Solucionario Fundamentos de Física 9na edición Capitulo 11

11Energy in Thermal Processes

CLICKER QUESTIONS

Question J2.01

Description: Introducing “heat” and identifying preconceptions about it.

Question

Which of the following phrases best describes heat?

1. The total energy possessed by a body 2. The fl ow of temperature to or from a body 3. The amount of energy dissipated by friction 4. The total energy fl owing between two bodies at different temperatures 5. The useful work that could be extracted from a body

Commentary

Purpose: To explore your preconceptions about the meaning of the word “heat,” and relate those to the formal physics term and concept.

Discussion: In colloquial usage, the word “heat” is often used to refer to temperature or to the amount of thermal energy stored in a body. In physics, “heat” means thermal energy fl owing into or out of a body. We don’t use the word to refer to the thermal energy possessed by a body. (Unfortunately, many texts use the redundant phrase “heat fl ow.)”

Sometimes people talk about friction converting kinetic energy into heat. This also is inaccurate; it may convert kinetic energy into thermal (as well as vibrational) energy, but that is only heat while it is fl owing between bodies.

Key Points:

• Heat refers to thermal energy fl owing into or out of a body, not thermal energy in general.

• Heat is not the same thing as temperature, though they are closely related.

For Instructors Only

Any time you introduce new physics vocabulary that involves words students already know from other contexts (such as everyday language), you should investigate to fi nd out what connotations those words already have for students. This accomplishes two things: it alerts you to potential misunderstandings and misconceptions, and makes students aware of their own associations so they can be defensive about distin-guishing them from the proper physics meaning.

525

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526 Chapter 11

Question J2.02

Description: Introducing and distinguishing heat capacity and specifi c heat.

Question

Two objects made from the same material have different masses and different initial temperatures as shown. If the bodies are placed in thermal contact, the fi nal equilibrium temperature is most nearly:

M

20° C 60° C

2

1

2M

1. 27°C 2. 33°C 3. 40°C 4. 47°C 5. None of the above 6. Cannot be determined

Commentary

Purpose: To understand heat capacity, distinguish it from specifi c heat, and apply it to thermal equilibration.

Discussion: When the objects are placed in thermal contact, heat will fl ow from the warmer one to the colder. As it does, the warmer one will cool down and the colder one will warm up. In this question, the warmer one is larger and has more material to cool down, so its temperature will not change as rapidly as the smaller one’s will.

The concept of “heat capacity” quantifi es this. If the two objects are made from the same material, they have the same specifi c heat: the amount of heat required to change the temperature of 1 g of the material by 1°C. However, the heat capacity of an object is the heat required to change the temperature of the entire object by 1°C. So, if the objects are made of the same material and object 2 has twice the mass, it must have twice the heat capacity.

This means that for every degree that object 2 cools, the heat transferred will warm object 1 by two degrees. The initial temperature difference is 40°C. Object 1 will warm up by 2�3 of that, and object 2 will cool by 1�3 of it. Thus, the fi nal temperature must be about 47°C: answer (4).

Key Points:

• When two objects are placed in thermal equilibrium, the thermal energy lost by one is gained by the other. It is not true that the temperature lost by one is gained by the other. Energy, not temperature, is conserved.

• Make sure you understand the difference between specifi c heat and heat capacity.


Energy in Thermal Processes 527

• An object’s heat capacity is the heat required to change its temperature by 1°C.

• A material’s specifi c heat is the heat required to change the temperature of 1 g of if by 1°C.

• The heat capacity of an object made of one material is its mass times the material’s specifi c heat.


This question is designed to confront and resolve confusion between heat capacity and specifi c heat. A related concept that you may want to connect to is “molar specifi c heat”, the heat required to change the temperature of one mole of a material by 1°C.

Answer (1) is the sum of the initial temperatures divided by three (the mass in units of M ), a misguided attempt at averaging.

Answer (2) indicates an error translating the conceptual representation to the algebraic: students have taken the factor of two for mass mass into account, but applied it to the wrong side of the equation (a common and well-documented error in forming algebraic representations of verbal statements).

Answer (3)—typically the most common—is the unweighted average of the starting temperatures. Students can arrive at this by simplistically splitting the difference, or by reasoning with the specifi c heat instead of the heat capacity. In either case, the are neglecting the difference in the objects’ masses.

The question does not explicitly state that the material’s specifi c heat is independent of temperature. We are assuming it is. If students do not detect and articulate this ambiguity, you can raise it yourself and ask how a temperature-dependent specifi c heat might affect the answer. (Even if you don’t wish to spend time discussing the question, you can mention it as an aside so that top-end students have something extra to keep engaged with. We fi nd it productive to throw the bright and easily bored an extra bone to chew on now and then.)

Question J2.03

Description: Understanding temperature and heat fl ow in a familiar context, and paying attention to environmental effects.

Question

To have your coffee be as hot as possible when you drink it later, when should you add room temperature cream?

1. As soon as the coffee is served 2. Just before you drink it 3. Either; it makes no difference. 4. It is impossible to determine.

Commentary

Purpose: To explore energy exchanges in everyday situations.

Discussion: When two liquids are mixed, such as the coffee and the cream in this situation, the hot coffee loses some energy and the cream gains the same amount of energy, so that they come to some fi nal temper-ature between the two initial temperatures. The specifi c fi nal temperature depends upon the masses, specifi c heats, and initial temperatures of the coffee and cream.


528 Chapter 11

If the coffee and cream existed in isolation, and no heat could leave the system, it wouldn’t matter when they were mixed. However, the hot coffee—before or after cream is added—is constantly losing heat to the surrounding air. The coffee will take a long time to reach room temperature, but we know that it will eventually occur. The rate at which energy is exchanged depends on the difference in temperature between the coffee and the air: The larger the difference in temperature, the larger the rate of energy exchange. If the cream is added early, the temperature difference between coffee and air is smaller than if it is not, so less heat escapes to the air. The coffee will be hotter if you add the cream right away.

Key Points:

• The rate at which heat fl ows between two objects or substances is proportional to the temperature difference between them.

• When two substances of different temperatures are mixed, the new, combined temperature is somewhere between the initial temperatures.

• Sometimes heat lost to or gained from a system’s environment is signifi cant in analyzing a situation.


Many students will neglect the effect of the environment on the temperature of the coffee. In some ways, they have been taught to neglect the environment. Some students will think that the answer is impossible to determine, because they recognize the importance of the environment but have not been taught how to compute its effect.

The correct answer will seem counterintuitive to some students. They simply will not believe that to keep something hot, you need to cool it down fi rst!

Question J2.04

Description: Reasoning about temperature and thermal equilibrium.

Question

Two identical thermodynamic systems, one at T1 and the other at T

2, are placed in thermal contact. When

they reach thermal equilibrium, what is true about the fi nal temperature?

1. T T Tfinal > +( )12 1 2

2. T T Tfinal = +( )12 1 2

3. T T Tfinal < +( )12 1 2

4. Not enough information

Commentary

Purpose: To develop your ability to reason about temperature and thermal equilibrium.

Discussion: You might think it is impossible to determine, because you are not told enough about the systems, but it turns out that you know all that you need to know. Since the systems are “identical,” it means that as they are exchanging energy by heat, their temperatures are changing by the exact same amount. Therefore, when they are done, their temperatures must also have changed by the exact same amount, which means they end up at the average of their temperatures: answer (2).



Let’s assume that T1 is smaller than T

2. Mathematically, the change in temperature of the cooler

system is 12 1 2 1

12 2 1( ) ( )T T T T T+ − = − . The change in temperature of the warmer system is

T T T T T212 1 2

12 2 1− − = −( ) ( ), the same result.


Some students might choose the correct response for the wrong reasons, so it is useful to fi nd out why students are choosing their answers.

Some of the more thoughtful students will think that it is impossible to determine, either because so little is said about the systems, or because the correct answer just seems too simple and they suspect a catch.

(Actually, we are implicitly assuming that the systems have a constant heat capacity. Since the heat lost by one system must equal the heat gained by the other, if heat capacity varies with temperature, the fi nal temperature is not necessarily midway between T

1 and T

1. Answer (4) is defensible with this reasoning.)

Question J9.01

Description: Understanding Stefan’s law.

Texts: Principles of Physics, Scientists & Engineers

Question

By what factor would the total power from a black body be modifi ed if the surface area were to decrease by a factor of two while the temperature was doubled?

1. 1�16 2. 1�8 3. 1�4 4. 1 5. 4 6. 8 7. None of the above 8. Cannot be determined

Commentary

Purpose: To check your understanding of the primary quantities upon which black-body radiation power depends.

Discussion: Along with conduction and convection, radiation is one of the important energy fl ow processes, and you should be familiar with the primary factors governing the rate at which energy is radiated. According to the Stefan-Boltzmann law, the power radiated from a black body is proportional to the surface area of the body and to the fourth power of the body’s Kelvin temperature. Thus, if the temperature is doubled and the surface area halved, the power radiated would change by 24�2 � 8.


530 Chapter 11

Key Points:

• The power radiated from a black body is proportional to the surface area of the body.

• The power radiated from a black body is proportional to the fourth power of the body’s Kelvin temperature.


Students frequently obtain the inverse of the correct response, obtaining 1�8 rather than 8. In general, it is important to ask students how they arrived at their answers in order to distinguish algebra mistakes from misremembered formulas or erroneous thinking.

It is useful to extend the discussion of this question by contrasting radiation to the other energy fl ow processes. Black body radiation is unique in that the power radiated depends upon the properties of the body itself, not to differences between the body and its environment.

QUICK QUIZZES

1. (a) Water, glass, iron. Because it has the highest specifi c heat ( )4 186 J kg °C⋅ , water has the smallest change in temperature. Glass is next ( )837 J kg °C⋅ , and iron ( )448 J kg °C⋅ is last. (b) Iron, glass, water. For a given temperature increase, the energy transfer by heat is proportional

to the specifi c heat.

2. (b). The slopes are proportional to the reciprocal of the specifi c heat, so larger specifi c heat results in a smaller slope, meaning more energy to achieve a given change in temperature.

3. (c). The blanket acts as a thermal insulator, slowing the transfer of energy by heat from the air into the cube.

4. (b). The rate of energy transfer by conduction through a rod is proportional to the difference in the temperatures of the ends of the rod. When the rods are in parallel, each rod experiences the full difference in the temperatures of the two regions. If the rods are connected in series, neither rod will experience the full temperature difference between the two regions, and hence neither will conduct energy as rapidly as it did in the parallel connection.

5. (a) 4. From Stefan’s law, the power radiated from an object at absolute temperature T is proportional to the surface area of that object. Star A has twice the radius and four times the surface area of star B. (b) 16. From Stefan’s law, the power radiated from an object having surface area A is proportional to the fourth power of the absolute temperature. Thus,

P PA B B4

B= = =σ σAe T AeT( ) ( )2 2 164 4 . (c) 64. When star A has both twice the radius and twice the absolute temperature of star B, the ratio of the radiated powers is

PP

A

B

A A4

B B4

A2

A4

B2

= =( )( )( )

σσ

σ πσ π

A eT

A eT

R T

R

4 1

4 1(( )=

( ) ( )= ( )( ) =

T

R T

R TB4

B B

B2

B4

2 22 2 64

2 4

2 4



ANSWERS TO MULTIPLE CHOICE QUESTIONS

1. From the mechanical equivalent of heat, 1 4 186cal J= . . Therefore,

3 50 10 3 50 104 186

13 3. .

.× = ×( )⎛⎝

⎞⎠cal cal

J

cal== ×1 47 104. J

and (b) is the correct choice for this question.

2. 7 80 10 7 80 101 15 5. .× = ×( )⎛

⎝⎜⎞⎠⎟

J Jcal

4.186 J

CCal

calCal310

186⎛⎝⎜

⎞⎠⎟

= , so (a) is the correct choice.

3. The required energy input is

Q mc T= ( ) = ( ) ⋅( ) −(∆ 5 00 128 327 20 0. .kg J kg °C °C °C)) = ×1 96 105. J

and the correct response is (e).

4. The energy which must be added to the 0°C ice to melt it, leaving liquid at 0°C, is

Q mL f15 52 00 3 33 10 6 66 10= = ( ) ×( ) = ×. . .kg J kg J

Once this is done, there is Q Q Q2 15 5 59 30 10 6 66 10 2 64 10= − = × − × = ×total J J. . . J of energy

still available to raise the temperature of the liquid. The change in temperature this produces is

∆T TQ

mcf= − = = ×( )0

2 64 10

4 1862

5

°CJ

2.00 kgwater

.

J kg °C°C

⋅( ) = 31 5.

so the fi nal temperature is Tf = + =0°C °C °C31 5 31 5. . and the correct choice is (c).

5. The rate of energy transfer by conduction through a wall of area A and thickness L is P = −( )kA T T Lh c , where k is the thermal conductivity of the material making up the wall, while T Th cand are the temperatures on the hotter and cooler sides of the wall, respectively. For the case given, the transfer rate will be

P =⋅ ⋅

⎛⎝

⎞⎠ ( ) −( )

×0 10 48 0

25 14. .

J

s m °Cm

°C °C

4.002

1101 3 10 1 3 10

23 3

−( ) = × = ×m

J s W. .

and the (d) is the correct answer.

6. The power radiated by an object with emissivity e, surface area A, and absolute temperature T,

in a location with absolute ambient temperature T0, is given by P = −( )σ Ae T T404

where σ = × ⋅−5 669 6 10 8. W m K2 4 is a constant. Thus, for the given spherical object

A r=( )4 2π , we have

P = × ⋅( ) ( ) ( )−5 669 6 10 4 2 00 0 450 4088 2. . .W m K m2 4 π K K( ) − ( )⎡⎣ ⎤⎦

4 4298

yielding P = ×2 54 104. W, so (e) is the correct choice.

7. The temperature of the ice must be raised to the melting point, ∆T = +20 0. °C, before it will start to melt. The total energy input required to melt the 2.00-kg of ice is

Q mc T mL f= ( ) + = ( ) ⋅( )ice kg J kg °C °∆ 2 00 2 090 20 0. . CC J kg J( ) + ×⎡⎣ ⎤⎦ = ×3 33 10 7 50 105 5. .

The time the heating element will need to supply this quantity of energy is

∆t

Q= = ××

= ⎛P

7 50 10

10750

5. J

1.00 J ss

1 min

60 s3 ⎝⎝⎞⎠ = 12 5. min

making (d) the correct choice.


532 Chapter 11

8. We use − =Q Qhot cold or − −( ) = −( )m c T T m c T Tx x f x i w w f w i, , to compute the specifi c heat of

the unknown material and fi nd

cm c T T

m T Tx

w w f w i

x f x i

=−( )

− −( ) =( ),

,

.0 400 4 186kg J kg °C °C °C

0.250 kg °C

⋅( ) −( )( ) −

36 0 20 0

36 0 9

. .

. 55 01 82 103

..

°CJ kg °C

( ) = × ⋅

which is a match for the specifi c heat of Beryllium, so (b) is the correct choice.

9. Since less energy was required to produce a 5°C rise in the temperature of the ice than was required to produce a 5°C rise in temperature of an equal mass of water, we conclude that the

specifi c heat of ice c Q m T= ( )( )∆ is less than that of water. Thus, choice (d) is correct.

10. With e eA B= , r rA B= 2 , and T TA B= 2 , the ratio of the power output of A to that of B is

PP

A

B

A A A

B B B

A A

B B

A

B

A e T

A e T

r T

r T

r

r= = =

σσ

ππ

4

4

2 4

2 4

4

4

⎛⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

= ( ) ( ) = ( ) =2 4

2 4 62 2 2 64T

TA

B

making (e) the correct choice.

11. By agitating the coffee inside this sealed, insulated container, the person is raising the internal energy of the coffee, which will result is a rise in the temperature of the coffee. However, doing this for only a few minutes, the temperature rise will be quite small. The correct response to this question is (d).

12. One would like the poker to be capable of absorbing a large amount of energy, but undergo a small rise in temperature. This means it should be made of a material with a high specifi c heat capacity. Also, it is desirable that energy absorbed by the end of the poker in the fi re be conducted to the person holding the other end very slowly. Thus, the material should have a low thermal conductivity. The correct choice is (d).

ANSWERS TO EVEN-NUMBERED CONCEPTUAL QUESTIONS

2. In winter the produce is protected from freezing. The specifi c heat of Earth is so high that soil freezes only to a depth of a few inches in temperate regions. Throughout the year the temperature will stay nearly constant day and night. Factors to be considered are the insulating properties of the soil, the absence of a path for energy to be radiated away from or to the vegetables, and the hindrance of the formation of convection currents in the small, enclosed space.

4. The high thermal capacity 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 produce froze solid. Evaporation of the water keeps the relative humidity high to protect foodstuffs from drying out.

6. Yes, if you know the specifi c heat of zinc and copper, you can determine the relative fraction of each by heating a known weight of pennies to a specifi c initial temperature, say 100° C, then dump them into a known quantity of water, at say 20° C. The equation for conservation of energy will be

m x c x T m cpennies water wa°⋅ + −( )[ ] −( ) =Cu Znc C1 100 tter T −( )20°C

The equilibrium temperature, T, and the masses will be measured. The specifi c heats are known, so the fraction of metal that is copper, x, can be computed.



8. Write m c V cwater water air airC C1 1° °( ) = ( ) ( )ρ , to fi nd

Vm c

c= =

×( )water water

air air

kg J kg

ρ1 0 10 4 1863. ⋅⋅( )

( ) × ⋅( ) = ×°

°

C

kg m J kg Cm

33

1 3 1 0 103 2 10

33

. ..

10. The black car absorbs more of the incoming energy from the Sun than does the white car, making it more likely to cook the egg.

12. Keep them dry. The air pockets in the pad conduct energy slowly. Wet pads absorb some energy in warming up themselves, but the pot would still be hot and the water would quickly conduct a lot of energy to your hand.

PROBLEM SOLUTIONS

11.1 As mass m of water drops from top to bottom of the falls, the gravitational potential energy given up (and hence, the kinetic energy gained) is Q mgh= . If all of this goes into raising the temperature, the rise in temperature will be

∆T

Q

mc

m gh

m c= = =

( )( )water water

2m s m9 80 807

4 18

.

661 89

J kg °C°C

⋅= .

and the fi nal temperature is T T Tf i= + = =∆ 15 0 16 9. .° °CC+1.89°C .

11.2 Q mc T= ( ) = ( ) ⋅( ) −(∆ 1 50 230 150 20 0. .kg J kg °C C °C° )) = × =4 49 10 44 94. .J kJ

11.3 The mass of water involved is

m V= = ⎛⎝

⎞⎠ ×( ) = ×ρ 10 4 00 10 4 00 103 11 14kg

mm k3

3. . gg

(a) Q mc T= ( ) = ×( ) ⋅( )(∆ 4 00 10 4 186 1 0014. .kg J kg C C° ° )) = ×1 67 1018. J

(b) The power input is P = = ×1 000 1 00 109 MW J s. , so,

t

Q= = ×× ×

⎛P

1 67 10

1 00 10

118

9

.

.

J

J s

yr

3.156 10 s7⎝⎝⎜⎞⎠⎟ = 52 9. yr

11.4 The change in temperature of the rod is

∆T

Q

mc= = ×

( )( ) =1

90031 7

.00 10 J

0.350 kg J kg C

4

°. °°C

and the change in the length is

∆ ∆L L T= ( )

= × ( )⎡⎣ ⎤⎦( )− −

α 0

6 124 10 20 0 31 7C cm° °. . CC cm mm( ) = × =−1 52 10 0 1522. .

11.5 ∆T TQ

mcf= − = = ( ) ⋅( )25750

0 168°C

cal

75 g cal g °C.== 60°C

so

Tf = + =25 60 85° ° °C C C


534 Chapter 11

11.6 (a) Q =⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞540

10 4 1863

Calcal

1 Cal

J

1 cal

.⎠⎠⎟

= ×2 3 106. J

(b) The work done lifting her weight mg up one stair of height h is W mgh1 = . Thus, the total work done in climbing N stairs is W Nmgh= , and we have W Nmgh Q= = or

NQ

mgh= = ×

( )( )( )=2 3 10

9 80 0 152

6.

. .

J

55 kg m s m2..8 104× stairs

(c) If only 25% of the energy from the donut goes into mechanical energy, we have

NQ

mgh

Q

mgh= =

⎛⎝⎜

⎞⎠⎟

= ×0 250 25 0 25 2 8 104.. . . stairss stairs( ) = ×7 0 103.

11.7 (a) W KE m fnet kg m s= = −( ) = ( ) ( ) −∆ 1

2

1

275 11 0 02

02 2v v .⎡⎡⎣ ⎤⎦ = × → ×4 54 10 4 5 103 3. .J J

(b) P = = × = × =W

tnet J

5.0 sJ s W

∆4 54 10

9 1 10 9103

2..

(c) If the mechanical energy is 25% of the energy gained from converting food energy, then W Qnet = ( )0 25. ∆ and P = 0 25. ( )∆ ∆Q t, so the food energy conversion rate is

∆∆Q

t= = ⎛

⎝⎞⎠

⎛⎝⎜

⎞⎠⎟

=P0 25

910 1

.

J s

0.25

Cal

4 186 J00 87. Cal s

(d) The excess thermal energy is transported by conduction and convection to the surface of the skin and disposed of through the evaporation of sweat.

11.8 (a) The instantaneous power is P = Fv , where F is the applied force and v is the instantaneous velocity.

(b) From Newton’s second law, F manet = , and the kinematics equation v v= +0 at with v0 0= , the instantaneous power expression given above may be written as

P = = ( ) +( )F ma atv 0 or P = ma t2

(c) at t

= = −−

= =∆∆

v v 0

0

11 0

5 002 20

.

..

m s

sm s2

(d) P = = ( )( ) = ⋅( ) ⋅ma t t2 275 0 2 20 363. .kg m s kg m s2 2 4 tt t= ( ) ⋅363 W s

(e) Maximum instantaneous power occurs when t t= =max .5 00 s, so

Pmax . .= ( )( ) = ×363 5 00 1 82 103J s s J s2

If this corresponds to 25.0% of the rate of using food energy, that rate must be

∆∆Q

t= = × ⎛Pmax

.

.

.0 250

1 82 10

0 250

1

4 186

3 J s Cal

J⎝⎝⎜⎞⎠⎟

= 1 74. Cal s

11.9 The mechanical energy transformed into internal energy of the bullet is Q KE m mi i i= ( ) = ( ) =1

212

12

2 14

2v v . Thus, the change in temperature of the bullet is

∆TQ

mc

m

m ci= = =

( )⋅( ) =

14

2 2300

4 1

v

lead

m s

28 J kg C°1176°C



11.10 The internal energy added to the system equals the gravitational potential energy given up by the 2 falling blocks, or Q PE m ghg b= =∆ 2 . Thus,

∆TQ

m c

m gh

m cw w

b

w w

= = =( )( )2 2 1 50 9 80 3 00. . .kg m s2 m

kg J kg °CC

( )( ) ⋅( ) =0 200 4 186

0 105.

. °

11.11 The quantity of energy transferred from the water-cup combination in a time interval of 1 minute is

Q mc mc T= ( ) + ( )⎡⎣ ⎤⎦( )

= ( )

water cup

kg

∆

0 800 4 186.JJ

kg Ckg

J

kg C⋅⎛⎝⎜

⎞⎠⎟

+ ( )⋅

⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢ ° °

0 200 900.⎤⎤

⎦⎥( ) = ×1 5 5 3 103. .°C J

The rate of energy transfer is

P = = × = =Q

t∆5 3 10

88 883. J

60 s

J

s W

11.12 (a) The mechanical energy converted into internal energy of the block is Q KE mi i= =0 85 0 85 1

22. ( ) . ( )v . The change in temperature of the block will be

∆TQ

mc

m

m ci= =

( )=

( )Cu Cu

m s

87

0 85 0 85 3 0

2 3

12

2 2. . .v

J kg CC

⋅( ) = × −

°°9 9 10 3.

(b) The remaining energy is absorbed by the horizontal surface on which the block slides.

11.13 From ∆ ∆L L T= ( )α 0 , the required increase in temperature is found, using Table 10.1, as

∆ ∆T

L

L= = ×

× ( )( )−

− −αsteel6

m

11 °C0

3

1

3 0 10

10 13

.

yyd

yd

3.0 ft

ft

1 m°C( )

⎛

⎝⎜⎞

⎠⎟⎛⎝⎜

⎞⎠⎟

=1 3 281

23.

The mass of the rail is

mw

g= =

( )( ) ⎛⎝⎜

70 13lb yd yd

9.80 m s

4.448 N

1 lb2

⎞⎞⎠⎟

= ×4 1 102. kg

so the required thermal energy (assuming that c csteel iron= ) is

Q mc T= ( ) = ×( ) ⋅( )(steel kg J kg °C °C∆ 4 1 10 448 232. )) = ×4 2 106. J

11.14 (a) From the relation between compressive stress and strain, F A Y L L= ( )∆ 0 , where Y is Young’s modulus of the material. From the discussion on linear expansion, the strain due to thermal expansion can be written as ( ) ( )∆ ∆L L T0 = α , where α is the coeffi cient of linear

expansion. Thus, the stress becomes F A Y T= ( )[ ]α ∆ .

(b) If the concrete slab has mass m, the thermal energy required to produce a change in temperature ∆T is Q mc T= ( )∆ where c is the specifi c heat of concrete. Using the result from part (a), the absorbed thermal energy required to produce compressive stress F A is

Q mcF A

Y= ⎛

⎝⎞⎠α or Q

mc

Y

F

A= ⎛

⎝⎞⎠α

continued on next page


536 Chapter 11

(c) The mass of the given concrete slab is

m V= = ×( ) ×( )( )−ρ 2 40 10 4 00 10 1 00 13 2. . . .kg m m m3 000 96 0m kg( )⎡⎣ ⎤⎦ = .

(d) If the maximum compressive stress concrete can withstand is F A = ×2 00 107. Pa, the maximum thermal energy this slab can absorb before starting to break up is found, using Table 10.1, to be

Qmc

Y

F

Amaxmax

.

.= ⎛

⎝⎞⎠ =

( ) ⋅( )α

96 0 880

2 1

kg J kg °C

××( ) × ( )( ) ×( ) =− −10 12 10

2 00 10 610 6 1

7

Pa °CPa. .77 106× J

(e) The change in temperature of the slab as it absorbs the thermal energy computed above is

∆TQ

mc= = ×

( ) ⋅( ) = °6 7 10

88079

6. J

96.0 kg J kg °CC

(f ) The rate the slab absorbs solar energy is

P Pabsorbed solar W J= = ×( ) = ×0 5 0 5 1 00 10 5 103 2. . . ss

so the time required to absorb the thermal energy computed in (d) above is

tQ= = ×

×⎛⎝⎜

⎞Pabsorbed

2

J

5 J s

h

3 600 s

6 7 10

10

16.⎠⎠⎟

∼ 4 h

11.15 When thermal equilibrium is reached, the water and aluminum will have a common temperature of Tf = 65 0. °C. Assuming that the water-aluminum system is thermally isolated from the environment, Q Qcold hot= − , so m c T T m c T Tw w f i w f i−( ) = − −( ), ,Al Al Al , or

mm c T T

c T Tw

f i

w f i w

=− −( )

−( ) =− ( )Al Al Al kg,

,

.1 85 9000 65 0

65 0

J kg °C °C 150°C

4 186 J kg °C

⋅( ) −( )⋅( )

.

. °°C °Ckg

−( ) =25 0

0 845.

.

11.16 If N pellets are used, the mass of the lead is Nmpellet . Since the energy lost by the lead must equal the energy absorbed by the water,

Nm c T mc Tpellet lead water∆ ∆( ) = ( )[ ]

or the number of pellets required is

Nm c T

m c Tw w w=

( )

=( )

∆∆pellet lead lead

kg0 500 4 18. 66 25 0 20 0

128

J kg C C C

1.00 10 kg-3

⋅( ) −( )×( )

° ° °. .

J kg C C C⋅( ) −( )=

° ° °200 25 0467

.



11.17 The total energy absorbed by the cup, stirrer, and water equals the energy given up by the silver sample. Thus,

m c m c m c T mc Tc s w w wAl Cu Ag+ +[ ]( ) = [ ]∆ ∆

Solving for the mass of the cup gives

mc

m cT

Tm c m cc

ws w w= ( ) ( ) − −

⎡

⎣⎢

⎤

⎦⎥

1

AlAg Ag

AgCu

∆∆

,

or

mc = ( )( ) −( )−( ) − ( )1

900400 234

87 32

32 2740 387g g (( ) − ( )( )⎡

⎣⎢

⎤⎦⎥ =225 4 186 80g g

11.18 The mass of water is

m Vw w w= = ( )( ) = =ρ 1 00 100 100 0 100. .g cm cm g kg3 3

For each bullet, the energy absorbed by the bullet equals the energy given up by the water, so m c T m c Tb b w w−( ) = −( )20 90°C °C . Solving for the fi nal temperature gives

Tm c m c

m c m cw w b b

w w b b

= ( ) + ( )+

90 20°C °C.

For the silver bullet, m cb b= × = ⋅−5 0 10 2343. kg and J kg °C, giving

Tsilver

°C=

( )( )( ) + ×( )( )−0 100 4 186 90 5 0 10 2343. . 220

0 100 4 186 5 0 10 23489 8

3

°C°C

( )( )( ) + ×( )( )

=−. ..

For the copper bullet, m cb b= × = ⋅−5 0 10 3873. kg and J kg °C, which yields

Tcopper

°C=

( )( )( ) + ×( )( )−0 100 4 186 90 5 0 10 3873. . 220

0 100 4 186 5 0 10 38789 7

3

°C°C

( )( )( ) + ×( )( )

=−. ..

Thus, the copper bullet wins the showdown of the water cups.

11.19 The total energy given up by the copper and the unknown sample equals the total energy absorbed by the calorimeter and water. Hence,

m c T m c T m c m c Tc w w wCu Cu Cu unk unk unk Al∆ ∆ ∆+ = +[ ]( )

Solving for the specifi c heat of the unknown material gives

cm c m c T m c T

m Tc w w w

unkAl Cu Cu Cu

unk unk

=+[ ]( ) −∆ ∆

∆, or

cunk g °C

g J kg °C g= ( )( ) ( ) ⋅( ) +1

70 80100 900 250(( ) ⋅( )⎡⎣ ⎤⎦( ){

− ( )

4 186 10

50

J kg °C °C

g 33 60 1 8 10387 J kg °C °C J kg °C⋅( )( )} = × ⋅.


538 Chapter 11

11.20 The energy absorbed by the water equals the energy given up by the iron and they come to thermal equilibrium at 100°F. Thus, considering cooling 1.00 kg of iron, we have

m c T m c Tw w w∆ ∆( ) = Fe Fe Fe or mc T

c Tww w

=( )

( )1 00. kg Fe Fe∆

∆

giving

mw =( ) ⋅( ) −( )1 00 448 500 100 1 9

5. kg J kg °C °F °F °C °FF

J kg °C °F °F °C °F

( )⋅( ) −( ) ( )

=4 186 100 75 1

1 795

. kkg

11.21 Since the temperature of the water and the steel container is unchanged, and neither substance undergoes a phase change, the internal energy of these materials is constant. Thus, all the energy given up by the copper is absorbed by the aluminum, giving m c T m c TAl Al Al Cu Cu Cu∆ ∆( ) = , or

mc

c

T

TmAl

Cu

Al

Cu

AlCu=

⎛⎝⎜

⎞⎠⎟ ( )

⎡

⎣⎢

⎤

⎦⎥

= ⎛

∆∆

387

900⎝⎝⎞⎠

−−

⎛⎝

⎞⎠ ( ) = ×85

25200 2 6 102° °

° °

C 25 C

C 5.0 Cg . gg kg= 0 26.

11.22 The kinetic energy given up by the car is absorbed as internal energy by the four brake drums (a total mass of 32 kg of iron). Thus, ∆ ∆KE Q m c T= = ( )drums Fe or

∆Tm

m ci= =

( )( )12

2 12

21 500 30

32car

drums Fe

kg m svkg J kg °C

°C( ) ⋅( ) =448

47

11.23 (a) Assuming that the tin-lead-water mixture is thermally isolated from the environment, we have

Q Qcold hot= − or m c T T m c T T m c T Tw w f i w f i f i−( ) = − −( ) − −, ,Sn Sn Sn Pb Pb ,,Pb( ) and since m m mSn Pb metal kg= = = 0 400. and T T Ti i, , .Sn Pb hot °C= = = 60 0 , this yields

Tm c T m c c T

m c m cfw w i w

w w

=+ +( )+

, metal Sn Pb hot

metal SSn Pb

kg J kg °C °C

+( )

=( ) ⋅( )( ) +

c

1 00 4 186 20 0 0 40. . . 00 227 128 60 0

1 00

kg J kg °C J kg °C °C

kg

( ) ⋅ + ⋅( )( ).

.(( ) ⋅( ) + ( ) ⋅ + ⋅4 186 0 400 227 128J kg °C kg J kg °C J kg. °°C( )

yielding Tf = 21 3. °C

(b) If an alloy containing a mass mSn of tin and a mass mPb of lead undergoes a rise in tempera-ture ∆T , the thermal energy absorbed would be Q Q Q= +Sn Pb, or

m m c T m c T m c TSn Pb alloy Sn Sn Pb Pb+( ) ( ) = ( ) + ( )∆ ∆ ∆ giving cm c m c

m malloySn Sn Pb Pb

Sn Pb

=++

If the alloy is a half-and-half mixture, so m mSn Pb= , this reduces to c c calloy Sn Pb= +( ) 2 and yields

calloy

J kg °C J kg °CJ kg °C= ⋅ + ⋅ = ⋅227 128

2178




(c) For a substance forming monatomic molecules, the number of atoms in a mass equal to the molecular weight of that material is Avogadro’s number, NA. Thus, the number of tin atoms in mSn kg g= =0 400 400. of tin with a molecular weight of MSn g mol= 118 7. is

Nm

MNASn

Sn

Sn

g

118.7 g mol=

⎛⎝⎜

⎞⎠⎟

=⎛⎝⎜

⎞⎠⎟

4006 0. 22 10 2 03 1023 24×( ) = ×−mol 1 .

and, for the lead,

Nm

MNAPb

Pb

Pb

g

207.2 g mol=

⎛⎝⎜

⎞⎠⎟

=⎛⎝⎜

⎞⎠⎟

4006 0. 22 10 1 16 1023 24×( ) = ×mol-1 .

(d) We have

N

NSn

Pb

= ××

=2 03 10

1 16 101 75

24

24

.

..

and observe that

c

cSn

Pb

J kg °C

J kg °C= ⋅

⋅=227

1281 77.

from which we conclude that the specific heaat of an element is proportional to thenumbber of atoms per unit mass of that element..

11.24 Assuming that the unknown-water-calorimeter system is thermally isolated from the environment, − =Q Qhot cold, or − − = − + −m c T T m c T T m c T Tx x f i x w w f i w f i( ) ( ) (, , ,Al Al All )

and, since T T Ti w i, , .= = =Al cold °C25 0 , we have c m c m c T T m T Tx w w f x i x f= + − −( )( ) ( ),Al Al cold

or

cx =( ) ⋅( ) + ( )0 285 4 186 0 150 900. .kg J kg °C kg J kgg °C °C

kg °C

⋅( )⎡⎣ ⎤⎦ −( )( ) −

32 0 25 0

0 125 95 0 32

. .

. . ..0°C( )

yielding cx = × ⋅1 18 103. J kg °C .

11.25 Remember that energy must be supplied to melt the ice before its temperature will begin to rise. Then, assuming a thermally isolated system, Q Qcold hot= − , or

m L m c T m c Tf f w fice ice water water°C °C+ −( ) = − −(0 25 )) and

Tm c m L

m m cfw f

w

=( ) −+( ) =water ice

ice water

C25 825° g J kg °C C g J kg( ) ⋅( )( ) − ( ) ×4 186 25 75 3 33 105° .(( )+( ) ⋅( )75 825 4 186g g J kg °C

yielding Tf = 16°C .


540 Chapter 11

11.26 The total energy input required is

Q = ( )+

energy to melt 50 g of ice

energy to warrm 50 g of water to 100 C°( )+ energy to vaporize 5.0 g water

50 g

( )

= ( ) LL c Lf + ( ) −( ) + ( )50 g 100 C 0 C 5.0 gwater ° ° v

Thus,

Q = ( ) ×⎛⎝⎜

⎞⎠⎟

0.050 kg J

kg3 33 105.

+( )0.050 kg J

k4 186

gg °C100°C 0°C

⋅⎛⎝⎜

⎞⎠⎟

−( )

++ ×( ) ×⎛⎝⎜

⎞⎠⎟

−5.0 10 kg J

kg3 62 26 10.

which gives Q = × =4 9 10 494. J kJ .

11.27 The conservation of energy equation for this process is

energy to melt ice energy to warm melted( ) + iice to energy to cool water toT T( ) = ( ) or

m L m c T m c Tf w w wice ice °C °C+ −( ) = −( )0 80

This yields

Tm c m L

m m cw w f

w w

=( ) −

+( )80°C ice

ice

so

T =( ) ⋅( )( ) − ( )1 0 4 186 80 0 100 3 3. . .kg J kg °C °C kg 33 10

1 1 4 18665

5×( )( ) ⋅( ) =

J kg

kg J kg °C°C

.

11.28 The energy required is the following sum of terms:

Q = ( )+

energy to reach melting point

ennergy to melt energy to reach boiling po( ) + iint

e

( )+ nnergy to vaporize energy to reach 110°C( ) + ( ))

Mathematically,

Q m c L c L cf w= − −( )[ ]+ + −( ) + +ice s°C 10°C °C °C0 100 0 v tteam °C °C110 100−( )⎡⎣ ⎤⎦

This yields

Q = ×( ) ⋅⎛⎝⎜

⎞⎠⎟

( ) + ×−40 10 2 090 10 3 333 kgJ

kg °C°C . 110

4 186

5 J

kg

J

kg

⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

+⋅°°C

°CJ

kg

J⎛⎝⎜

⎞⎠⎟

( ) + ×⎛⎝⎜

⎞⎠⎟

+100 2 26 10 2 0106.kkg °C

°C⋅

⎛⎝⎜

⎞⎠⎟

( )⎤⎦⎥10

or

Q = × =1 2 10 0 125. .J MJ



11.29 Assuming all work done against friction is used to melt snow, the energy balance equation is f s m L f⋅ = snow . Since f m gk= ( )µ skier , the distance traveled is

sm L

m gf

k

= ( ) =( ) ×( )snow

skier

kg J kg

µ1 0 3 33 105. .

00.20 kg m sm km

275 9 802 3 10 2 33

( )( ) = × =.

. .

11.30 (a) Observe that the equilibrium temperature will lie between the two extreme temperatures −( )10 0. °C and +30.0°C of the mixed materials. Also, observe that a water-ice change

of phase can be expected in this temperature range, but that neither aluminum nor ethyl alcohol undergoes a change of phase in this temperature range. The thermal energy transfers we can anticipate as the system come to an equilibrium temperature are:

ice at °C to ice at 0°C; ice at 0°C to− 10 0. liquid water at 0°C; water at 0°C towater at ; aluminum at 20.0°C to aluminum atT T ;; ethyl alcohol at 30.0°Cto ethyl alcohol aat .T

(b) Q m (kg) c J kg °C⋅( ) L J kg( ) Tf °C( ) Ti °C( ) Expression

Qice 1.00 2 090 0 −10 0. m cice ice °C0 10 0− −( )[ ].

Qmelt 1.00 3 33 105. × 0 0 m L fice

Qwater 1.00 4 186 T 0 m c Tice water −( )0

QAl 0.500 900 T 20.0 m c TAl Al °C−[ ]20 0.

Qalc 6.00 2 430 T 30.0 m c Talc alc °C−[ ]30 0.

(c) m c m L m c T mfice ice ice ice water Al°C10 0 0.( )+ + −( )+ cc T m c TAl alc alc°C °C−[ ]+ −[ ] =20 0 30 0 0. .

(d) Tm c m c m c

=( ) + ( ) −Al Al alc alc ice ice°C °C20 0 30 0. . 110 0. °C

ice water Al Al alc alc

( ) +⎡⎣ ⎤⎦+ +

L

m c m c m cf

Substituting in numeric values from the table in (b) above gives

T =( )( )( ) + ( )( )( ) −0 500 900 20 0 6 00 2 430 30 0 1 0. . . . . 00 2 090 10 0 3 33 10

1 00 4 186 0

5( ) ( )( ) + ×⎡⎣ ⎤⎦( )( ) +

. .

. .. .500 900 6 00 2 430( )( ) + ( )( )

and yields T = 4 81. °C .


542 Chapter 11

11.31 Assume that all the ice melts. If this yields a result T > 0, the assumption is valid, otherwise the problem must be solved again based on a different premise. If all ice melts, energy conservation Q Qcold hot= −( ) yields

m c L c T m cf w wice ice °C 78°C °C0 0− −( )[ ] + + −( )⎡⎣ ⎤⎦ = − ww m c T+( ) −( )cal Cu °C25

or

Tm c m c m c Lw w f=

+( )( ) − ( ) +⎡⎣ ⎤cal Cu ice ice°C 78°C25 ⎦⎦+( ) +m m c m cw wice cal Cu

With m m m cw w= = =0 560 0 080 0 040. . .kg, g, g,cal ice == ⋅4 186 J kg °C,

J kg °C J kg °C, and Cu icec c L= ⋅ = ⋅ =387 2 090 3, f ..33 105× J kg

this gives

T =( )( ) + ( )( )⎡⎣ ⎤⎦( ) −0 560 4 186 0 080 387 25 0 04. . .°C 00 2 090 3 33 10

0 560 0 040 4

5( ) ( )( ) + ×⎡⎣ ⎤⎦+( )

78°C .

. . 1186 0 080 387( ) + ( ).

or T = 16°C and the assumption that all ice melts is seen to be valid.

11.32 At a rate of 400 kcal h, the excess internal energy that must be eliminated in a half-hour run is

Q = ×⎛⎝

⎞⎠

⎛⎝

⎞⎠ ( )400

4 1860 50010

cal

h

J

1 calh3 .

. == ×8 37 105. J

The mass of water that will be evaporated by this amount of excess energy is

mQ

Levaporated 6

J

2.5 10 J kg= = ×

×=

v

8 37 100 33

5.. kkg

The mass of fat burned (and thus, the mass of water produced at a rate of 1 gram of water per gram of fat burned) is

mproduced

kcal h h

9.0 kcal gram=

( )( )400 0 500.

oof fatg kg= = × −22 22 10 3

so the fraction of water needs provided by burning fat is

fm

m= = × =

−produced

evaporated

kg

kg

22 10

0 330

3

..0066 or 6.6%



11.33 The mass of 2.0 liters of water is m Vw = = ( ) ×( ) =−ρ 10 2 0 10 2 03 3kg m m kg3 3. . .

The energy required to raise the temperature of the water (and pot) up to the boiling point of water is

Q m c m c Tw wboil Al Al= +( )( )∆

or

Qboil kgJ

kgkg

J= ( )⎛⎝⎜

⎞⎠⎟

+ ( )2 0 4 186 0 25 900. .kkg

°C °C J⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥ −( ) = ×100 20 6 9 105.

The time required for the 14 000 Btu h burner to produce this much energy is

tQ

boilboil

Btu h

J

Btu h= = ×

14 000

6 9 10

14 000

15. BBtu

1.054 10 Jh min3×

⎛⎝

⎞⎠ = × =−4 7 10 2 82. .

Once the boiling temperature is reached, the additional energy required to evaporate all of the water is

Q m Lwevaporate kg J kg= = ( ) ×( ) = ×v 2 0 2 26 10 4 56. . . 1106 J

and the time required for the burner to produce this energy is

tQ

boilevaporate

Btu h

J

B= = ×

14 000

4 5 10

14 000

6.

ttu h

Btu

1.054 10 Jh min3

10 31 18

×⎛⎝

⎞⎠ = =.

11.34 In 1 hour, the energy dissipated by the runner is

∆E t= ⋅ = ( )( ) = ×P 300 3 600 1 08 106J s s J.

Ninety percent, or Q = ×( ) = ×0 900 1 08 10 9 72 106 5. . .J J, of this is used to evaporate bodily fl uids. The mass of fl uid evaporated is

mQ

L= = ×

×=

v

9 72 10

100 403

5

6

..

J

2.41 J kg kg

Assuming the fl uid is primarily water, the volume of fl uid evaporated in 1 hour is

Vm= = = ×( )−

ρ0 403

4 03 10104

6..

kg

1 000 kg mm

cm3

333

33

mcm

1403

⎛⎝⎜

⎞⎠⎟

=


544 Chapter 11

11.35 The energy required to melt 50 g of ice is

Q m L f1 0 050 333 16 7= = ( )( ) =ice kg kJ kg kJ. .

The energy needed to warm 50 g of melted ice from 0°C to 100°C is

Q m c Tw2 0 050 4 100= ( ) = ( ) ⋅( )ice kg .186 kJ kg °C °∆ . CC kJ( ) = 20 9.

(a) If 10 g of steam is used, the energy it will give up as it condenses is

Q m Ls3 0 010 2 22 6= = ( )( ) =v . . kg 260 kJ kg kJ

Since Q Q3 1> , all of the ice will melt. However, Q Q Q3 1 2< + , so the fi nal temperature is less than 100°C. From conservation of energy, we fi nd

m L c T m L c Tf w wice steam°C °C+ −( )⎡⎣ ⎤⎦ = + −( )[ ]0 100v

or

Tm L c m L

m mw f=

+ ( )⎡⎣ ⎤⎦ −+(

steam ice

ice steam

°Cv 100

))cw

giving

T =( ) × + ( )( )⎡⎣ ⎤⎦ − ( )10 2 26 10 4 186 100 50 3 336g g. . ××( )

+( )( ) =10

50 10 4 18640

5

g g°C

(b) If only 1.0 g of steam is used, then ′ = =Q m Ls v3 2 26. kJ. The energy 1.0 g of condensed steam can give up as it cools from 100°C to 0°C is

Q m c Ts w431 0 10 4 186 100= ( ) = ×( ) ⋅( )−∆ . .kg kJ kg °C °°C kJ( ) = 0 419.

Since ′ +Q Q3 4 is less than Q1, not all of the 50 g of ice will melt, so the fi nal temperature will

be 0°C . The mass of ice which melts as the steam condenses and the condensate cools to 0°C is

mQ Q

L f

= ′ += +( )

= × −3 4 32 268 0 10

..

0.419 kJ

333 kJ kgkg g= 8 0.

11.36 First, we use the ideal gas law (with V = = × −0 600 0 600 10 3. .L m3 and T = =37 0 310. °C K) to determine the quantity of water vapor in each exhaled breath:

PV nRT nPV

RT= ⇒ = =

×( ) ×( −Pa m33 20 10 0 600 103 3. . ))⋅( )( ) = × −

8 31 3107 45 10 4

..

J mol K Kmol

or

m nM= = ×( )⎛

⎝⎜⎞

⎠⎟−

water molg

mol

1 k7 45 10 18 04. .

gg

gkg310

1 34 10 5⎛

⎝⎜⎞

⎠⎟= × −.

The energy required to vaporize this much water, and hence the energy carried from the body with each breath is

Q mL= = ×( ) ×( ) =−v 1 34 10 2 26 10 30 35 6. . . kg J kg J

The rate of losing energy by exhaling humid air is then

P = ⋅( ) = ⎛⎝

⎞⎠Q

J

breath

brespiration rate 30 3 22 0. .

rreaths

min

min

60 sW⎛

⎝⎞⎠

⎛⎝

⎞⎠ =1

11 1.



11.37 (a) The bullet loses all of its kinetic energy as it is stopped by the ice. Also, thermal energy is transferred from the bullet to the ice as the bullet cools from 30.0°C to the fi nal temperature. The sum of these two quantities of energy equals the energy required to melt part of the ice. The final temperature is 0°C because not all of the ice melts.

(b) The total energy transferred from the bullet to the ice is

Q KE m c m mi i= + − = +bullet lead bullet°C °C0 30 01

22. v bbullet lead °C

kg

c 30 0

3 00 102 40 10

32

.

..

( )

= ×( ) ×− m s J kg °C °C

( )+ ⋅( )( )

⎡

⎣⎢⎢

⎤

⎦⎥⎥

=2

2128 30 0 97 9. . JJ

The mass of ice that melts when this quantity of thermal energy is absorbed is

mQ

L f

= ( ) =×

= × −

water

J

J kgk

97 9

3 33 102 94 105

4.

.. gg

10 g

1 kgg

3⎛⎝⎜

⎞⎠⎟

= 0 294.

11.38 (a) The rate of energy transfer by conduction through a material of area A, thickness L, with thermal conductivity k, and temperatures T Th c> on opposite sides is P = −( )k A T T Lh c . For the given windowpane, this is

P =⋅ ⋅

⎛⎝⎜

⎞⎠⎟ ( )( )⎡⎣ ⎤⎦

−0 84 1 0 2 0

25. . .

J

s m °C m m

°C 00

106 8 10 6 8 103 3°C

0.62 m J s W2

( )×

= × = ×− . .

(b) The total energy lost per day is

E t= ⋅ = ×( ) ×( ) = ×P ∆ 6 8 10 8 64 10 5 9 103 4 8. . .J s s J

11.39 The thermal conductivity of concrete is k = ⋅ ⋅1 3. J s m °C, so the energy transfer rate through the slab is

P =−( )

=⋅ ⋅

⎛⎝⎜

⎞⎠⎟

( ) (kA

T T

Lh c 1 3 5 0

20. .

J

s m °Cm

°C2 ))×

= × = ×−12 mJ s W210

1 1 10 1 1 103 3. .

11.40 (a) The R value of a material is R L k= , where L is its thickness and k is the thermal conductivity. The R values of the three layers covering the core tissues in this body are as follows:

Rskin2m

0.020 W m Km K W= ×

⋅= × ⋅

−−1 0 10

5 0 103

2..

Rfat2 m

0.20 W m K m K W= ×

⋅= × ⋅

−−0 50 10

2 5 102

2..

and

Rtissue2m

0.50 W m Km K W= ×

⋅= × ⋅

−−3 0 10

6 0 102

2..

so the total R value of the three layers taken together is

R Rii

total

2m K

W= + +( ) × ⋅ = ×

=∑ =

−

1

3

5 0 2 5 6 0 10 142. . . 110 0 142− ⋅ = ⋅m K

W

m K

W

2 2

.



546 Chapter 11

(b) The rate of energy transfer by conduction through these three layers with a surface area of A = 2 0. m2 and temperature difference of ∆T = −( ) = =37 0 37 37°C °C K is

P =( ) =

( )( )⋅

= ×A T

R

∆

total

2

2

m K

m K W

2 0 37

0 145 3

.

.. 1102 W

11.41 P = ⎛⎝⎜

⎞⎠⎟kA

T

L

∆, with k =

⋅ ⋅⎛⎝⎜

⎞⎠⎟

0 20010 4 186

..cal

cm °C s

cm

1 m

J

1 c

2

aal

J

s m °C⎛⎝

⎞⎠ =

⋅ ⋅83 7.

Thus, the energy transfer rate is

P =⋅ ⋅

⎛⎝

⎞⎠ ( )( )[ ] ° −

83 7 8 00 50 0200 2

. . .J

s m °Cm m

C 00 0

1 50 10

4 02 10 402

2

8

.

.

.

°C

m

J

sMW

×⎛⎝

⎞⎠

= × =

−

11.42 The total surface area of the house is

A A A A A= + + +side walls end walls gables roof

where

Aside walls2m m m= ( ) × ( )[ ] =2 5 00 10 0 100. .

Aend walls2m m m= ( ) × ( )[ ] =2 5 00 8 00 80 0. . .

A base altitudegables m= ( ) × ( )[ ] = ( ) ×2 2 8 0012

12 . 44 00 37 0 24 1. tan . .m ° m2( )[ ] =

Aroof2m m ° m= ( ) × ( )⎡⎣ ⎤⎦ =2 10 0 4 00 37 0 100. . cos .

Thus,

A = + + + =100 80 0 24 1 100 304m m m m m2 2 2 2 2. .

With an average thickness of 0.210 m, average thermal conductivity of 4 8 10 4. × ⋅− kW m °C, and a 25.0°C difference between inside and outside temperatures, the energy transfer from the house to the outside air each day is

E tk A T

Lt= ( ) =

( )⎡⎣⎢

⎤⎦⎥( ) =

× ⋅( )−

P ∆∆

∆4 8 10 4. kW m °C 3304 25 0

86 400 m °C

0.210 m s

2( )( )⎡

⎣⎢⎢

⎤

⎦⎥⎥( ).

or

E = × = ×1 5 10 1 5 106 9. .kJ J

The volume of gas that must be burned to replace this energy is

VE

heat of combustion= = ×

(J

kcal m3

1 5 10

9 300

9.

))( ) =4 186

39J kcal

m3

11.43 R R R R Ri= = + +Σ outsideair film

shingles sheathing ++ + +R R Rcellulose dry wall insideair film

R = + + + ( ) + +[ ] ⋅0 17 0 87 1 32 3 3 70 0 45 0 17. . . . . .

ft °F2

BBtu h

ft °F

Btu h

2

= ⋅14



11.44 The rate of energy transfer through a compound slab is

P = ( )=A T

RR L ki i

∆ Σ, where

(a) For the thermopane, R R R R R R= + + = +pane trapped air pane pane trapped2 aair.

Thus,

R = ×⋅

⎛⎝⎜

⎞⎠⎟

+ ×− −

20 50 10 1 0 102 2. .m

0.84 W m °C

m

0.00234 W m °C

m °C

W

2

⋅= ⋅

0 44.

and

P =( )( )

⋅=

1 0 23

0 4452

.

.

m °C

m °C WW

2

2

(b) For the 1.0 cm thick pane of glass:

R = ×⋅

= × ⋅−−1 0 10

1 2 102

2..

m

0.84 W m °C

m °C

W

2

so

P =( )( )× ⋅

= × =−

1 0 23

11 9 10 13

..

m °C

.2 10 m °C W W

2

2 2 ..9 kW , 37 times greater

11.45 When the temperature of the junction stabilizes, the energy transfer rate must be the same for each of the rods, or P PCu Al= . The cross-sectional areas of the rods are equal, and if the temperature of the junction is 50°C, the temperature difference is ∆T = 50°C for each rod.

Thus,

P PCu CuCu

AlAl

Al=⎛⎝⎜

⎞⎠⎟

=⎛⎝⎜

⎞⎠⎟

=k AT

Lk A

T

L

∆ ∆,

which gives

Lk

kLAl

Al

CuCu

W m °C

97 W m °C=

⎛⎝⎜

⎞⎠⎟

= ⋅⋅

⎛⎝⎜

⎞⎠

238

3 ⎟⎟ ( ) =15 9 0cm cm.

11.46 The energy transfer rate is

P = = =( ) ×( )∆

∆ ∆Q

t

m L

tfice kg J kg

8.0 h

5 0 3 33 105. .

(( )( ) =3 600

58 s 1 h

W

Thus, P = k A T L( )∆ gives the thermal conductivity as

kL

A T= ⋅

( ) =( ) ×( )

( ) −

−P∆

58 2 0 10

0 80 25

2 W m

m °C2

.

. 55 07 2 10 2

..

°C W m °C

( )= × ⋅−


548 Chapter 11

11.47 The absolute temperature of the sphere is T = 473 K and that of the surroundings is T0 295= K. For a perfect black-body radiator, the emissivity is e = 1. The net power radiated by the sphere is

Pnet

2 4 W

m K

= −( )

= ×⋅

⎛⎝⎜

⎞⎠⎟

−

σ

π

Ae T T404

85 67 10 4 0. .0060 473 22 4 4

m K 95 K( )⎡⎣ ⎤⎦ ( ) − ( )⎡⎣ ⎤⎦

or

Pnet W kW= × =1 1 10 0 112. .

11.48 Since 97.0% of the incident energy is refl ected, the rate of energy absorption from the sunlight is Pabsorbed = × ⋅( ) = ⋅( )3 00 0 0300. % .I A I A , where I is the intensity of the solar radiation.

Pabsorbed2W m m= ×( ) ×( )0 0300 1 40 10 1 00 103 3 2

. . . == ×4 20 107. W

Assuming the sail radiates equally from both sides (so A = ( ) = ×2 1 00 2 00 102 6. .km m2), the rate at which it will radiate energy to a 0 K environment when it has absolute temperature T is

Prad 2 4

W

m K= −( ) = ×

⋅⎛⎝

⎞⎠ ×−σ Ae T 4 80 5 669 6 10 2 00 10. . 66 4 3 40 03 3 40 10m

W

K2

4( )( ) ⋅ = ×⎛⎝

⎞⎠ ⋅−. .T T

At the equilibrium temperature, where P Prad absorbed= , we then have

3 40 10 4 20 103 4 7. .×⎛⎝

⎞⎠ ⋅ = ×− W

KW4 T or T = ×

×⎡⎣⎢

⎤⎦⎥ =−

4 20 10

3 40 10333

7

3

1 4.

.

W

W KK4

11.49 The absolute temperatures of the two stars are T TX Y= =6 000 12 000K and K. Thus, the ratio of their radiated powers is

PP

Y

X

Y

X

Y

X

AeT

AeT

T

T= =

⎛⎝⎜

⎞⎠⎟

= ( ) =σσ

4

4

44

2 16

11.50 The net power radiated is Pnet = −( )σ Ae T T404 , so the temperature of the radiator is

T T

Ae= +⎡

⎣⎢⎤⎦⎥0

4

14Pnet

σ

If the temperature of the surroundings is T0 22 295= =°C K,

T = ( ) +× ⋅( ) ×− −295

25

5 67 10 2 5 104

8 5K

W

W m K m2 4. . 22

K °C

( )( )⎡

⎣⎢⎢

⎤

⎦⎥⎥

= × = ×

0 90

2 1 10 1 8 10

1

4

3 3

.

. .

11.51 At a pressure of 1 atm, water boils at 100°C. Thus, the temperature on the interior of the copper kettle is 100°C and the energy transfer rate through it is

P = ⎛⎝⎜

⎞⎠⎟ =

⋅⎛⎝⎜

⎞⎠⎟ ( )⎡⎣ ⎤⎦kA

T

L

∆397 0 10

2

W

m °C mπ .

1102 100

2 0 10

1 2 10 12

3

4

°C °C

m

W k

−×

⎛⎝⎜

⎞⎠⎟

= × =

−.

. WW



11.52 The mass of the water in the heater is

m V= = ⎛⎝

⎞⎠ ( )⎛

⎝⎜⎞ρ 10 50 0

3 7863 kg

mgal

L

1 gal3 ..

⎠⎠⎟⎛⎝⎜

⎞⎠⎟

=1

101893

m

Lkg

3

The energy required to raise the temperature of the water from 20.0°C to 60.0°C is

Q mc T= ( ) = ( )( ) −( ) =∆ 189 4 186 60 0 20 0kg J kg °C °C. . 33 17 107. × J

The time required for the water heater to transfer this energy is

tQ= = × ⎛

⎝⎜⎞⎠⎟

=P

3 17 10 11 83

7..

J

4 800 J s

h

3 600 s h

11.53 The energy conservation equation is

m c m L m m c m cf w wPb Pb ice ice cup C°C °C98 12−( ) = + +( ) + uu °C °C⎡⎣ ⎤⎦ −( )12 0

This gives

m Pb

J

kg °C°C kg128 86 0 040 3 33 10

⋅⎛⎝⎜

⎞⎠⎟

( ) = ( ) ×. . 55

0 24

J kg

kg

( )+ .(( ) ⋅( ) + ( ) ⋅( )⎡⎣ ⎤⎦4 186 0 100 3J kg °C kg 57 J kg °C. 112°C( )

or m Pb kg= 2 3. .

11.54 The energy needed is

Q mc T V c T= ( ) = ( ) ( )

= ⎛⎝

⎞⎠ ( )⎡

⎣⎢

∆ ∆ρ

10 1 003 kg

mm3

3. ⎤⎤⎦⎥( )( ) = ×4 186 40 0 1 67 108J kg °C J. .

The power input is P = ( )( ) = ×550 6 00 3 30 103 W m m J s2 2. . , so the time required is

tQ= = ×

×⎛⎝⎜

⎞⎠⎟

=P

1 67 10

3 30 10

11

8

3

.

.

J

J s

h

3 600 s44 1. h

11.55 The conservation of energy equation is

m c m c T m c Tw w +( ) −( ) = −( )cup glass Cu Cu°C °C27 90

This gives

Tm c m c m c

m cw w

w w

=( ) + +( )( )

+Cu Cu cup glass°C °C90 27

mm c m ccup glass Cu Cu+

or

T =( )( )( ) + ( )( ) + ( )0 200 387 90 0 400 4 186 0 300 8. . .°C 337 27

0 400 4 186 0 300 837 0

( )⎡⎣ ⎤⎦( )( )( ) + ( )( ) +

°C

. . ..200 38729

( )( ) = °C


550 Chapter 11

11.56 (a) The energy delivered to the heating element (a resistor) is transferred to the liquid nitrogen, causing part of it to vaporize in a liquid-to-gas phase transition. The total energy delivered to the element equals the product of the power and the time interval of 4.0 h.

(b) The mass of nitrogen vaporized in a 4.0 h period is

mQ

L

t

Lf f

= =⋅( ) =

( )( )( )×

P ∆ 25 4 0 3

2 01

J s h s h.

.

600

1101 85 J kg

kg= .

11.57 Assuming the aluminum-water-calorimeter system is thermally isolated from the environment, Q Qcold hot= − , or

m c T T m c T T m c Tf i w w f i w fAl Al Al cal cal−( ) = − −( ) − −, , TTi ,cal( )

Since Tf = 66 3. °C and T Ti i w, , .cal °C= = 70 0 , this gives

cm c m c T T

m T T

w w i w f

f i

Al

cal cal

Al Al

=+( ) −( )

−( ),

,

or

cAl

kgJ

kg °Ckg

J

=( )

⋅⎛⎝⎜

⎞⎠⎟

+ ( )0 400 4 186 0 040 630. .kkg °C

°C

kg

⋅⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥ −( )

( )

70 0 66 3

0 200 66

. .

. .. ..

3 27 08 00 102

−( ) = ×⋅°C

J

kg °C

The variation between this result and the value from Table 11.1 is

% %=⎛⎝⎜

⎞⎠⎟

× = −variation

accepted value100

800 9000

900100 11 1

J kg °C

J kg °C

⋅⋅

⎛⎝⎜

⎞⎠⎟

× =% . %

which is within the 15% tolerance.

11.58 (a) With a body temperature of T = + =37 273 310°C K and surroundings at temperature T0 24 273 297= + =°C K, the rate of energy transfer by radiation is

Prad

2 4

W

m K

= −( )= ×

⋅⎛⎝

⎞⎠

−

σ Ae T T404

85 669 6 10 2 0. . mm K K W2( )( ) ( ) − ( )⎡⎣ ⎤⎦ = ×0 97 310 297 1 6 104 4 2. .

(b) The rate of energy transfer by evaporation of sweat is

Psweatsweat

3 kg kJ= = =

( ) ×Q

t

mL

t∆ ∆v,

. .0 40 2 43 10 kkg J kJ

3 600 s W

( )( )= ×

102 7 10

32.

(c) The rate of energy transfer by evaporation from the lungs is

Plungs

kJ

h

h

3 600 s

J

1 kJ= ⎛

⎝⎞⎠

⎛⎝⎜

⎞⎠⎟

⎛⎝

381 103

⎜⎜⎞⎠⎟

= 11 W

(d) The excess thermal energy that must be dissipated is

P Pexcess metabolic

kJ

h= = ×⎛

⎝⎞⎠0 80 0 80 2 50 103. . .

11 105 6 10

32h

3 600 s

J

1 kJW

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

= ×.

so the rate energy must be transferred by conduction and convection is

P P P P Pc c& . .= − + +( ) = − −excess rad sweat lungs 5 6 1 6 2.. . .7 11 10 1 2 102 2−( ) × = ×W W



11.59 The rate at which energy must be added to the water is

P = =⎛⎝⎜

⎞⎠⎟

= ⎛⎝⎜

⎞⎠⎟

∆∆

∆∆

Q

t

m

tLv 0 500

1.

kg

min

min

600 s

J

kg⎛⎝⎜

⎞⎠⎟

⎡⎣⎢

⎤⎦⎥

×⎛⎝⎜

⎞⎠⎟

= ×2 26 10 1 88 106 4. . W

From P = −k A T L( 100°C) , the temperature of the bottom surface is

TL

k A= + ⋅ = +

×( ) × −

100 1001 88 10 0 500 104 2

°C °C W P . . mm

238 W m °C m°C

( )⋅( ) ( )⎡⎣ ⎤⎦

=π 0 120

1092

.

11.60 The energy added to the air in one hour is

Q t= ( ) = ( )⎡⎣ ⎤⎦( ) = ×Ptotal W s 10 200 3 600 7 20 106. JJ

and the mass of air in the room is

m V= = ( ) ( )( )( )[ ] =ρ 1 3 6 0 15 0 3 0 3 5. . . . .kg m m m m3 ×× 102 kg

The change in temperature is

∆TQ

mc= = ×

×( ) ⋅( ) =7 2 10

3 5 10 83725

6

2

.

.

J

kg J kg °C°°C

giving T T T= + = + =0 20 25 45∆ °C °C °C .

11.61 In the steady state, P PAu Ag= , or

k AT

Lk A

T

LAu Ag

°C °C80 0 30 0. .−⎛⎝

⎞⎠ = −⎛

⎝⎞⎠

This gives

Tk k

k k=

( ) + ( )+

= (Au Ag

Au Ag

°C °C °C80 0 30 0 314 80 0. . . )) + ( )+

=427 30 0

314 42751 2

..

°C°C

11.62 (a) The rate work is done against friction is

P = ⋅ = ( )( ) = × =f v 50 40 2 0 10 2 03 N m s J s kW. .

(b) In a time interval of 10 s, the energy added to the 10-kg of iron is

Q t= ⋅ = ×( )( ) = ×P 2 0 10 10 2 0 103 4. . J s s J

and the change in temperature is

∆TQ

mc= = ×

( ) ⋅( ) =2 0 10

4484 5

4..

J

10 kg J kg °C°C


552 Chapter 11

11.63 (a) The energy required to raise the temperature of the brakes to the melting point at 660°C is

Q mc T= ( ) = ( ) ⋅( ) −( ) =∆ 6 0 900 660 20 3. .kg J kg °C °C °C 446 106× J

The internal energy added to the brakes on each stop is

Q KE m i12 21

2

1

21 500 25 4 69 1= = = ( )( ) = ×∆ car kg m sv . 005 J

The number of stops before reaching the melting point is

NQ

Q= = ×

×=

1

6

5

3 46 10

4 69 107

.

.

J

Jstops

(b) This calculation assumes no energy loss to the surroundings and that all internal energy generated stays with the brakes. Neither of these will be true in a realistic case.

11.64 When liquids 1 and 2 are mixed, the conservation of energy equation is

mc mc1 217 10 20 17°C °C °C °C−( ) = −( ), or c c2 1

7

3= ⎛

⎝⎞⎠

When liquids 2 and 3 are mixed, energy conservation yields

mc mc3 230 28 28 20°C °C °C °C−( ) = −( ), or c c c3 2 1428

3= = ⎛

⎝⎞⎠

Then, mixing liquids 1 and 3 will give mc T mc T1 310 30−( ) = −( )°C °C

or

Tc c

c c=

( ) + ( )+

=+ ( )( )

+1 3

1 3

10 30 10 28 3 30

1

°C °C °C °C

228 328( ) = °C

11.65 (a) The internal energy ∆Q added to the volume ∆V of liquid that fl ows through the calorimeter in time ∆t is ∆ ∆ ∆ ∆ ∆Q m c T V c T= =( ) ( ) ( ) ( )ρ . Thus, the rate of adding energy is

∆∆

∆ ∆∆

Q

tc T

V

t= ( )⎛

⎝⎜⎞⎠⎟

ρ

where ∆ ∆V t is the fl ow rate through the calorimeter.

(b) From the result of part (a), the specifi c heat is

cQ t

T V t= ( )( ) = ( )( )

∆ ∆∆ ∆ ∆ρ

40

0 72 5 8 3

J s

g cm °C3. . .55

2 710

23

cm s

J

g °C

g

1 kg

3( )

=⋅

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

=. .77 103× ⋅ J kg °C



11.66 (a) The surface area of the stove is A A A r rhstove ends cylindricalside

= + = ( ) + ( )2 22π π , or

Astove m m m= ( ) + ( )( ) =2 0 200 2 0 200 0 500 02π π. . . .8880 m2

The temperature of the stove is Ts = − = =59 400 32 0 204 477( . )°F °F °C K while that of

the air in the room is Tr = − = =59 70 0 32 0 21 1 294( . . ) .°F °F °C K. If the emissivity of the

stove is e = 0 920. , the net power radiated to the room is

P = −( )= × ⋅( )−

σ A e T Ts rstove

2 4 W m K

4 4

85 67 10 0 880. . m K K2( )( ) ( ) − ( )⎡⎣ ⎤⎦0 920 477 2944 4

.

or

P = ×2 03 103. W

(b) The total surface area of the walls and ceiling of the room is

A A A= + = ( )( )[ ] +4 4 8 00 25 0 25wall ceiling ft ft. . .00 1 43 102 3ft ft2( ) = ×.

If the temperature of the room is constant, the power lost by conduction through the walls and ceiling must equal the power radiated by the stove. Thus, from thermal conduction equation, P = −( )A T T Rh c iΣ , the net R value needed in the walls and ceiling is

ΣRA T T

ih c=

−( )=

×( ) −( )P

1 43 10 70 0 32 03. . . ft °F °F2

22.03 10 J s

J

Btu

h

3 600 s3×⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠

1 054

1

1⎟⎟

or

ΣRi = ⋅ ⋅7 84. ft °F h Btu2

11.67 A volume of 1.0 L of water has a mass of m V= = ( ) ×( ) =−ρ 10 1 0 10 1 03 3kg m m kg3 3. . .

The energy required to raise the temperature of the water to 100°C and then completely evaporate it is Q mc T mL= ( ) +∆ v, or

Q = ( ) ⋅( ) −( ) + (1 0 4 186 100 20 1 0. .kg J kg °C °C °C kg)) ×( ) = ×2 26 10 2 59 106. .6 J kg J

The power input to the water from the solar cooker is

P = ( ) = ( )( ) ( )efficiency IA 0 50 600

0 50

4

2

..

W m m2 π⎡⎡

⎣⎢⎢

⎤

⎦⎥⎥

= 59 W

so the time required to evaporate the water is

tQ= = × = ×( )⎛

⎝P2 59 10

594 4 10

64.

. J

J s s

1 h

3 600 s⎜⎜⎞⎠⎟

= 12 h


554 Chapter 11

11.68 (a) From the thermal conductivity equation, P = −( )⎡⎣ ⎤⎦k A T T Lh c , the total energy lost by conduction through the insulation during the 24-h period will be

Qk A

L= ( ) + ( ) = −( ) +P P1 212 0 12 0 37 0 23 0. . . . h h °C °C 337 0 16 0 12 0. . .°C °C h−( )⎡⎣ ⎤⎦( )

or

Q =⋅( )( )

+0 012 0 0 490

0 095 014 0 21

. .

.. .

J s m°C m

m°C

2

00 12 03 600

9 36 104°C hs

1 hJ[ ]( )⎛

⎝⎞⎠ = ×. .

The mass of molten wax which will give off this much energy as it solidifi es (all at 37°C) is

mQ

L f

= = ××

=9 36 100 457

4..

J

205 10 J kgkg3

(b) If the test samples and the inner surface of the insulation is preheated to 37.0°C during the assembly of the box, nothing undergoes a temperature change during the test period. Thus, the masses of the samples and insulation do not enter into the calculation. Only the duration of the test, inside and outside temperatures, along with the surface area, thickness, and thermal conductivity of the insulation need to be known.

11.69 The energy m kilograms of steam give up as it (i) cools to the boiling point of 100°C, (ii) condenses into a liquid, and (iii) cools on down to the fi nal temperature of 50.0°C is

Q mc T mL mc T

m

m = ( ) + + ( )

=

steam liquidwater

2.0

∆ ∆1 2v

11 10 J kg °C °C °C J kg3× ⋅( ) −( ) + ×⎡⎣ 130 100 2 26 106.

J kg °C °C+ ⋅( ) −4 186 100 500 0

2 53 106

.

.

°C

J kg

( )⎤⎦= ×( )m

The energy needed to raise the temperature of the 200-g of original water and the 100-g glass container from 20.0°C to 50.0°C is

Q m c m c Tw w g gneeded kg J kg °C= +( ) = ( ) ⋅∆ 0 200 4 186. (( ) + ( ) ⋅( )⎡⎣ ⎤⎦( )

=

0 100 837 30 0

2 76

. .

.

kg J kg °C °C

×× 104 J

Equating the energy available from the steam to the energy required gives

m 2 53 10 2 76 106 4. .×( ) = ×J kg J or m = ××

= =2 76 10

2 53 100 010 9 10 9

4

6

.

.. .

J

J kgkg g



11.70 We approximate the latent heat of vaporization of water on the skin (at 37°C) by asking how much energy would be needed to raise the temperature of 1.0 kg of water to the boiling point and evaporate it. The answer is

L c T Lv v37 100 4 186 100°C

water°C J kg °C °≈ ( ) + = ⋅( )∆ CC °C J kg−( ) + ×37 2 26 106.

or

Lv37 62 5 10°C J kg≈ ×.

Assuming that you are approximately 2.0 m tall and 0.30 m wide, you will cover an area of A = ( )( ) =2 0 0 30 0 60. . .m m m2 of the beach, and the energy you receive from the sunlight in one hour is

Q IA t= ( ) = ( )( )( ) = ×∆ 1 000 0 60 3 600 2 2 10W m m s2 2. . 66 J

The quantity of water this much energy could evaporate from your body is

mQ

L= ≈ ×

×=

v37

6

6

2 2 10

2 5 100 9°C

J

J kg kg

.

..

The volume of this quantity of water is

Vm= = ≈ =−

ρ0 9. kg

10 kg m10 m 1 L3 3

3 3

Thus, you will need to drink almost a liter of water each hour to stay hydrated. Note, of course, that any perspiration that drips off your body does not contribute to the cooling process, so drink up!

11.71 During the fi rst 50 minutes, the energy input is used converting m kilograms of ice at 0°C into liquid water at 0°C. The energy required is Q mL mf1

53 33 10= = ×( ). J kg , so the constant power input must be

P = ( ) =×( )Q

t

m1

1

53 33 10

50∆. J kg

min

During the last 10 minutes, the same constant power input raises the temperature of water having a total mass of m +( )10 kg by 2.0°C. The power input needed to do this is

P = ( ) =+( ) ( )

( ) =+( )Q

t

m c T

t

m2

2 2

10 10 4 186

∆∆

∆ kg kg J kg °C °C

10 min

⋅( )( )2 0.

Since the power input is the same in the two periods, we have

m m3 33 10

50

10 4 1865. ×( )

=+( ) ⋅(J kg

min

kg J kg °C))( )2 0. °C

10 min

which simplifi es to 8 0 10.( ) = +m m kg, or

m = =101 4

kg

7.0kg.


556 Chapter 11

11.72 (a) First, energy must be removed from the liquid water to cool it to 0°C. Next, energy must be removed from the water at 0°C to freeze it, which corresponds to a liquid-to-solid phase transition. Finally, once all the water has frozen, additional energy must be removed from the ice to cool it from 0°C to –8.00°C.

(b) The total energy that must be removed is

Q Q Q Q= + +cool waterto 0°C

freezeat 0°C

cool icetto 8.00°C

ice°C °C−

= − + + −m c T m L m c Tw w i w f w f0 0

or

Q = ×( ) ⋅⎛⎝⎜

⎞⎠⎟

− +−75 0 10 4 186 20 0 3 333. . .kgJ

kg °C°C ×× +

⋅⎛⎝⎜

⎞⎠⎟

−⎡

⎣⎢

⎤

⎦⎥

=

10 2 090 8 00

3 2

5 J

kg

J

kg °C°C.

. 55 10 32 54× =J kJ.

11.73 (a) In steady state, the energy transfer rate is the same for each of the rods, or

P PAl Fe= . Thus, k AT

Lk A

T

LAl Fe

°C °C100 0−⎛⎝

⎞⎠ = −⎛

⎝⎞⎠

giving

Tk

k k=

+⎛⎝⎜

⎞⎠⎟

( ) =+

⎛⎝

⎞⎠

Al

Al Fe

°C100238

238 79 510

.00 75 0°C °C( ) = .

(b) If L A= =15 5 0cm and cm2. , the energy conducted in 30 min is

Q t= ⋅ =⋅

⎛⎝⎜

⎞⎠⎟ ×( ) −−PAl

2 W

m °C m

°C238 5 0 10

100 74.55 0

0 151 800

3 6 104

.

.

.

°C

m s

J

⎛⎝⎜

⎞⎠⎟

⎡⎣⎢

⎤⎦⎥( )

= × = 336 kJ


Science

Solucionario Fundamentos de Física 9na edición Capitulo 11