body as the water evaporates can be estimate as

Q mL= = × × =−1 34 10 30 35. . kg 2.26 10 J kg J6e j .

Your rate of energy loss is P = = FHG

IKJFHGIKJ =

Qt∆

30 3 22 0 111 1

. .min

.J

breathbreath min

60 s W . Note that a dog does

not perspire. instead, the dog pants, maximizes energy loss through the pathway considered here.

Chapter 17 469

Section 17.4 Work in Thermodynamic Processes

P17.19 W PdVifi

f

= −z

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

ifi

f

f i

f i

= − = − −

= = =

zα α2 3 313

2 2 1 00 2 00

e j

e j. . m m3 3

P

O V

1.00 m 3 2.00 m3

P = V α 2

f

i

FIG. P17.19

Wif = − × −LNM

OQP = −1

35 00 1 013 10 2 00 1 00 1 185 3 3. . . . . atm m Pa atm m m MJ6 3 3e je j e j e j

P17.20 (a) W PdV= −z

W

Wi f

= − × − +

− × − +

− × −

= −→

6 00 10 2 00 1 00

4 00 10 3 00 2 00

2 00 10 4 00 3 00

12 0

6

6

6

. . .

. . .

. . .

.

Pa m

Pa m

Pa m

MJ

3

3

3

e ja fe ja fe ja f

(b) W f i→ = +12 0. MJ

FIG. P17.20

P17.21 W P V PnRP

T T nR Tf i= − = − FHGIKJ − = − = − = −∆ ∆d i a fa fa f0 200 8 314 280 466. . J

P17.22 W PdV P dV P V nR T nR T Ti

f

i

f

= − = − = − = − = − −z z ∆ ∆ 2 1b g The negative sign 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.

Section 17.5 The First Law of Thermodynamics P17.23 ∆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.


P17.24 (a) Q W= − = Area of triangle

Q = =12

4 00 6 00 12 0. . . m kPa kJ3e ja f

(b) Q W= − = −12 0. kJ

FIG. P17.24 P17.25 Q W ∆Eint BC – 0 – Q E WBC= =∆ int since 0b g CA – + – ∆E W Qint and so < > <0 0 0,b g AB + – + W E E B C A Q< > < → → >0 0 0 0, ;∆ ∆int int since for so b g

Section 17.6 Some Applications of the First Law of Thermodynamics

P17.26 (a) W nRTV

VP V

V

Vf

if f

f

i= −

FHGIKJ = −

FHGIKJln ln

so V VW

P Vi ff f

= +FHG

IKJ =

−

×

L

NMM

O

QPP =exp . exp

. ..0 025 0

3 000

0 025 0 1 013 100 007 65

5b ge j

m3

(b) TP V

nRff f= =

×

⋅=

1 013 10 0 025 0

1 00305

5. .

.

Pa m

mol 8.314 J K mol K

3e jb g

P17.27 (a) ∆ ∆E Q P Vint

3 kJ kPa 3.00 m kJ= − = − − =12 5 2 50 1 00 7 50. . . .a f

(b) VT

VT

1

1

2

2=

TVV

T22

11

3 001 00

300 900= = =..

K Ka f

P17.28 (a) W P V P V T

W

= − = −

= − × × °×

FHG

IKJ °

LNMM

OQPP

= −

− −

∆ ∆3

1 013 10 3 24 0 101 00

18 0

48 6

5 6 1

α

. ..

.

.

N m C kg

2.70 10 kg mC

mJ

23 3e j e j a f

(b) Q cm T= = ⋅° ° =∆ 900 1 00 18 0 16 2 J kg C kg C kJb gb ga f. . .

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

Chapter 17 471

P17.29 W P V P V VP nRT

PPs w= − = − − = − +L

NMM

O

QPP∆ b g a f

e je j18 0

106

. g

1.00 g cm cm m3 3 3

W

Q mL

E Q W

v

= − ⋅ + ×FHG

IKJ = −

= = × =

= + =

1 00 8 314 373 1 013 1018 0

3 10

0 018 0 40 7

37 6

5. . ..

.

. .

.

mol J K mol K N m g

10 g m kJ

kg 2.26 10 J kg kJ

kJ

26 3

6

int

a fb ga f e j

e j∆

P17.30 (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 PVDA AB BC CD

i i i i i i i i

= + + +

= − − + − − + = −3 0 3 3 0 4b g b g

(b) The initial and final 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. . .a fa fa f kJ

FIG. P17.30

P17.31 (a) PV P V nRTi i f f= = = ⋅ = ×2 00 300 4 99 103. . mol 8.314 J K mol K Jb ga f

VnRT

P

VnRTP

V

ii

ff

i

= = ×

= = × = =

4 99 10

4 99 10 13

0 041 0

3

3

.

..

J0.400 atm

J1.20 atm

m3

(b) W PdV nRTV

Vf

i= − = −

FHGIKJ = − × F

HGIKJ = +z ln . ln .4 99 10

13

5 483e j kJ

(c) ∆E Q Wint = = +0 Q = −5 48. kJ

P17.32 ∆ ∆E EABC ACint, int, = (conservation of energy)

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

QABC = + =800 500 1 300 J J J (b) W P VCD C CD= − ∆ , ∆ ∆V VAB CD= − , and P PA C= 5

Then, W P V WCD A AB AB= = − =15

15

100∆ 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 J= − = − − = −800 500 1 300

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

FIG. P17.32


Section 17.7 Molar Specific Heats of Ideal Gases P17.33 We use the tabulated values for Cp and CV .

(a) Q nC Tp= = ⋅ − =∆ 1 00 28 8 420 300 3 46. . . mol J mol K K kJb ga f

(b) ∆ ∆E nC TVint . . .= = ⋅ =1 00 20 4 120 2 45 mol J mol K K kJb ga f

(c) W Q E= − + = − + = −∆ int . .3 46 2 45 1 kJ kJ .01 kJ

P17.34 E nRTint = 32

∆ = ∆ = ⋅ =E nR Tint . . . .32

32

3 00 8 314 2 00 74 8 mol J mol K K Ja fb ga f

P17.35 n Ti= =1 00 300. mol, K

(b) Since V = constant, W = 0 .

(a) ∆ = + = + =E Q Wint 209 0 209 J J

(c) ∆ = ∆ = FHGIKJ∆E nC T n R TVint

32

so

∆ =

∆=

⋅=

= + ∆ = + =

TEnR

T T Ti

23

2 2093 1 00 8 314

16 8

300 16 8 317

int

. ..

.

J mol J mol K

K

K K K

a fa fb g

P17.36 (a) C RV = = ⋅FHG

IKJ = ⋅ = ⋅5

252

8 3141 000 0289

719 0 719...

. J mol K mol

kg J kg K kJ kg Kb g

(b) m Mn MPVRT

= = FHGIKJ

m =×

⋅

FHGG

IKJJ =0 0289

8 3140 811.

.. kg mol

200 10 Pa 0.350 m

J mol K 300 K kg

3 3

b g e jb ga f

(c) We consider a constant volume process where no work is done.

Q mC TV= ∆ = ⋅ − =0 811 700 300 233. kg 0.719 kJ kg K K K kJb ga f (d) We now consider a constant pressure process where the internal energy of the gas is

increased and work is done.

Q mC T m C R T m

RT m

CT

Q

P VV= ∆ = + ∆ = FHG

IKJ∆ = FHG

IKJ∆

= ⋅LNM

OQP =

b g

b g

72

75

0 81175

0 719 400 327. . ( ) kg kJ kg K K kJ

Chapter 17 473

P17.37 Consider 800 cm3 of (flavored) water at 90.0 °C mixing with 200 cm3 of diatomic ideal gas at 20.0°C: Q Qcold hot= − or

m c T T m c T

Tm c T T

m c

V c

V c

P f i w w w

wP f i

w w

P

w w w

air air air

air air air air air C C

, ,

, , ,( )( ) ( ) . .

( )

− = − ∆

∆ =− −

=− ° − °

d i a fa fρ

ρ90 0 20 0

where we have anticipated that the final temperature of the mixture will be close to 90.0°C. The

molar specific heat of air is C RP , air = 72

. So the specific heat per gram is

cRM

T

P

w

, ..

..

( )[( . )( )]( .

[( . )( )]( .

air

3 3

3 3

J mol K mol

g J g C

g cm cm J g C)(70.0 C) g cm cm J kg C)

= = ⋅FHG

IKJ = ⋅°

∆ = −× ⋅° °

⋅°

FHGGIKJJ

−

72

72

8 3141 00

28 91 01

1 20 10 200 1 011 00 800 4 186

3

b g

or ( ) .∆ ≈ − × °−T w 5 05 10 3 C . The change of temperature for the water is

between C and 10 C10 3 2− −° ° .

*P17.38 (a) The air is far from liquefaction so it behaves as an ideal gas. From PV nRT= we have

PVmM

RT= , PMmV

RT RT= = ρ . For the samples of air in the balloon at 10°C (cold) and at

the elevated temperature (hot) we have PM RTc c= ρ and PM RTh h= ρ . Then ρ ρh h c cT T=

and ρ ρh

c c

h

TT

= . For equilibrium of the balloon on the point of rising,

F ma B F F

Vg Vg mg

VT

TV m

T

T

T

y y g g

c h

cc c

h

h

h

h

∑ = + − − =

+ − − =

+ − − =

− − =

=

= =

hot air cargo

3 3 3 kg m m kg m K

m kg

kg kg K

K K

0

0

0

1 25 400 1 25283

400 200 0

300 500283

500300

283 472

3

ρ ρ

ρ ρ

. .e j e j

The quantity of air that must be warmed is given by PV n RTh h= , nPVRTh

h= . The heat input

required is

Q nC TPVRT

R T Tph

h c= = − − =× −

= ×∆ 72

101 10 400 7 472 283472 2

5 66 103 3

7b g a fN m K Km K

J2 .

(b) Q mH= , mQH

= = ××

=5 66 105 03 10

1 127

7.

..

J J kg

kg


P17.39 Q nC T nC TP V= ∆ + ∆( ) ( )isobaric isovolumetric

In the isobaric process, V doubles so T must double, to 2Ti .

In the isovolumetric process, P triples so T changes from 2Ti to 6Ti .

Q n R T T n R T T nRT PVi i i i i= FHGIKJ − + FHG

IKJ − = =7

22

52

6 2 13 5 13 5b g b g . .

Section 17.8 Adiabatic Processes for an Ideal Gas

P17.40 (a) PV P Vi i f fγ γ= so

V

VPP

f

i

i

f=FHGIKJ = FHG

IKJ =

1 5 71 0020 0

0 118/ /.

..

γ

(b) T

T

P V

PV

P

P

V

Vf

i

f f

i i

f

i

f

i= =

FHGIKJFHGIKJ = 20 0 0 118. .a fa f , T

Tf

i= 2 35.

(c) Since the process is adiabatic, Q = 0 .

Since γ = = =+

1 40.CC

R CC

P

V

V

V, C RV = 5

2 and ∆ = − =T T T Ti i i2 35 1 35. .

∆ = ∆ = FHGIKJ ⋅ =E nC TVint . . .0 0160

52

8 314 1 35 300 135 mol J mol K K Ja f b g a f

and W Q E= − + ∆ = + = +int 0 135 135 J J .

P17.41 (a) PV P Vi i f f

γ γ=

P PVVf i

i

f=FHGIKJ = F

HGIKJ =

γ

5 0012 030 0

1 391.40

...

. atm atm

(b) TPVnRii i= =

× ×

⋅=

−5 00 1 013 10 12 0 10366

5 3. . . Pa m

2.00 mol 8.314 J mol K K

3e je jb g

TP V

nRff f= =

× ×

⋅=

−1 39 1 013 10 30 0 10253

5 3. . . Pa m

2.00 mol 8.314 J mol K K

3e je jb g

(c) The process is adiabatic: Q = 0

γ = = =+

1 40.CC

R CC

P

V

V

V, C RV = 5

2

∆ = ∆ = ⋅F

HGIKJ − = −

= ∆ − = − − = −

E nC T

W E Q

Vint

int

. . .

. .

2 0052

8 314 253 4 66

4 66 4 66

mol J mol K K 365 K kJ

kJ 0 kJ

b g a f

Chapter 17 475

P17.42 T

TVV

f

i

i

f=FHGIKJ = FHG

IKJ

−γ 1 0 40012

.

If Ti = 300 K , then Tf = 227 K .

P17.43 (a) See the diagram at the right.

(b) P V P VB B C C

γ γ=

3PV PVi i i Cγ γ=

V V V VC i i i= = =3 3 2 191 5 7/ / .γe j e j

VC = =2 19 4 00 8 77. . . L La f

(c) P V nRT PV nRTB B B i i i= = =3 3

T TB i= = =3 3 300 900 K Ka f

(d) After one whole cycle, T TA i= = 300 K

FIG. P17.43

(e) In AB, Q nC V n R T T nRTAB V i i i= ∆ = FHGIKJ − =5

23 5 00b g a f. , QBC = 0 as this process is adiabatic

P V nRT P V nRTC C C i i i= = =2 19 2 19. .b g a f so T TC i= 2 19. ,

Q nC T n R T T nRTCA P i i i= ∆ = FHGIKJ − = −7

22 19 4 17. .b g a f

For the whole cycle,

Q Q Q Q nRT nRT

E Q W

W Q nRT PV

W

ABCA AB BC CA i i

ABCA ABCA ABCA

ABCA ABCA i i i

ABCA

= + + = − =

∆ = = +

= − = − = −

= − × × = −−

5 00 4 17 0 829

0

0 829 0 829

0 829 1 013 10 4 00 10 3365 3

. . .

. .

. . .

int

a f a fb g

a f a fa fe je j Pa m J3


P17.44 (a) The work done on the gas is

W PdVab V

V

a

b= −z

For the isothermal process,

W nRTV

dV

W nRTVV

nRTVV

ab a V

V

ab ab

a

a

b

a

b′

′′

′

= − FHGIKJ

= −FHGIKJ =

FHGIKJ

′z 1

ln ln

Thus,

W

W

ab

ab

′

′

= ⋅

=

5 00 293 10 0

28 0

. ln .

.

mol 8.314 J mol K K

kJ

b ga f a f

FIG. P17.44

(b) For the adiabatic process, we must first find the final temperature, Tb . Since air consists

primarily of diatomic molecules, we shall use γ air = 1 40. and

CR

V ,.

.air J mol K= = = ⋅52

58 314

220 8 . Then, for the adiabatic process

T TVVb a

a

b=FHGIKJ = =

−γ 10 400293 736 K 10.0 Ka f . .

Thus, the work done on the gas during the adiabatic process is

W Q E nC T nC T Tab ab V ab V b a= − + ∆ = − + ∆ = −intb g b g b g0

or Wab = ⋅ − =5 00 736 293 46 0. . mol 20.8 J mol K K kJb ga f .

(c) For the isothermal process, we have P V P Vb b a a′ ′ = . Thus,

P PVVb a

a

b′

′

′=FHGIKJ = =1 00 10 0. . atm 10.0 atma f . For the adiabatic process, we have P V P Vb b a a

γ γ= .

Thus, P PVVb a

a

b=FHGIKJ = =

γ

1 00 25 11.40. . atm 10.0 atma f .

Chapter 17 477

P17.45 We suppose the air plus burnt gasoline behaves like a diatomic ideal gas. We find its final absolute pressure:

21 0 400

21 0 1 14

7 5 7 5

7 5

.

. .

/ /

/

atm 50.0 cm cm

atm18

atm

3 3e j e j=

= FHGIKJ =

P

P

f

f

Now Q = 0 and W E nC T TV f i= ∆ = −int d i

FIG. P17.45

∴ = − = −

= −×F

HGIKJ

= −

−

W nRT nRT P V PV

W

W

f i f f i i52

52

52

52

1 14 21 01 013 10

110

150

56

d i

e j e j e j. ..

atm 400 cm atm 50.0 cm N m

atmm cm

J

3 32

3 3

The output work is − = +W 150 J. The time for this stroke is 14

1 606 00 10 3min

2500 s

1 min sF

HGIKJFHGIKJ = × −.

P = − =×

=−Wt∆

15025 0

J6.00 10 s

kW3 .

Section 17.9 Molar Specific Heats and the Equipartition of Energy

P17.46 (1) E Nfk T

fnRTB

int = FHGIKJ =FHGIKJ2 2

(2) Cn

dEdT

fRV = FHGIKJ =

1 12

int

(3) C C R f RP V= + = +12

2b g

(4) γ = =+C

Cf

fP

V

2


P17.47 The heat capacity at constant volume is nCV . An ideal gas of diatomic molecules has three degrees of freedom for translation in the x, y, and z directions. If we take the y axis along the axis of a molecule, then outside forces cannot excite rotation about this axis, since they have no lever arms. Collisions will set the molecule spinning only about the x and z axes. (a) If the molecules do not vibrate, they have five degrees of freedom. Random collisions put

equal amounts of energy 12

k TB into all five kinds of motion. The average energy of one

molecule is 52

k TB . The internal energy of the two-mole sample is

N k T nN k T n R T nC TB A B V52

52

52

FHGIKJ =

FHGIKJ =FHGIKJ = .

The molar heat capacity is C RV = 52

and the sample’s heat capacity is

nC n R

nC

V

V

= FHGIKJ = ⋅F

HGIKJ

=

52

252

8 314

41 6

mol J mol K

J K

.

.

b g

For the heat capacity at constant pressure we have

nC n C R n R R nR

nC

P V

P

= + = +FHG

IKJ = = ⋅F

HGIKJ

=

b g b g52

72

272

8 314

58 2

mol J mol K

J K

.

.

(b) In vibration with the center of mass fixed, both atoms are always moving in opposite

directions with equal speeds. Vibration adds two more degrees of freedom for two more terms in the molecular energy, for kinetic and for elastic potential energy. We have

nC n RV = FHGIKJ =

72

58 2. J K and nC n RP = FHGIKJ =

92

74 8. J K .

Section 17.10 Energy Transfer Mechanisms in Thermal Processes

P17.48 P = ∆k A

TL

kL

A T=

∆=

°= × ⋅°−P 10 0

1 20 15 02 22 10 2.

. ..

W 0.0400 m m C

W m C2

a fa f

Chapter 17 479

P17.49 In the steady state condition, P PAu Ag= so that

k ATx

k ATxAu Au

AuAg Ag

Ag

∆∆FHGIKJ = ∆

∆FHGIKJ

In this case

A A

x x

T T

T T

Au Ag

Au Ag

Au

Ag

=

∆ = ∆

∆ = −∆ = −

80 0

30 0

.

.

a fa f

FIG. P17.49

where T is the temperature of the junction. Therefore, k T k TAu Ag80 0 30 0. .− = −a f a f . And

T = °51 2. C .

P17.50 We suppose that the area of the transistor is so small that energy flow by heat from the transistor

directly to the air is negligible compared to energy conduction through the mica.

P

P

=−

= + = ° +×

⋅° ×= °

−

−

kAT T

L

T TL

kA

h c

h c

b g

e jb ga f35 0

1 50

0 0753 8 25 6 2567 96.

.

. . ..C

W 0.0852 10 m

W m C 10 mC

3

2

P17.51 P = = × ⋅ ×LNM

OQPσ πAeT 4 4 8 2 45 4 6 96 10 0 986 5 800.67 10 W m K m K-8 2e j e j a fa f. .

P = ×3 85 1026. W

P17.52 P =−

= ⋅° ° − °FHG

IKJ

= =FHG

IKJFHG

IKJ =

kAT T

Lh cb g b ge j

b g

0 21037 0 34 0

35 3 35 33600

30 3

.. .

. . .

W m C 1.40 mC C

0.0250 m

W J s1 kcal4186 J

s1 h

kcal h

2

Since this is much less than 240 kcal/h, blood flow is essential to cool the body. P17.53 The net rate of energy loss from his skin is

Pnet2 4 2 W m K m K K W= − = × ⋅ − =−σAe T T4

04 8 4 45 67 10 1 50 0 900 308 293 125e j e je ja f a f a f. . . .

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 104b ga f . .

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


Section 17.11 Context ConnectionEnergy Balance for the Earth P17.54 We suppose the earth below is an insulator. The square meter must radiate in the infrared as much

energy as it absorbs, P = σAeT 4 . Assuming that e = 1 00. for blackbody blacktop:

1000 W W m K m2 4 2= × ⋅−5 67 10 1 00 1 008 4. . .e je ja fT

T = × =1 76 10 36410 1 4.

/ K K4e j (You can cook an egg on it.)

P17.55 The sphere of radius R absorbs sunlight over the area of its day hemisphere, projected as a flat circle

perpendicular to the light: π R2 . It radiates in all directions, over area 4 2π R . Then, in steady state,

P Pin out

2 W m

=

=e R e R T1370 42 2 4e j e jπ σ π

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

T =× ⋅

L

NMM

O

QPP = = °

−

1370

4 5 67 10279 6

8

1 4

W / m

W m K K C

2

2 4.

/

e j.

P17.56 (a) Among atmospheric layers, PV γ = constant and PVT

= constant . Then P V1 /γe j = constant

and PV

T P V1/γe j= constant. Then T P T P1 1 1/ /γ γ γb g b ge j e j− −= = constant .

(b) We differentiate the equation T P 1 1/γb ge j− = constant with respect to y:

T PdPdy

PdTdy

TP

dPdy

dTdy

dTdy

TP

dPdy

11 0

11 0

11

1 1 1 1 1

γ

γ

γ

γ γ−FHGIKJFH IK + =

−FHGIKJ + =

= −FHGIKJ

− − −/ /b g b ge j

(c) Substituting dPdy

g= −ρ gives dTdy

TP

g= −FHGIKJ −1

1γ

ρb g. Now ρ = =m

VnMV

sample . So

dTdy

TnMgPV

MgR

= − −FHGIKJ = − −

FHGIKJ1

11

1γ γ

Chapter 17 481

Additional Problems

*P17.57 (a) The heat leaving the box during the day is given by P =−

=kAT T

LQ

tH cb g

∆

Q =°

° − ° FHG

IKJ = ×0 012 0 49

37 230 045

123 600

17 90 104. .

..

Wm C

mC C

m h

s h

J2 .

The heat lost at night is

Q =°

° − ° FHG

IKJ = ×0 012 0 49

37 160 045

123 600

11 19 105. .

..

Wm C

mC C

m h

s h

J2 .

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=

mQL

= = ××

=1 98 10205 10

0 9645

3.

. J

J kg kg or more

(b) The test samples and the inner surface of the insulation can be preheated to 37°C as the box

is filled. 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.

P17.58 The increase in internal energy required to melt 1.00 kg of snow is

∆Eint . . .= × = ×1 00 3 33 10 3 33 105 5 kg J kg Jb ge j .

The force of friction is

f n mg= = = =µ µ 0 200 75 0 9 80 147. . . kg m s N2b ge j .

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 .= = = ×147 3 33 105 N Ja f

and

∆r = ×2 27 103. m .

P17.59 (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− = − −d i d i

9 00 387 16 0 14 0 234 30 0

30 0 17 0

. . . .

. .

g J kg C C g J kg C C

C C

Ag

Ag

b gb ga f b gb gd id i

⋅° ° = − ⋅° − °

− ° = − °

T

T

f

f

so Tf , . Ag C= °13 0 .

continued on next page


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

m cdTdtAg Ag

AgCu Cu

Cu

FHGIKJ = − F

HGIKJ

dTdt

m cm c

dTdt

dTdt

FHGIKJ = − F

HGIKJ = −

⋅°⋅°

+ °

FHGIKJ = − ° ⇒

Ag

Cu Cu

Ag Ag Cu

Ag

g 387 J kg C

g 234 J kg CC s

C s negative sign decreasing temperature

9 00

14 00 500

0 532

.

..

.

b gb g b g

b g

P17.60 (a) Before conduction has time to become important, the energy lost by the rod equals the

energy gained by the helium. Therefore,

mL mc Tvb g c hHe Al= ∆

or ρ ρVL Vc Tvb g c hHe Al= ∆

so VVc T

LvHe

Al

He

=ρ

ρ∆c hb g

V

V

He

3 3

3

He3

g cm cm cal g C C

g cm J kg cal 4.186 J kg 1 000 g

cm liters

=⋅° °

×

= × =

2 70 62 5 0 210 295 8

0 125 2 09 10 1 00 1 00

1 68 10 16 8

4

4

. . . .

. . . .

. .

e je jb ga fe je jb gb g

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

is given by

P = FHGIKJ = ⋅ ⋅ F

HGIKJ =kA

dTdx

31 0 2 50295 8

917. ..

J s cm K cmK

25.0 cm W2b ge j

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

P

ρLv=

×= =

917 10

0 125 2 09 10351 0 351

3

4

W g kg

g cm J kg cm s L s

33

a fe je je j. .

.

P17.61 Q mc T V c T= =∆ ∆ρb g so that when a constant temperature difference ∆T is maintained,

the rate of adding energy to the liquid is P = = FHGIKJ =dQ

dtdVdt

c T Rc Tρ ρ∆ ∆

and the specific heat of the liquid is cR T

= Pρ ∆

.

Chapter 17 483

P17.62 (a) Work done by the gas is the negative of the area under the PV curve

W PV

VPV

ii

ii i= − −FHG

IKJ = +

2 2.

(b) In this case the area under the curve is W PdV= −z . Since the

process is isothermal,

PV PV PV

nRTi i ii

i= = FHGIKJ =4

4

and WdVV

PV PVVV

PVi iV

V

i ii

ii i

i

i

= − FHGIKJ = −

FHGIKJ =z b g

4 44ln ln

= +1 39. PVi i

a

b c

P

V

FIG. P17.62

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

*P17.63 The initial moment of inertia of the disk is

12

12

12

12

8 920 28 1 2 1 033 102 2 2 2 4 10MR VR R tR= = = = × ⋅ρ ρπ π kg m m m kg m3 2e j a f . .

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

I I

MR MR MR T

T

i i f f

i i f f i f

f i

ω ω

ω ω α ω

ω ωα

=

= = −

=−

=− × ° °

=−

12

12

12

1

1

125

1

1 17 10 83025 720 7

2 2 2 2

2 6 2

∆

∆

c h

c h e j rad s

1 C C rad s.

(a) The kinetic energy increases by

12

12

12

12

12

12

1 033 10 25 0 720 7 9 31 10

2 2 2

10 10

I I I I If f i i i i f i i i i f iω ω ω ω ω ω ω ω− = − = −

= × ⋅ = ×

d ib g. . . kg m rad s rad s J2

(b) ∆ ∆E mc Tint kg 387 J kg C C C J= = × ⋅° ° − ° = − ×2 64 10 20 850 8 47 107 12. .b ga f

(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.


*P17.64 The loss of mechanical energy is

12

12

6706 67 10

6 57 10 4 20 10 1 08 10

2 2 11

10 10 11

mvGM m

RiE

E+ = × +

× ××

= × + × = ×

−

kg 1.4 10 m s Nm 5.98 10 kg 670 kgkg 6.37 10 m

J J J

42 24

2 6e j .

. . .

One half becomes extra internal energy in the aluminum: ∆Eint .= ×5 38 1010 J . To raise the

temperature of the aluminum to the melting point requires energy

mc T∆ =°

− − ° = ×670 900 660 15 4 07 108 kg J

kg CC Ja fc h . .

To melt it, mL = × = ×670 3 97 10 2 66 105 8 kg J kg J. . . To raise it to the boiling point,

mc T∆ = − = ×670 1 170 2 450 660 1 40 109b gb gJ J. . To boil it, mL = × = ×670 1 14 10 7 64 107 9 kg J kg J. . . Then

5 38 10 9 71 10 670 1 170 2 450

5 87 10

10 9

4

. .

.

× = × + − ° °

= × °

J J C J C

C

b gd iT

T

f

f

P17.65 The power incident on the solar collector is

Pi IA= = =600 0 300 1702 W m m W2e j a fπ . .

For a 40.0% reflector, 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 10 1 30 106 6. . . . kg 4 186 J kg C C J kg Jb ga f .

The time interval required is ∆tQ

c= = × =P

1 30 105 31

6..

J67.9 W

h .

FIG. P17.65

P17.66 (a) Fv = =50 0 40 0 2 000. . N m s Wa fb g

(b) Energy received by each object is 1 000 10 10 2 3894 W s J calb ga f = = . The specific heat of iron

is 0 107. cal g C⋅° , so the heat capacity of each object is 5 00 10 0 107 535 03. . .× × = ° cal C.

∆T =°

= °2 389

4 47cal

535.0 cal CC.

Chapter 17 485

P17.67 (a) The block starts with K mvi i= = =12

12

1 60 2 50 5 002 2. . . kg m s Jb gb g

All this becomes extra internal energy in ice, melting some according to “ ”Q m L f= ice . Thus,

the mass of ice that melts is

mQL

KLf

i

fice 5

J3.33 10 J kg

kg mg= = =×

= × =−“ ” .. .

5 001 50 10 15 05 .

For the block: Q = 0 (no energy flows 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 (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= ∆ : ∆TQmc

= =⋅°

= × °−“ ” ..

5 00387

4 04 10 3J2 1.60 kg J kg C

Cb gb g

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.


P17.68 L Adx

dtkA

Tx

ρ = FHGIKJ

∆

L xdx k T dt

Lx

k T t

t

t

t

ρ

ρ

4.00

8 00

0

2

4.00

8 00

52 2

4

2

3 33 10 9170 080 0 0 040

22 00 10 0

3 66 10 10 2

.

.

.. .

. .

. .

z z=

=

×−F

HGG

IKJJ = ⋅° °

= × =

∆

∆ ∆

∆

∆

∆

J kg kg m m 0 m

W m C C

s h

3e je j b g b g b ga f

P17.69 A A A A A= + + +end walls ends of attic side walls roof

A

A

kA TL

= × + × × × °LNM

OQP

+ × +°

FHG

IKJ

=

= =× ⋅° °

= =−

2 8 00 2 212

4 00 37 0

2 10 0 2 10 04 00

304

4 80 10 304 25 0

0 21017 4 4 15

4

. . tan .

. ..

. .

.. .

m 5.00 m m 4.00 m

m 5.00 m m m

cos37.0

m

kW m C m C

m kW kcal s

2

2

a f a f

a f a f

e je ja fP

∆

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

The gas needed to replace this loss is 3 59 10

9 30038 6

5..

×=

kcal day kcal m

m day33 .


or Q Q QAl water calo= − +b g

m c T T m c m c T T

c

c

f i w w c c f i wAl Al Al

Al

Al

kg C kg 4 186 J kg C kg 630 J kg C C

J7.86 kg C

J kg C

− = − + −

+ ° = − ⋅° + ⋅° − °

= ×⋅°

= ⋅°

d i b gd ib g a f b g b g a f0 200 39 3 0 400 0 040 0 3 70

6 29 10800

3

. . . . .

.

P17.71 (a) Pf = 100 kPa , Tf = 400 K

VnRT

P

E nR T

W P V nR T

Q E W

ff

f= =

⋅

×= =

∆ = ∆ = ⋅ =

= − ∆ = − ∆ = − ⋅ = −

= ∆ − = + =

2 00 400

1000 0665 66 5

3 50 3 50 2 00 100 5 82

2 00 100 1 66

5 82 1 66 7 84

.. .

. . . .

. .

. . .

int

int

mol 8.314 J mol K K

10 Pa m L

mol 8.314 J mol K K kJ

mol 8.314 J mol K K kJ

kJ kJ kJ

33b ga f

a f a fb ga fa fb ga f


Chapter 17 487

(b) Tf = 400 K , V VnRT

Pf ii

i= = =

⋅

×= =

2 00 300

100 100 0499 49 93

.. .

mol 8.314 J mol K K

Pa m L3b ga f

P PT

Tf if

i=FHGIKJ =

FHGIKJ =100 133 kPa

400 K300 K

kPa , W PdV= − =z 0 since V = constant

∆ =Eint .5 82 kJ as in part (a) Q E W= ∆ − = − =int . .5 82 5 82 kJ 0 kJ

(c) Pf = 120 kPa , Tf = 300 K , V VPPf i

i

f=FHGIKJ =

FHG

IKJ =49 9 41 6. . L

100 kPa120 kPa

L ,

∆ = ∆ =E nR Tint .3 50 0a f since T = constant

W PdV nRTdVV

nRTV

VnRT

PP

W

Q E W

i if

iiV

V i

fi

f= − = − = −FHGIKJ = −

FHGIKJ

= − ⋅ FHG

IKJ = +

= ∆ − = − = −

z z ln ln

. . ln

int

2 00 8 314 300100

909

0 909 909

mol J mol K K kPa

120 kPa J

J J

a fb ga f

(d) Pf = 120 kPa , γ = =+

= + = =CC

C RC

R RR

P

V

V

V

3 503 50

4 503 50

97

..

.

.

P V PVf f i iγ γ= : so V V

PPf i

i

f=FHGIKJ = F

HGIKJ =

1 7 9

49 9120

43 3/ /

. .γ

L100 kPa

kPa L

T TP V

PV

E nR T

Q

W Q E

f if f

i i=FHGIKJ =

FHG

IKJFHG

IKJ =

∆ = ∆ = ⋅ =

=

= − + ∆ = + = +

30043 3

312

3 50 3 50 2 00 8 314 12 4 722

0

0 722 722

K120 kPa100 kPa

L49.9 L

K

mol J mol K K J

(adiabatic process)

J J

.

. . . . .int

int

a f a fb ga f

*P17.72 (a) PV kr = . So, W PdV kdVV

P V PV

i

f

ri

ff f i i= − = − =

−−z z γ 1

(b) dE dQ dWint = + and dQ = 0 for an adiabatic process.

Therefore, W E nC T TV f i= + = −∆ int d i.

To show consistency between these two equations, consider that γ =C

Cp

V and C C Rp V− = .

Therefore, 1

1γ −=

CRV .

Using this, result found on part (a) becomes

W P V PVCRf f i iV= −d i .

Also, for an ideal gas PVR

nT= so that W nC T TV f i= −d i.


*P17.73 (a) W nC T TV f i= −d i

− = ⋅ −

=

2 500 132

8 314 500

300

J mol J mol K K

K

. T

T

f

f

d i

(b) PV P Vi i f f

γ γ=

PnRT

PP

nRT

Pii

if

f

f

FHGIKJ =FHGIKJ

γ γ

T P T Pi i f fγ γ γ γ1 1− −=

T

P

T

Pi

i

f

f

γ γ γ γ− −

=1 1b g b g

P PT

Tf if

i=FHGIKJ

−γ γ 1b g

P PT

Tf if

i=FHGIKJ = F

HGIKJ =

5 3 3 2 5 2

3 60300500

1 00b gb g

. . atm atm

P17.74 The ball loses energy 12

12

12

0 142 47 2 42 5 29 92 2 2 2mv mvi f− = − =. . . . kg m s J2 2b g a f a f .

The air volume is V = =π 0.037 0 m m ma f a f2 319 4 0 083 4. .

and its quantity is nPVRT

= =×

⋅=

1 013 10

8 3143 47

5.

..

Pa 0.083 4 m

J mol K 293 K mol

3e jb ga f .

The air absorbs energy according to Q nC Tp= ∆ . So

∆ = =⋅

= °TQ

nCP

29 98 314

0 296.

..

J3.47 mol J mol K

C72 b g .

P17.75 (a) nPVRT

= =× ×

⋅=

−1 013 10 5 00 10

8 314 3000 203

5 3. .

..

Pa m

J mol K K mol

3e je jb ga f

(b) T TPPB A

B

A=FHGIKJ =

FHGIKJ =300 900 K

3.001.00

K

T T

V VTT

C B

C AC

A

= =

=FHGIKJ =

FHGIKJ =

900

5 00 15 0

K

L900300

L. .

(c) E nRTA Aint, . .= = ⋅32

32

0 203 8 314 300 mol J mol K Ka fb ga f

FIG. P17.75

=

= = = ⋅ =

760

32

32

0 203 8 314 900 2 28

J

mol J mol K K kJE E nRTB C Bint, int, . . .a fb ga f


Chapter 17 489

(d) P (atm) V(L) T(K) Eint (kJ) A 1.00 5.00 300 0.760 B 3.00 5.00 900 2.28 C 1.00 15.00 900 2.28 (e) For the process AB, lock the piston in place and put the cylinder into an oven at 900 K. For

BC, keep the sample in the oven while gradually letting the gas expand to lift a gradually decreasing load on the piston as far as it can. For CA, carry the cylinder back into the room at 300 K and let the gas cool without touching the piston yourself.

(f) For AB: W = 0 , ∆ = − = − =E E EB Aint int, int, . . .2 28 0 760 1 52a f kJ kJ

Q E W= ∆ − =int .1 52 kJ

For BC: ∆ =Eint 0 , W nRTVVB

C

B= −

FHGIKJln

W

Q E W

= − ⋅ = −

= ∆ − =

0 203 8 314 900 3 00 1 67

1 67

. . ln . .

.int

mol J mol K K kJ

kJ

a fb ga f a f

For CA: ∆ = − = − = −E E EA Cint int, int, . . .0 760 2 28 1 52a f kJ kJ

W P V nR T

Q E W

= − ∆ = − ∆ = − ⋅ − =

= ∆ − = − − = −

0 203 8 314 600 1 01

1 52 1 01

. . .

. .int

mol J mol K K kJ

kJ kJ 2.53 kJ

a fb ga f

(g) We add the amounts of energy for each process to find them for the whole cycle.

Q

W

E

ABCA

ABCA

ABCA

= + + − =

= − + = −

∆ = + + − =

1 52 1 67 2 53 0 656

0 1 67 1 01 0 656

1 52 0 1 52 0

. . . .

. . .

. .int

kJ kJ kJ kJ

kJ kJ kJ

kJ kJb g

*P17.76 (a) The chemical energy input becomes partly work output and partly internal energy. The

energy flow each second is described by

400 60 4004 186 1

3 600465

465 60 405

4052 26 10

3 6001

0 6456

kcal h J s kcal hJ

1 kcal h

s W

W W J s

J s J kg

s h

kg h

= + = FHG

IKJFHG

IKJ =

= − =

=×

FHG

IKJ =

mLt

mt

L

mt

∆

∆

∆ ..

(b) The rate of fat burning is ideally 4009

0 044 4kcal h

kcal g kg h= . . The 0 044 4. kg h of water

produced by metabolism is this fraction of the water needed for cooling: 00

0 068 9 6 89%.0 444 kg h.645 kg h

= =. . . Moral: drink plenty of fruit juice while you exercise.


ANSWERS TO EVEN PROBLEMS P17.2 0 105. °C P17.4 1 78 104. × kg P17.6 88 2. W P17.8 (a) 0 435. cal g C⋅° ; (b) beryllium

P17.10 m c m c T m c T

m c m c m cc w c h w h

c w h w

Al Al

Al Al

+ ++ +

b g

P17.12 12 9. g steam P17.14 0 414. kg P17.16 3 40 10 4. × m P17.18 11 1. W P17.20 (a) −12 0. MJ; (b) +12 0. MJ P17.22 − −nR T T2 1b g P17.24 (a) 12 0. kJ ; (b) −12 0. kJ P17.26 (a) 0 007 65. m3 ; (b) 305 K P17.28 (a) −48 6. mJ ; (b) 16 2. kJ ; (c) 16 2. kJ P17.30 (a) −4PVi i ; (b) 4PVi i ; (c) −9 08. kJ P17.32 (a) 1 300 J ; (b) 100 J; (c) −900 J; (d) −1 400 J P17.34 74 8. J P17.36 (a) 0 719. kJ kg K⋅ ; (b) 0 811. kg ; (c) 233 kJ ; (d) 327 kJ P17.38 (a) 5 66 107. × J ; (b) 1 12. kg

P17.40 (a) 0.118; (b) 2.35; (c) 0, 135 J, +135 J P17.42 227 K P17.44 (a) 28.0 kJ; (b) 46.0 kJ; (c) 10.0 atm, 25.1 atm P17.46 see the solution P17.48 2 22 10 2. × ⋅°− W m C P17.50 67 9. °C P17.52 30 3. kcal h P17.54 364 K P17.56 (a) see the solution; (b) see the solution;

(c) see the solution P17.58 2 27 103. × m P17.60 (a) 16 8. liters ; (b) 0 351. L s

P17.62 (a) +PVi i

2; (b) +1 39. PVi i ; (c) 0

P17.64 5 87 104. × °C P17.66 (a) 2 000 W ; (b) 4 47. °C P17.68 10 2. h P17.70 800 J kg C⋅° P17.72 (a) see the solution; (b) see the solution P17.74 0 296. °C P17.76 (a) 0 645. kg h ; (b) 0 068 9.

energy in thermal processes: the first law of...

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