Solucionario Fundamentos de Física 9na edición Capitulo 11

11
Energy 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

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:
2
1
2M
M
20° C 60° C
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 23 of that, and object 2 will
cool by 13 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 T2, are placed in thermal contact. When
they reach thermal equilibrium, what is true about the fi nal temperature?
1. Tfinal 1 (T + T )
2 1 2
2. T = 1 ( T + T )
final2 1 2
3. T 1 ( T + T )
final2 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 T2. Mathematically, the change in temperature of the cooler
system is 1
(T + T ) − T = 2 (T2 − T1). The change in temperature of the warmer system is
2 1 2 1 1
1
2 2 1 − ( − ) = ( − ), the same result.
T 1
T T T T 2
2 1 2
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 T1 and T1. 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. 116
2. 18
3. 14
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 242 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 18 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 P A B B
=σ Ae (2T )4 = 24 (σ AeT 4
) = 16 . (c) 64. When star A has both twice the radius and twice
B the absolute temperature of star B, the ratio of the radiated powers is
P
P
A
B
4
( )( )
= A A
= 4
2 B B
4 2
1
4
( A
2 ) ( 2
) ) = = ( )( ) =
4 1( A
R T
R T B
( )
B
σ
σ
σ π
σ π
A eT
A eT
R T
R
T
4
B B
2
B
4
B
2 2 64
2 4
2 4

ANSWERS TO MULTIPLE CHOICE QUESTIONS
1. From the mechanical equivalent of heat, 1 cal = 4.186 J. Therefore,
. × = × ( )⎛⎝
3 50 10 3 50 10
4 186
1
. 3 . 3
⎞⎠
cal cal
J
cal
= 1.47 × 104 J
and (b) is the correct choice for this question.
5 5 1 1 . × = ( . × )⎛
2. 7 80 10 7 80 10
⎝ ⎜
⎞
⎠ ⎟
J J
cal
4.186 J
Cal
cal
186
Cal 103
⎛
⎝ ⎜
⎞
⎠ ⎟
= , so (a) is the correct choice.
3. The required energy input is
Q = mc(ΔT ) = (5.00 kg)(128 J kg ⋅°C)(327°C − 20.0°C) = 1 96 × 10. 5 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
= = (2.00 kg)(3.33 × 105 J kg) = 6.66 × 105 J
Q mLf
1
= − = 9 30 × 105 − 6 66 × 105 = 2 64 × 105 total . J . J . J of energy
Once this is done, there is Q Q Q 2 1
still available to raise the temperature of the liquid. The change in temperature this produces is
Q
mc f = − = = ×
ΔT T
2 64 10
( ) 0
4 186
2
5
°C
J
2.00 kg water
.
J kg °C
°C
( ⋅ ) = 31 5 .
so the fi nal temperature is Tf= 0°C + 31.5°C = 31.5°C 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 ) L h c , where k is the thermal conductivity of the material making up the wall, while
T T h c and are the temperatures on the hotter and cooler sides of the wall, respectively. For the case
given, the transfer rate will be
P =
J
⋅ ⋅
⎛⎝
⎞⎠
( ) ( − )
( − )= × = ×
×
0 10 48 0
25 14
. .
s m °C
m
°C °C
4.00
2
10
3 3
1 3 10 1 3 10 2
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 T, is given by P =σ Ae(T 4 − T 4
)
00
where σ = 5.669 6 × 10−8 W m2 ⋅K4 is a constant. Thus, for the given spherical object
(A = 4π r2 ), we have
P = (5 669 6 ×10−8 ⋅ )4 (2 00 )2 (0 450) 408 . . . W m K m 2 4 π K K ( ) − ( ) ⎡⎣
⎤⎦
4 4 298
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, ΔT = +20 0 . °C, before it will start
to melt. The total energy input required to melt the 2.00-kg of ice is
= ( ) + = ( ) ( ⋅ ) ice kg J kg °C ° Δ 2 00 2 090 20 0 . . C J kg J ( ) + × ⎡⎣
Q mc T mLf
⎤⎦
3.33 105 = 7.50 × 105
The time the heating element will need to supply this quantity of energy is
Δt
Q = = ×
×
= ⎛
P
7 50 10
10
750
. 5 J
1.00 J s
s
1 min
3 ⎝ 60 s
⎞⎠
= 12 5. min
making (d) the correct choice.

532 Chapter 11
8. We use −Q = Q hot cold or −m c (T − T ) = m c (T − T ) x x f x,i w w f w,i to compute the specifi c heat of
the unknown material and fi nd
c
m c T T
m T T x
w w f wi
x f xi
=
( − )
, ( )
= − ( − ) ,
( ⋅ )( − )
( ) −
0.400 kg 4 186 J kg °C °C °C
J kg °C ( ) = × ⋅
0.250 kg °C
36 0 20 0
36 0 9
. .
. 5 0
1 82 103
.
.
°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(ΔT )) is less than that of water. Thus, choice (d) is correct.
10. With e e A B = , r r A B = 2 , and T T A B = 2 , the ratio of the power output of A to that of B is
P
P
A
B
4
A A A
B B B
A A
B B
A
B
A e T
A e T
r T
r T
r
r
σ
σ
π
π
= = =
4
2 4
2 4
4
4
⎛
⎝ ⎜
⎞
⎠ ⎟
⎛
⎝ ⎜
⎞
⎠ ⎟
= ( ) ( ) = ( ) =
2 4
2 4 6 2 2 2 64
T
T
A
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 xc x T m c pennies water wa [ ⋅ + ( − ) ]( ° − ) = Cu Zn 1 c 100 C ter (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 mwatercwater (1°C) = (ρairV )cair (1°C), to fi nd
V
( × ) water water
m c
= =
c
air air
kg J kg
ρ
1.0 103 (4 186 ⋅⋅ °
C
)
( )( ×
× ⋅ °
)= m 3
kg m J kg C
3
1 3 1 0 10
3 2 10 3
3
. .
.
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
ΔT
= = = (
Q
mc
mgh
mc
)( )
water water
9 .
80 m s2 807 m
4 18
6
1 89
J kg °C
°C
⋅
= .
and the fi nal temperature is T T T f i = + Δ = 15.0°C+1.89°C = 16.9°C .
11.2 Q = mc(ΔT ) = (1.50 kg)(230 J kg ⋅°C)(150°C − 20.0°C) = 4.49 × 104 J = 44.9 kJ
11.3 The mass of water involved is
ρ 103 kg (4 00 × 1011 ) = 4 00 × 1014
m V = =⎛⎝
⎞⎠
m
. 3 . g
m k 3
(a) Q = mc(ΔT ) = (4.00 × 1014 kg)(4 186 J kg ⋅°C)(1.00°C) = 1.67 × 1018 J
(b) The power input is P = 1 000 MW= 1.00 ×109 J s , so,
t
Q = = ×
18 ⎛
1
9
× ×
P
1 .
67 10
1 .
00 10
J
J s
yr
3.156 10 s 7 ⎝ ⎜
⎞
⎠ ⎟= 52 9 . yr
11.4 The change in temperature of the rod is
ΔT
= = ×
Q
mc
4
1
( )( ) = 900
31 7
.00 10 J
°
0.350 kg J kg C
. °C
and the change in the length is
ΔL = L (ΔT )
= × ( ) ⎡⎣
− − ( )
⎤⎦
α 0
6 1 24 10 °C 20.0 cm (31.7°C) = 1.52 × 10−2 cm = 0.152 mm
Q
mc f = − = =(
11.5 ΔT T
750
0 168
25
°C
)( ⋅ ) cal
75 g . cal g °C
= 60°C
so
Tf= 25°C + 60°C = 85°C

534 Chapter 11
11.6 (a) Q =
⎛
103 4 186
⎞ 540
⎝ ⎜
⎞
⎠ ⎟
⎛
⎝ ⎜
Cal
cal
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 mgh 1 = . Thus, the total
work done in climbing N stairs is W = Nmgh, and we have W = Nmgh = Q or
N
= = ×
Q
mgh
2 3 10
( )( = 9 80 )( 0 15
) 2
. 6
J
. .
55 kg m s2 m .8 × 104 stairs
(c) If only 25% of the energy from the donut goes into mechanical energy, we have
N
0 25 0 25 2 8 104 .
0 25
Q
= × mgh
Q
mgh
= =
⎛
. . ( . stairs) = 7.0 × 103 stairs
⎝ ⎜
⎞
⎠ ⎟
net = Δ = ( − ) = ( kg) ( m s) − 1
2 2 v v . ⎡⎣
11.7 (a) W KE mf
2
1
2
2 75 11 0 0
0
⎤⎦
= 4.54 × 103 J → 4.5 × 103 J
W
× J
(b) P= net = = × = t
5.0 s
J s W
Δ
4 54 10
9 1 10 910
3
. 2
.
(c) If the mechanical energy is 25% of the energy gained from converting food energy, then
W Q net = 0.25(Δ ) and P = 0.25 (ΔQ) Δt, so the food energy conversion rate is
Δ
Q
t
Δ
P =
0 25
= =⎛⎝
⎞⎠
⎛
⎝ ⎜
⎞
⎠ ⎟
910 1
.
J s
0.25
Cal
4 186 J
0.87 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 ma net = , and the kinematics equation v = v + 0 at
with v0 = 0, the instantaneous power expression given above may be written as
P = Fv = (ma)(0 + at ) or P = ma2t
(c) a
= = −
Δ v v 0
= 11 0
=
Δ
t t
−
0
5 00
2 20
.
.
.
m s
s
m s2
(d) P = ma2t = ( )( ) t = ( ⋅ )⋅ 2 75.0 kg 2.20 m s2 363 kg m2 s4 t = (363 W s)⋅ t
(e) Maximum instantaneous power occurs when t = t = max 5.00 s, so
Pmax = (363 J s2 )(5.00 s) = 1.82 × 103 J s
If this corresponds to 25.0% of the rate of using food energy, that rate must be
Δ
Δ
Q
t
= P = × ⎛ max
.
1 .
82 10
0 250 .
3 J s Cal
0 250
1
4 186
J ⎝ ⎜
⎞
⎠ ⎟
= 1.74 Cal s
11.9 The mechanical energy transformed into internal energy of the bullet is
Q KE m m i i i = ( ) = ( ) = 12
12
12
v v2. Thus, the change in temperature of the bullet is
2 14
ΔT
i = = = (
Q
mc
m
mc
)
( ⋅ ) =
14
2 2 300
4 1
v
lead
m s
28 J kg °C
176°C

11.10 The internal energy added to the system equals the gravitational potential energy given up by the
2 falling blocks, or Q = ΔPEg = 2mbgh. Thus,
ΔT
2 2 1.50 kg)(9.80 m s2 ) 3.00 m
= = = (
Q
m c
m gh
b
m c w w
w w
kg J kg °C
C
( )
( )( ⋅ ) =
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
)
Δ
0.800 4 186
J
kg C
0.200 900
kg
J
⎛
⋅ kg C
⎛
⎝ ⎜
⎞
⎠ ⎟
+ ( ) ⋅
⎝ ⎜
⎞
⎠ ⎟
⎡
⎣ ⎢
° °
⎤
⎦ ⎥
(1.5°C) = 5.3 × 103 J
The rate of energy transfer is
P= = × = = Q
Δt
5 3 10
88 88
. 3 J
60 s
J
s
W
11.12 (a) The mechanical energy converted into internal energy of the block is
Q KE m i i = = 0 85 0 85 12
. ( ) . ( v2 ). The change in temperature of the block will be
ΔT
i = = ( ) = ( )
Q
mc
0 85 m
0 85 3 0
mc
Cu Cu
m s
87
2 3
12
2 2 . v . .
J kg C
C
( ⋅ )= × −
°
9.9 10 3 °
(b) The remaining energy is absorbed by the horizontal surface on which the block slides.
11.13 From ΔL =α L (ΔT ) 0 , the required increase in temperature is found, using Table 10.1, as
Δ Δ
T
= = ×
L
L
−
3 0 10
10 13
( × ( ) )
α − 6
− steel
m
0 11 °C
3
1
.
( )
°C yd
yd
3.0 ft
ft
1 m
⎛
⎝ ⎜
⎞
⎠ ⎟
⎛
⎞
⎟
⎠ ⎜⎝ =
1 3 281
23
.
The mass of the rail is
m
= =( )( ) ⎛
w
g
⎝ ⎜
70 lb yd 13 yd
9.80 m s
4.448 N
2 1 lb
⎞
⎠ ⎟
= 4.1 × 102 kg
so the required thermal energy (assuming that c c steel iron = ) is
Q = mc ( T ) = ( × )( ⋅ )( steel Δ 4.1 102 kg 448 J kg °C 23°C) = 4.2 × 106 J
11.14 (a) From the relation between compressive stress and strain, F A = Y (ΔL 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 (ΔL L ) (ΔT ) 0 =α , where α is the coeffi cient of linear
expansion. Thus, the stress becomes F A = Y [α (ΔT )] .
(b) If the concrete slab has mass m, the thermal energy required to produce a change in
temperature ΔT is Q = mc(ΔT ) 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 mc
F 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 1 3 2 . . . . kg m m m 3 00 96 0 m 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
F
A max
Q
mc
Y
max
.
.
= ⎛⎝
⎞⎠
= ( )( ⋅ )
α
96 0 880
2 1
kg J kg °C
( )( ) ( × 7
) = × 10 12 × 10− ( ) −
2 00 10 6 10 6 1
Pa °C
. Pa .7 × 106 J
(e) The change in temperature of the slab as it absorbs the thermal energy computed above is
ΔT
= = ×
Q
mc
6 7 10
( )( ) = 79
° 880
⋅ . 6 J
96.0 kg J kg °C
C
(f ) The rate the slab absorbs solar energy is
P P absorbed solar = 0.5 = 0.5(1.00 × 103 W) = 5 × 102 J s
so the time required to absorb the thermal energy computed in (d) above is
t
Q = = ×
. 6 1
×
⎛
⎝ ⎜
⎞
Pabsorbed
2
J
5 Js
h
3 600 s
6 7 10
10
⎠ ⎟
∼ 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 Q cold hot = − , so mc T T m c T T w w f iw f i ( − ) = − ( − ) , Al Al ,Al , or
m
Al Al ,Al −( kg)
m c T T
c T T w
f i
w f iw
=
− ( − )
( − ) =
,
( ⋅ )( − )
1.85 900 65 0
J kg °C °C 150°C
65 0
( 4 186 J kg ⋅ °C
)
.
. °C °C
kg
( − ) =
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 T pellet lead water (Δ ) = [ (Δ )]
or the number of pellets required is
N
= w w w ( )
Δ
Δ pellet lead lead
0.500 kg 4 186 25 0 20 0
m c T
m c T
= ( )
( J kg ⋅ C )( C − C
)
128
( 1.00 × 10-3 kg
)
° . ° . °
( J kg⋅ C)( C− C) =
° 200° 25 0°
467
.

11.17 The total energy absorbed by the cup, stirrer, and water equals the energy given up by the silver
sample. Thus,
mccAl mscCu mwcw T w mc T Ag [ + + ](Δ ) = [ Δ ]
Solving for the mass of the cup gives
m
= ( m c
)( c
)− −
s w w T
T
m c m c c
w
⎡
⎣ ⎢
⎤
⎦ ⎥
1
Al
Ag Ag
Ag
Cu
Δ
Δ
,
or
mc = ( )( ) ( − )
g 40 g (387) − ( )( ) ⎡
( − ) 1 − ( )
900
400 234
87 32
32 27
⎣ ⎢
⎤
225 g 4 186 = 80 g
⎦ ⎥
11.18 The mass of water is
m V w w w = ρ = (1.00 g cm3 )(100 cm3 ) = 100 g = 0.100 kg
For each bullet, the energy absorbed by the bullet equals the energy given up by the water, so
m c T m c T b b w w ( − 20°C) = (90°C − ). Solving for the fi nal temperature gives
T
= m c ( 90°C ) + m c
( 20°C
)
w w b b
+
m c m c
w w b b
.
For the silver bullet, m c b b = 5.0 × 10−3 kg and = 234 J kg ⋅°C, giving
Tsilver
°C
=
(0.100)(4 186)(90 ) + (5.0 × 10−3 )(234) 20
89 8 3
0 100 4 186 5 0 10 234
°C
°C
( )
( )( ) + ( × )( ) = . . −
.
For the copper bullet, m c b b = 5.0 × 10−3 kg and = 387 J kg ⋅°C, which yields
Tcopper
°C
=
(0.100)(4 186)(90 ) + (5.0 × 10−3 )(387) 20
89 7 3
0 100 4 186 5 0 10 387
°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 mc m c T Cu Cu Cu unk unk unk c Al w w w Δ + Δ = [ + ](Δ )
Solving for the specifi c heat of the unknown material gives
c
m c m c T m c T
c w w w
m T
unk
Al Cu Cu Cu
unk unk
=
[ + ](Δ ) − Δ
Δ
, or
{ ( )
= g J kg °C g ( )( ) ( )( ⋅ ) + 1
cunk g °C
70 80
100 900 250 ( ) ⋅ ( ) ⎡⎣
⎤⎦
− ( )
4 186 10
50
J kg °C °C
g (387 J kg ⋅°C)(60°C)} = 1.8 × 103 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
mwcw (ΔT )w = m c ΔT Fe Fe Fe or m
1.00 kg c Δ
T
Fe Fe = ( )
( )
c T w
Δ
w w
giving
mw =
( )
(1 00 )(448 ⋅ )(500 − 100 ) 1 9
5 . kg J kg °C °F °F °C °F
( =
4 186 J kg ⋅ °C )( 100 °F − 75 °F ) ( 1
°C °F
) 1 7
9
5
. kg
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 T Al Al Al Cu Cu Cu (Δ ) = Δ , or
m
c
c
T
T
=
Cu
Cu m Al
Al
Cu
Al
⎛
⎝ ⎜
⎞
⎠ ⎟
( )
⎡
⎣ ⎢
⎤
⎦ ⎥
= ⎛
Δ
Δ
387
⎝ 900
⎞⎠
−
−
⎛⎝
200 2 6 102 ° °
° °
( )= × 85
25
⎞⎠
C 25C
C 5.0 C
g . g = 0.26 kg
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, ΔKE = Q = m c (ΔT ) drums Fe or
ΔT
1 v ( kg )( m s
)
= i = m
2
m c
2 1 500 30
32
2 1
2
( )( ) =
°C ⋅ car
drums Fe
448
kg J kg °C
47
11.23 (a) Assuming that the tin-lead-water mixture is thermally isolated from the environment,
we have
Q Q cold hot = − or m c T T m c T T m c T T w w f iw f i f i ( − ) = − ( − )− − , Sn Sn ,Sn Pb Pb ,Pb ( )
and since m m m Sn Pb metal = = = 0.400 kg and T T T i, i, . Sn Pb hot = = = 60 0°C, this yields
T
m c T m c c T
w w iw
( + )
m c m c f
w w
=
+ ( + )
+
, metal Sn Pb hot
c
metal Sn Pb
= ( )( ⋅ )( ) +
1.00 kg 4 186 J kg °C 20.0 °C
0.400 227 128 60 0
1 00
kg J kg °C J kg °C °C
kg
( )( ⋅ + ⋅ )( . )
( . )(4 186 J kg ⋅°C) + (0.400 kg)(227 J kg ⋅°C + 128 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
ΔT, the thermal energy absorbed would be Q = Q +Q Sn Pb, or
m m c T m c T m c T Sn Pb alloy Sn Sn Pb Pb ( + ) ( Δ ) = ( Δ )+ ( Δ ) giving c
m c m c
Sn Sn Pb Pb
m m alloy
Sn Pb
=
+
+
If the alloy is a half-and-half mixture, so m m Sn Pb = , this reduces to c c c alloy Sn Pb = ( + ) 2 and
yields
calloy
= 227 J kg ⋅ °C + 128
J kg ⋅ °C
= J kg ⋅ °C 2
178

(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 = 0.400 kg = 400 g of tin with a molecular weight of MSn = 118.7 g mol is
N
m
M
Sn
NSn A
Sn
g
118.7 g mol
=
⎛
⎝ ⎜
⎞
⎠ ⎟
=
⎛
⎝ ⎜
⎞
⎠ ⎟
400
(6.02 × 1023 mol−1 ) = 2.03 × 1024
and, for the lead,
N
m
M
Pb
NPb A
Pb
g
207.2 g mol
=
⎛
⎝ ⎜
⎞
⎠ ⎟
=
⎛
⎝ ⎜
⎞
⎠ ⎟
400
(6.02 × 1023 mol-1 ) = 1.16 × 1024
(d) We have
N
N
Sn
Pb
= ×
= 2 03 10
1 16 10
×
1 75
24
24
.
.
.
and observe that
c
c
Sn
Pb
= ⋅
227
J kg °C
= 128
J kg °C
⋅
1.77
from which we conclude that the specific heat of an element is proportional to the
number of atoms per unit mass of that element.
11.24 Assuming that the unknown-water-calorimeter system is thermally isolated from the
environment, −Q = Q hot cold, or −m c T − T = m c T − T + m c T − T x x f ix ww f iw f i ( ) ( ) ( , , Al Al ,Al )
and, since T T T i,w i, = = = . Al cold 25 0°C, we have c m c m c T T m T T x w w f x i x f = ( + )( − ) ( − ) Al Al cold ,
or
cx =
⋅ ( ) ⎡⎣
(0.285 kg)(4 186 J kg ⋅°C) + (0.150 kg) 900 J kg °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 Q cold hot = − , or
m L m c T m c T ice f ice water f w water f + ( − 0°C) = − ( − 25°C)
and
T
25°C (825 g )(4 186 J kg ⋅°C)(25°C) − (75 g )(3.33 × 105 J kg)
m c m L
w f
m m c f
w
=
( ) −
water ice
( + ) = ice water
(75 g + 825 g )(4 186 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 warm 50 g of water to 100°C)
+ energy to vaporize 5.0 g water
50 g
( )
= ( ) L c L f + (50 g) (100 C − 0 C) + (5.0 g) water ° ° v
Thus,
⎛
Q = ( ) ×
⎝ ⎜
⎞
⎠ ⎟
0.050 kg
J
kg
3.33 105
+(0.050 kg)
J
k
4 186
g °C
100°C 0°C
⋅
⎛
⎝ ⎜
⎞
⎠ ⎟
( − )
⎛
+ ( × ) ×
⎝ ⎜
⎞
⎠ ⎟
5.0 10− kg
J
kg
3 2.26 106
which gives Q = 4.9 × 104 J = 49 kJ .
11.27 The conservation of energy equation for this process is
(energy to melt ice) + (energy to warm melted ice to T ) = (energy to cool water to T )
or
m L m c T m c T ice f ice w w w + ( − 0°C) = (80°C − )
This yields
T
m c m L
w w f
m m c
w w
=
( ) −
( + )
80°C ice
ice
so
T =
(1.0 kg)(4 186 J kg ⋅°C)(80°C) − (0.100 kg) 3.33 10
1 1 4 186
65
( × 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
(energy to melt) + energy to reach boiling point
+
( )
+ ( e
nergy to vaporize) + (energy to reach 110°C)
Mathematically,
Q m c L c L c f w = [ − (− )]+ + ( − ) + + ice s 0°C 10°C 100°C 0°C v team °C °C 110 100 − ( ) ⎡⎣⎤⎦
This yields
⎛
Q = ( × ) ⋅
40 10−3 kg 2 090 (10 ) + 3 33 ×
⎝ ⎜
⎞
⎠ ⎟
J
kg °C
°C . 10
⎛ J
4 186
5 J
kg
J
kg
⎛
⎝ ⎜
⎞
⎠ ⎟
⎡
⎣ ⎢
+
⋅°C
100 °C
2.26 106 + 2 010
J
kg
⎝ ⎜
⎞
⎠ ⎟
( ) + ⎛ ×
⎝ ⎜
⎞
⎠ ⎟
kg °C
°C
⋅
⎛
⎝ ⎜
⎞
⎠ ⎟
( )⎤
⎦ ⎥
10
or
Q = 1.2 × 105 J = 0.12 MJ

11.29 Assuming all work done against friction is used to melt snow, the energy balance equation is
⋅ = snow . Since f = μk (m g) skier , the distance traveled is
f s m Lf
s
m L
f
m g
= ( ) =
k
( )( × ) snow
skier
kg J kg
μ
1.0 3.33 105
0.20 kg m s
( )( )= × =
2 3 103 2 3
m km 75 9 .
80 2
. .
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 − 10.0°C to ice at 0°C; ice at 0°C to liquid water at 0°C; water at 0°C to
water at T; aluminum at 20.0°C to aluminum at T; ethyl alcohol at 30.0°C
to ethyl alcohol at 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 c ice ice [0 − (−10.0°C)]
Qmelt 1.00 3.33 × 105 0 0 m Lice f
Qwater 1.00 4 186 T 0 m c T ice water ( − 0)
QAl 0.500 900 T 20.0 m c T Al Al [ − 20.0°C]
Qalc 6.00 2 430 T 30.0 m c T alc alc [ − 30.0°C]
(c) m c m L m c T m ice ice ice f ice water Al (10.0°C)+ + ( − 0)+ c T m c T Al alc alc [ − 20.0°C]+ [ − 30.0°C] = 0
(d) T
( ) + ( ) − Al Al alc alc ice ice 20.0°C 30.0°C 10.0°C
( ) + ⎡⎣
m c m c m c
=
ice water Al Al alc alc
⎤⎦
+ +
L
m c m c m c
f
Substituting in numeric values from the table in (b) above gives
T =
(0.500)(900)(20.0) + (6.00)(2 430)(30.0) − 1.00 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
(Qcold = −Q)
yields
hot m c L c T mc ice ice f w w °C 78°C °C 0 0 − − ( ) [ ] + + − ( ) ⎡⎣
⎤⎦
= − w ( + m c )(T − ) cal Cu 25°C
or
T
+ ( )( ) − ( ) + ⎡⎣
⎤ cal Cu ice ice 25°C 78°C ⎦
(m + m )c + m c w ice w cal Cu
m c m c m c L w w f =
With m m m c w w = 0.560 kg, = 0.080 g, = 0.040 g, cal ice = 4 186 J kg ⋅°C,
J kg °C J kg °C, and Cu ice c = 387 ⋅ , c = 2 090 ⋅ L = 3 f .33 ×105 J kg
this gives
T =
( )( ) + ( )( ) ⎡⎣
0.560 4 186 0.080 387 (25°C) − 0.040 2 090 3 33 10
⎤⎦
0 560 0 040 4
5 ( ) ( )( ) + × ⎡⎣
⎤⎦
( + )
78°C .
. . ( 186) + 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 = × ⎛⎝
3 h .
⎞⎠
⎛⎝
⎞⎠
4 186
cal
h
J
400 10 ( 0 500
)
1 cal
. = 8.37 × 105 J
The mass of water that will be evaporated by this amount of excess energy is
m
= = ×
Q
L evaporated 6
J
×
2.5 10 J kg
=
v
8 37 10
0 33
. 5
. kg
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
(400 kcal h )(0.500 h
)
= 9.0 kcal gram
of fat
= 22 g = 22 × 10−3 kg
so the fraction of water needs provided by burning fat is
f
= = × =
m
m
−
produced
evaporated
kg
kg
22 10
0 33
0
3
.
.066 or 6.6%

11.33 The mass of 2.0 liters of water is mw = ρV = (103 kg m3 )(2.0 × 10−3 m3 ) = 2.0 kg.
The energy required to raise the temperature of the water (and pot) up to the boiling point of
water is
Q mc mc T boil w w Al Al = ( + )(Δ )
or
= ( )⎛
J 2.0 4 186 + (0.25 ) 900
Qboil kg
J
kg
kg
⎝ ⎜
⎞
⎠ ⎟
kg
°C °C J
⎛
⎝ ⎜
⎞
⎠ ⎟
⎡
⎣ ⎢
⎤
⎦ ⎥
(100 − 20 ) = 6.9 × 105
The time required for the 14 000 Btu h burner to produce this much energy is
t
Q
boil
boil
Btu h
. 5 J
1 Btu
Btu h
= = ×
14 000
6 9 10
14 000
h min × 3
1.054 10 J
⎛⎝
⎞⎠
= 4.7 × 10−2 = 2.8
Once the boiling temperature is reached, the additional energy required to evaporate all of the
water is
Q mL evaporate w = = ( kg)( × J kg) = × v 2.0 2.26 106 4.5 106 J
and the time required for the burner to produce this energy is
t
Q
boil
evaporate
Btu h
J
. 6
B
= = ×
14 000
4 5 10
14 000
tu h
Btu
h min 3
1
1.054 10 J
0 31 18
×
⎛⎝
⎞⎠
= . =
11.34 In 1 hour, the energy dissipated by the runner is
ΔE = P ⋅t = (300 J s)(3 600 s) = 1.08 ×106 J
Ninety percent, or Q = 0.900(1.08 × 106 J) = 9.72 × 105 J, of this is used to evaporate bodily
fl uids. The mass of fl uid evaporated is
m
= = ×
Q
L
×
=
v
9 72 10
10
0 403
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
V
. 6
0 403
= m = = ( 4 03 × 10
− )
ρ
4 10
.
kg
1 000 kg m
m
cm
3
3
3
3
3
m
cm
1
403
⎛
⎝ ⎜
⎞
⎠ ⎟
=

544 Chapter 11
11.35 The energy required to melt 50 g of ice is
Q m Lf
1 = = (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 T 2 w = ( ) = (0 050 )(4 ⋅ ) 100 ice Δ . kg .186 kJ kg °C ( °C) = 20.9 kJ
(a) If 10 g of steam is used, the energy it will give up as it condenses is
Q mL3 s = = (0 010 )(2 ) = 22 6 v . kg 260 kJ kg . kJ
Since Q Q 3 1 , all of the ice will melt. However, Q Q Q 3 1 2 + , so the fi nal temperature is less
than 100°C. From conservation of energy, we fi nd
0 = [ + (100 − )] v
m L c T m L c T ice f w steam w °C °C + − ( ) ⎡⎣
⎤⎦
or
T
+ ( 100
°C ) ⎡⎣
v m L c m L
steam ice
w f =
m m
⎤⎦
−
( +
ice steam
)cw
giving
T =
10 g 2.26 106 4 186 100 − (50 g)(3.33 × )
( ) × + ( )( ) ⎡⎣
⎤⎦
10
( + )( ) =
50 10 4 186
40
5
g g
°C
(b) If only 1.0 g of steam is used, then Q′ = m L = 3 s v 2.26 kJ. The energy 1.0 g of condensed
steam can give up as it cools from 100°C to 0°C is
= (Δ ) = (1.0 × 10−3 kg)(4.186 kJ kg ⋅°C)(100°C) = 0.419 kJ
Q mc T 4 s w
Since Q′ +Q 3 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
m
= ( + ) 3 4 2 26 = × −3
= ′ +
Q Q
Lf
8 0 10
.
.
0.419 kJ
333 kJ kg
kg = 8.0 g
11.36 First, we use the ideal gas law (with V = 0.600 L = 0.600 × 10−3 m3 and T = 37.0°C = 310 K) to
determine the quantity of water vapor in each exhaled breath:
PV nRT n
PV
RT
= ⇒ = =
(3.20 × 103 Pa)(0.600 × 10−3 m3 )
( ⋅ )( ) = × −
8 31 310
7 45 10 4
.
.
J mol K K
mol
or
m = nM = ( × )⎛
⎝ ⎜
⎞
⎠ ⎟
−
water mol
g
mol
⎛
1 k
1 34 10 5 7.45 10 4 18.0
g
g
kg 103
⎝ ⎜
⎞
⎠ ⎟
= . × −
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 5 kg 2.26 106 J kg 30.3 J
The rate of losing energy by exhaling humid air is then
P = ⋅( ) =⎛⎝
⎞⎠
Q
J
breath
b
respiration rate 30.3 22.0
reaths
min
= 1
min
60 s
W ⎛⎝
⎞⎠
⎛⎝
⎞⎠
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
1
2
Q KE = i + mbulletclead 0°C− 30 0°C = mbullet i + m
. v2 bullet lead °C
kg
c 30 0
3 00 10
J kg °C °C ( ) + ( ⋅ )( )
⎡
⎣
2 40 10 3
2
.
.
.
( )
= ( × ) × − m s
⎢⎢
⎤
⎦
⎥⎥
=
2
2
128 30.0 97.9 J
The mass of ice that melts when this quantity of thermal energy is absorbed is
m
Q
Lf
= ( ) =
×
. 4
= × −
water
J
J kg
k
97 9
3 33 10
2 94 10 5
.
. g
10 g
1 kg
g
⎛ 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 T h c on opposite sides is P = kA(T − T ) L h c .
For the given windowpane, this is
P =
J
⋅ ⋅
⎛⎝ ⎜
⎞⎠ ⎟
( )( ) ⎡⎣
⎤⎦
( −
)
0 84 1 0 2 0
25
. . .
s m °C
m m
°C 0
10
6 8 103 6 8 103 °C
= × = × − . .
J s W 2
×
0.62 m
(b) The total energy lost per day is
E = P ⋅ Δt = (6.8 × 103 J s)(8.64 × 104 s) = 5.9 × 108 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
h c J
1 3 ( ( 5 0
) ⋅ ⋅
⎛
⎝ ⎜
⎞
⎠ ⎟
T T
L
20
. .
s m °C
m
°C 2 )
= × = × 12 − m
×
1.1 103 1.1 103
J s W 10 2
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:
Rskin
= ×
m m 2
K W ⋅
0.020 W m K
= × ⋅
−
1 0 10 −
5 0 10
3
. 2
.
Rfat
= ×
m m 2
K W ⋅
0.20 W m K
= × ⋅
−
0 50 10 −
2 5 10
2
. 2
.
and
Rtissue
= ×
m m 2
K W ⋅
0.50 W m K
= × ⋅
−
3 0 10 −
6 0 10
2
. 2
.
so the total R value of the three layers taken together is
= 3
( + + ) × ⋅ = ×
R Ri
i
total
m2 K
W
= Σ
=
−
1
m 2 ⋅K 2
⋅
5.0 2.5 6.0 10 2 14 10−2 = 0 14 W
m K
W
.

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 ΔT = (37 − 0)°C = 37°C = 37 K is
P = ( ) = ( )( )
2 0 37
0 14
m K
m KW
= × A T
R
⋅
Δ
total
2
2
5 3
.
.
. 102 W
11.41 P = ⎛⎝ ⎜
⎞⎠ ⎟
kA
Δ
T
L
, with k =
⋅ ⋅
⎛
⎝ ⎜
⎞
⎠ ⎟
0 200
10 4 186
.
cal .
cm °C s
cm
1 m
J
1 c
2
al
J
s m °C
⎛⎝
⎞⎠
=
⋅ ⋅
83 7 .
Thus, the energy transfer rate is
P =
J
⋅ ⋅
⎛⎝
⎞⎠
[( )( )] ° −
83 7 8 00 50 0
200 2
. . .
s m °C
m m
C 0 0
1 50 10
4 02 10 402
2
8
.
.
.
°C
m
J
s
MW
×
⎛⎝
⎞⎠
= × =
−
11.42 The total surface area of the house is
A = A + A + A + A side walls end walls gables roof
where
Aside walls
= 2[(5.00 m) × (10.0 m)] = 100 m2
Aend walls
= 2[(5.00 m) × (8.00 m)] = 80.0 m2
A = 2[1 ( base ) × ( altitude )] = 2 (8 00 m) ×
gables 2
1
2 [ . (4.00 m) tan 37.0°] = 24.1 m2
Aroof
= 2 ( 10.0 m ) × ( 4.00 m cos37.0 ° ) = 100
m 2 ⎡⎣
⎤⎦
Thus,
A = 100 m2 + 80.0 m2 + 24.1 m2 + 100 m2 = 304 m2
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
= ( ) = t ⎡ ( )
E t
kA T
L
⎣ ⎢
⎤
⎦ ⎥( ) =
( × − ⋅ )
P Δ
Δ
Δ
4.8 10 4 kW m °C 304 25 0
.
( 86 400
) m °C
0.210 m
s
⎡ ( 2 )( )
⎣ ⎢⎢
⎤
⎦ ⎥⎥
or
E = 1.5 × 106 kJ = 1.5 × 109 J
The volume of gas that must be burned to replace this energy is
V
= E
= ×
heat of combustion
(
J
1 5 10
kcal m3
9 300
. 9
)( ) =
4 186
39
J kcal
m3
11.43 R R R R R i = Σ = + + outside
air film
shingles sheathing + R + R + R cellulose dry wall inside
air film
R = [ + + + ( ) + + ] ⋅
0.17 0.87 1.32 3 3.70 0.45 0.17
ft2 °F
Btu h
2
= ⋅
ft °F
Btu h
14

11.44 The rate of energy transfer through a compound slab is
P = ( ) = A T
Δ , where Σ
R
R L k i i
(a) For the thermopane, R = R + R + R = R + R pane trapped air pane pane trapped 2 air.
Thus,
R = ×
+ × − −
⋅
⎛
⎝ ⎜
⎞
⎠ ⎟
2
0.50 10 2 m 1.0 10 2
0.84 W m °C
m
0.0234 W m °C
2
m °C
W
⋅
= ⋅
0.44
and
P = ( )( )
⋅
=
1 0 23
0 44
52
.
.
m °C
m °CW
W
2
2
(b) For the 1.0 cm thick pane of glass:
R = ×
= × ⋅ −
1 0 10 −
⋅
1 2 10
2
. 2
.
m
0.84 W m °C
2
m °C
W
so
( 1 0 )( 23
)
P = = × = −
× ⋅
1
.
m °C
1 .
9 103 1 .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 P Cu Al = . The cross-sectional areas of the rods are equal, and if the temperature of
the junction is 50°C, the temperature difference is ΔT = 50°C for each rod.
Thus,
Δ Δ
k A =
Al =
P P Cu Cu
Cu
Al
Al
⎛
⎝ ⎜
⎞
⎠ ⎟
=
⎛
⎝ ⎜
⎞
⎠ ⎟
T
L
k A
T
L
,
which gives
L
k
k
Al
L Al
Cu
Cu
W m °C
97 W m °C
=
⎛
⎝ ⎜
⎞
⎠ ⎟
= ⋅
⋅
⎛
⎝ ⎜
⎞
⎠
238
3 ⎟ ( ) = 15 cm 9.0 cm
11.46 The energy transfer rate is
P= = = (Δ )( × )
Q
t
5.0 3.33 105
( )( ) =
ice f kg J kg
m L
Δ Δ
t
8.0 h
3 600
58
s 1 h
W
Thus, P = kA(ΔT L) gives the thermal conductivity as
k
= ⋅
L
( ) =
A T
( )( × )
( ) −
− P
Δ
W .
2 m
m2 °C
58 2 0 10
0 80 25
W m °C ( )= × − ⋅
. 5 0
7 2 10 2
.
.
°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
W
m K
2 4
= ( − )
= ×
⋅
⎛⎝ ⎜
⎞⎠ ⎟
−
σ
π
Ae T 4 T
4
0
5.67 10 8 4 0.060 473 2 2 4 4 m K 95 K ( ) ⎡⎣
⎤⎦
( ) − ( ) ⎡⎣
⎤⎦
or
Pnet = 1.1×102 W= 0.11 kW
11.48 Since 97.0% of the incident energy is refl ected, the rate of energy absorption from the sunlight is
Pabsorbed= 3.00%× (I ⋅ A) = 0.0300(I ⋅ A), where I is the intensity of the solar radiation.
Pabsorbed
= 0 0300(1 40 × 103 W m2 )(1 00 × 103 m)2 . . . = 4.20 × 107 W
Assuming the sail radiates equally from both sides (so A = 2(1 00 ) = 2 00 × 10 . km 2 . 6 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
= ( − ) = ×
⋅
⎛⎝
⎞⎠
W
K
4 ( )( )⋅ = × ⎛⎝
σ Ae T 4 0 5.669 6 10−8 2.00 × 106 m 2
0 03 4 3 40 10 3 4
⎞⎠
. T . − ⋅T
At the equilibrium temperature, where P P rad absorbed = , we then have
− W ⋅ = ×
3 40 10 4 20 10 3 4 7 . . × ⎛⎝
⎞⎠
K
W 4 T or T = ×
×
⎡
⎣ ⎢
⎤
= −
⎦ ⎥
4 20 10
3 40 10
333
7
3
1 4
.
.
W
W K
K 4
11.49 The absolute temperatures of the two stars are T T X Y = 6 000 K and = 12 000 K. Thus, the ratio
of their radiated powers is
P
P
Y
X
σ Y
= ( ) =
σ
X
Y
X
AeT
AeT
T
T
= =
⎛
⎝ ⎜
⎞
⎠ ⎟
4
4
4
4 2 16
11.50 The net power radiated is Pnet=σ Ae(T 4 − T )
4 , so the temperature of the radiator is
0
T T
Ae
= ⎡ +
⎣ ⎢
⎤
⎦ ⎥
4
0
1
P 4 net
σ
If the temperature of the surroundings is T0 = 22°C = 295 K,
T = ( ) +
25
W
295
( 5 67 × 10 − ⋅ ) 2 5 × 10
− 4
8 5 K
. W m2 K4 . m2
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 = ⎛⎝ ⎜
⎞⎠ ⎟
=
W
m °C
⋅
⎛⎝ ⎜
⎞⎠ ⎟
( ) ⎡⎣
⎤⎦
kA
Δ
T
L
397 0 10 2
π . m
102 100
2 0 10
1 2 10 12
3
4
°C °C
m
W k
−
×
⎛
⎝ ⎜
⎞
⎠ ⎟
= × =
. −
. W

11.52 The mass of the water in the heater is
ρ ⎞ 10 50 0
m V = =⎛⎝
3 kg 3 786
m
⎞⎠
( )⎛
⎝ ⎜
gal
L
.
3 1 gal .
⎠ ⎟
⎛
⎝ ⎜
⎞
= 1
10
189 3
⎠ ⎟
m
L
kg
3
The energy required to raise the temperature of the water from 20.0°C to 60.0°C is
Q = mc(ΔT ) = (189 kg)(4 186 J kg)(60.0°C− 20.0°C) = 3.17 ×107 J
The time required for the water heater to transfer this energy is
t
Q = = × ⎛
⎝ ⎜
⎞
⎠ ⎟
=
P
3 17 10 1
1 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 c Pb Pb ice f ice w w cup C (98°C − 12°C) = + ( + ) + u °C °C ⎡⎣
⎤⎦
(12 − 0 )
This gives
mPb
J
kg °C
128 86°C 0 040 kg 3 33 10
⋅
⎛
⎝ ⎜
⎞
⎠ ⎟
( ) = ( . ) . × 5
0 24
J kg
kg
( )
+ . ( ) ⋅ ( ) + ( ) ⋅ ( ) ⎡⎣
⎤⎦4 186 J kg °C 0.100 kg 357 J kg °C (12°C)
or mPb = 2.3 kg .
11.54 The energy needed is
Q = mc( T ) = ( V )c( T )
= ⎛⎝
⎡ ( )
⎞⎠
⎣ ⎢
Δ ρ Δ
103 1 00 kg
m
. 3 ⎤
m 3
⎦ ⎥
(4 186 J kg)(40.0°C) = 1.67 × 108 J
The power input is P = (550 W m2 )(6.00 m2 ) = 3.30 ×103 J s, so the time required is
t
Q = = ×
×
⎛
⎝ ⎜
⎞
⎠ ⎟
=
P
1 67 10
3 30 10
1
1
8
3
.
.
J
J s
h
3 600 s
4 1 . h
11.55 The conservation of energy equation is
m c m c T m c T w w ( + )( − ) = ( − ) cup glass Cu Cu 27°C 90°C
This gives
T
Cu Cu cup glass 90°C 27°C
m c m c m c
m c
w w
w w
=
( ) + ( + )( )
+
m c m c cup glass Cu Cu +
or
T =
(0.200)(387)(90°C) + (0.400)(4 186) + (0.300) 837 27
( ) ⎡⎣
29 ( )( ) = °C
0 400 4 186 0 300 837 0
⎤⎦
( )
( )( ) + ( )( ) +
°C
. . .200 387

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
m
Q
L
⋅( ) = ( )( )( )
P Δ t
25 4 0 3
= =
L f f
J s . h s h
.
×
2 01
600
10
= . kg
1 8 5 J kg
11.57 Assuming the aluminum-water-calorimeter system is thermally isolated from the environment,
Q Q cold hot = − , or
m c T T mc T T m c T Al Al f i Al w w f i w cal cal f ( − ) = − ( − )− − , , Ti,cal ( )
Since Tf = 66 3 . °C and T T i, i,w . cal = = 70 0°C, this gives
c
m c m c T T
,
w w iw f
m T T
f i
Al
cal cal
Al Al
=
( + )( − )
( − )
,
or
cAl
0.400 kg
4 186 + (0.040 ) 630
J
kg °C
kg
J
=
⎛
( ) ⋅
⎝ ⎜
⎞
⎠ ⎟
kg °C
°C
kg
⋅
⎛
⎝ ⎜
⎞
⎠ ⎟
⎡
⎣ ⎢
⎤
⎦ ⎥
( − )
( )
70 0 66 3
( − ) = ×
0 200 66
. .
. . .
.
3 270
8 00 102
J
kg °C
°C ⋅
The variation between this result and the value from Table 11.1 is
⎛
variation × = −
% = %
⎝ ⎜
⎞
⎠ ⎟
accepted value
100
800 900
900
100 11 1
J kg °C
J kg °C
⋅
⋅
⎛
⎝ ⎜
⎞
⎠ ⎟
× % = . %
which is within the 15% tolerance.
11.58 (a) With a body temperature of T = 37°C + 273 = 310 K and surroundings at temperature
T0 = 24°C + 273 = 297 K, the rate of energy transfer by radiation is
Prad
W
m K
2 4
= ( − )
= ×
⋅
⎛⎝
⎞⎠
−
σ Ae T 4 T
4
0
8 5 669 6 10 2 0 . . m K K W 2 ( )( ) ( ) − ( ) ⎡⎣
⎤⎦
0 97 310 297 = 1 6 × 10 . 4 4 . 2
(b) The rate of energy transfer by evaporation of sweat is
Psweat
W ( )( )= ×
Q ) ×
t
0.40 2.43 10 kg J kJ
v, sweat
kg 3 kJ
= = = (
mL
Δ Δt
3 600 s
10
2 7 10
3
. 2
(c) The rate of energy transfer by evaporation from the lungs is
Plungs
kJ
h
h
3 600 s
J
1 kJ
=⎛⎝
⎞⎠
⎛
⎝ ⎜
⎞
⎠ ⎟
⎛
⎝
38
1 103
⎜
⎞
⎠ ⎟
= 11W
(d) The excess thermal energy that must be dissipated is
P P excess metabolic
kJ
h
= = × ⎛⎝
⎞⎠
0.80 0.80 2.50 103
1 10
5 6 10
3
h 2
3 600 s
J
1 kJ
W
⎛
⎝ ⎜
⎞
⎠ ⎟
⎛
⎝ ⎜
⎞
⎠ ⎟
= . ×
so the rate energy must be transferred by conduction and convection is
P P P P P cc = − ( + + ) = . − . − excess rad sweat lungs (5 6 1 6 2.7 − .11) × 102 W = 1.2 × 102 W

11.59 The rate at which energy must be added to the water is
P= =
⎛
⎝ ⎜
⎞
⎠ ⎟
= ⎛⎝ ⎜
⎞⎠ ⎟
Δ
Δ
Δ
Δ
Q
t
m
t
Lv 0 500
1
.
kg
min
min
60 s
J
kg
⎛⎝ ⎜
⎞⎠ ⎟
⎡
⎣ ⎢
⎤
⎦ ⎥
2.26 ×
106 = 1.88 ×104 W
⎛
⎝ ⎜
⎞
⎠ ⎟
From P = kA(T −100°C) L, the temperature of the bottom surface is
T
°C ( )
P . W . m
= + ⋅ = +
L
k A
( × ) × −
100 100
1 88 104 0 500 10 2
°C °C
109 2 .
⎡⎣
( 238 W m ⋅ °C ) ( m
) ⎤⎦
=
π 0 120
11.60 The energy added to the air in one hour is
Q t = ( ) = ( ) ⎡⎣
P ( ) = × total 10 200 W 3 600 s 7.2 0 106 J
⎤⎦
and the mass of air in the room is
m = ρV = (1.3 kg m3 )[(6.0 m)(15.0 m)(3.0 m)] = 3.5 × 102 kg
The change in temperature is
ΔT
= = ×
Q
mc
7 2 10
( = 3 5 × 10 )( 837
⋅ ) 25
6
2
.
.
J
kg J kg °C
°C
giving T = T + T = + = 0 Δ 20°C 25°C 45°C .
11.61 In the steady state, P P Au Ag = , or
k A
80 . 0 °C − T
30 . 0 °C ⎛⎝
L
= T
k A
⎛⎝
− ⎞⎠
L Au Ag
⎞⎠
This gives
T
80.0°C 30.0°C 314 80.0°C) + ( )
= Au Ag (
k k
k k
=
( ) + ( )
+
Au Ag
+
= 427 30 0
314 427
51 2
.
.
°C
°C
11.62 (a) The rate work is done against friction is
P = f ⋅ v = (50 N)(40 m s) = 2.0 ×103 J s = 2.0 kW
(b) In a time interval of 10 s, the energy added to the 10-kg of iron is
Q = P ⋅t = (2.0 ×103 J s)(10 s) = 2.0 ×104 J
and the change in temperature is
ΔT
= = ×
Q
mc
2 0 10
( )( ) = 448
⋅ 4 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(ΔT ) = (6.0 kg)(900 J kg ⋅°C)(660°C − 20°C) = 3.46 × 106 J
The internal energy added to the brakes on each stop is
1 2 2
2
Q KE mi
1
1
2
= Δ = = (1 500 )(25 ) = 4 69 ×1 car v kg m s . 05 J
The number of stops before reaching the melting point is
N
= = ×
Q
Q
×
=
1
6
5
3 46 10
4 69 10
7
.
.
J
J
stops
(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
7
3
=⎛⎝
⎞⎠
mc mc 1 2 (17°C−10°C) = (20°C−17°C), or c c 2 1
When liquids 2 and 3 are mixed, energy conservation yields
28
3
= =⎛⎝
⎞⎠
mc mc 3 2 (30°C− 28°C) = (28°C− 20°C), or c c c 3 2 1 4
Then, mixing liquids 1 and 3 will give mc T mc T 1 3 ( −10°C) = (30°C− )
or
T
= ( ) + ( )
10 30 10 28 3 30
°C °C °C °C
c c
+
c c
=
+ ( )( )
+
1 3
1 3
1
( ) = 28 °C
28 3
11.65 (a) The internal energy ΔQ added to the volume ΔV of liquid that fl ows through the calorimeter
in time Δt is ΔQ = (Δm)c (ΔT ) = ρ(ΔV)c (ΔT ). Thus, the rate of adding energy is
Δ
Δ
Δ Δ
Δ
Q
t
c T
V
t
= ( )⎛
⎝ ⎜
⎞
⎠ ⎟
ρ
where ΔV Δt is the fl ow rate through the calorimeter.
(b) From the result of part (a), the specifi c heat is
c
Δ Δ
Q t
T V t
J s
= = ρ ( Δ )( Δ Δ
) ( )( )
40
. g cm3 . °C .5
0 72 5 8 3
. = .7 ×103 J kg ⋅°C
2 7
10
2
3
cm s
J
g °C
g
1 kg
( 3 )
=
⋅
⎛
⎝ ⎜
⎞
⎠ ⎟
⎛
⎝ ⎜
⎞
⎠ ⎟

= + = 2(π 2 )+ (2π ), or
11.66 (a) The surface area of the stove is A A A r rh stove ends cylindrical
side
Astove = 2 (0 200 m) + 2 (0 200 m)(0 500 m) = 0 2 π . π . . .880 m2
The temperature of the stove is Ts= − = = 59
(400°F 32.0°F) 204°C 477 K while that of
the air in the room is Tr= 5 − = =
9 (70.0°F 32.0°F) 21.1°C 294 K. If the emissivity of the
stove is e = 0.920, the net power radiated to the room is
P = σA eT ( 4 − T 4
)
stove s r
= ( × − ⋅ )
⎡⎣
5 . 67 10 8 W m2 K4
( 0 . 880 m 2 )( ) ( K ) − ( K ) ⎤⎦
0 920 477 294 4 4 .
or
P = 2.03 × 103 W
(b) The total surface area of the walls and ceiling of the room is
A = 4A + A = 4[(8 00 )(25 0 )] + 25 wall ceiling . ft . ft .0 1 43 10 ( ft)2 = . × 3 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 ) R h c i Σ , the net R value needed in the walls and ceiling is
ΣR
( − ) =
A T T
i
= h c
( × )( − )
P
1.43 103 ft2 70.0°F 32.0°F
2.03 10 J s
J
Btu
⎝ ⎜⎞
h
⎛⎝ ⎜
⎞⎠ ⎟⎛
× 3 3 600 s
⎠
1 054
1
1
⎟
or
ΣRi= 7.84 ft2 ⋅ °F ⋅ h Btu
11.67 A volume of 1.0 L of water has a mass of m = ρV = (103 kg m3 )(1.0 × 10−3 m3 ) = 1.0 kg.
The energy required to raise the temperature of the water to 100°C and then completely evaporate
it is Q = mc(ΔT )+ mLv, or
Q = (1.0 kg)(4 186 J kg⋅°C)(100°C− 20°C) + (1.0 kg)(2.26 ×106 J kg) = 2.59 ×106 J
The power input to the water from the solar cooker is
P = ( ) = ( )( ) ( ) efficiency IA 0 50 600
0 50
4
2
.
.
W m
m 2 ⎡π
⎣ ⎢⎢
⎤
⎦ ⎥⎥
= 59 W
so the time required to evaporate the water is
t
= Q = 2 . 59 × 10
6
J
= ( 4 .
4 × 10
4
s
)⎛
P 59
J s
⎝
1 h
⎜ 3 600 s
⎞
⎠ ⎟
= 12 h

554 Chapter 11
11.68 (a) From the thermal conductivity equation, P = − ( ) ⎡⎣
⎤⎦
kA T T L h c , the total energy lost by
conduction through the insulation during the 24-h period will be
Q
kA
L
= P ( )+P ( ) = ( − )+ 1 2 12 0 12 0 37 0 23 0 . . . . h h °C °C 37 0 16 0 12 0 . . . °C °C h − ( ) ⎡⎣
⎤⎦
( )
or
Q =
( 0 012 0 ⋅ )( 0 490
) +
0 095 0
⎛⎝
[ 14 0 21
]( )J . .
.
. .
J s m°C m
m
°C
2
0 120
3 600
s
°C h 9 36 104
1 h
⎞⎠
. = . ×
The mass of molten wax which will give off this much energy as it solidifi es (all at 37°C) is
m
= = ×
Q
Lf
= 9 36 10
×
0 457
kg 3
. 4
.
J
205 10 J kg
(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
m = ( )+ + ( )
=
Q mc T mL mc T
m
steam liquid
water
2.0
Δ Δ 1 v 2
( 1 × 10 3 J kg ⋅ °C ) ( 130 °C − 100 °C )+ 2.26 × 106
J ⎡⎣
kg +(4 186 J kg ⋅°C) 100°C− 50 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 mc mc T needed w w g g kg J kg °C = + ( ) = ( ) ⋅ Δ 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 × 106 J kg) = 2.76 × 104 J or m = ×
= = 2 76 10
2 53 10
×
0 010 9 10 9
4
6
.
.
. .
J
J kg
kg 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
≈ (Δ )+ °C = ( J kg ⋅°C)( °C− 37°C)+ 2.26 ×106 J kg
37°C 100 4 186 100
L c T L v water
v
or
37°C ≈ 2.5 ×106 J kg
Lv
Assuming that you are approximately 2.0 m tall and 0.30 m wide, you will cover an area of
A = (2.0 m)(0.30 m) = 0.60 m2 of the beach, and the energy you receive from the sunlight in one
hour is
Q = IA(Δt ) = (1 000 W m2 )(0.60 m2 )(3 600 s) = 2.2 × 106 J
The quantity of water this much energy could evaporate from your body is
m
= ≈ ×
Q
L
2 2 10
2 5 10
0 9 °C
×
=
37
v
6
6
J
J kg
kg
.
.
.
The volume of this quantity of water is
V
0 9. kg
10 kg m
= m = ≈ − 3 3
=
ρ
10 m 1 L 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
= = (3.33 × 105 J kg), so the constant
liquid water at 0°C. The energy required is Q mL m 1 f
power input must be
Q ( × )
t
P = ( ) =
m
1
1
3 33 105
. J kg
Δ 50
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
Q ( + )
t
( ⋅ )(2.0 )
kg kg J kg °C °C
10 10 4 186
m 2
P = ( ) =
( m + ) c ( T
)
( ) =
t
2 2
Δ
Δ
Δ
10 min
Since the power input is the same in the two periods, we have
10 4 186 ( . × 5 ) =
m 3 33 10 m
50
J kg ( + )( ⋅
min
kg J kg °C)(2.0°C)
10 min
which simplifi es to (8.0)m = m + 10 kg, or
m= = 10
1 4
kg
7.0
. kg

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 water
to 0°C
freeze
at 0°C
cool ice
to 8.00°C
ice °C °C
−
= m c − T + m L + m c T − w w i w f w f 0 0
or
⎛
Q = ( × ) ⋅
75.0 10−3 kg 4 186 −20.0 + 3.33
⎝ ⎜
⎞
⎠ ⎟
J
kg °C
°C × +
⋅
⎛
⎝ ⎜
⎞
⎠ ⎟
−
⎡
⎣ ⎢
⎤
⎦ ⎥
=
10 2 090 8 00
3 2
5 J
kg
J
kg °C
. °C
. 5 × 104 J = 32.5 kJ
11.73 (a) In steady state, the energy transfer rate is the same for each of the rods, or
P P Al Fe = . Thus, k A
100 °C − T
0 °C ⎛⎝
L
= T
k A
⎛⎝
− ⎞⎠
L Al Fe
⎞⎠
giving
T
k
k k
=
+
⎛
⎝ ⎜
⎞
⎠ ⎟
( ) =
+
⎛⎝
⎞⎠
Al
Al Fe
100°C
238
238 79 5
10
.
( 0°C) = 75.0°C
(b) If L = 15 cm and A = 5.0 cm2, the energy conducted in 30 min is
P ( × − ) − Al
Q = ⋅ t =
⋅
⎛⎝ ⎜
⎞⎠ ⎟
4 100 7 .
2
W
m °C
m
°C
238 5 0 10
5 0
0 15
1 800
3 6 104
.
.
.
°C
m
s
J
⎛⎝ ⎜
⎞⎠ ⎟
⎡
⎣ ⎢
⎤
⎦ ⎥
( )
= × = 36 kJ

Solucionario Fundamentos de Física 9na edición Capitulo 11

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Solucionario Fundamentos de Física 9na edición Capitulo 11