Chapter 11
11
Energy in Thermal Processes
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
Q
mcwater
T
and the final temperature is Tf
mc
T
11.2
Q
11.3
The mass of water involved is
m
(a) Q
(b)
11.4
103
T
kg
m3
4.00
The power input is
P
P
1.89°C
4 186 J kg °C
Ti
T
15.0°C+1.89°C
4.00
1011 m 3
16.9°C .
20.0°C
4.00
4.49
1 000 MW
1.00
104 J
44.9 kJ
1014 kg
1014 kg 4 186 J kg °C 1.00°C
1.67
1018 J
109 J s , so,
1.67 1018 J
1 yr
1.00 109 J s 3.156 10 7 s
Q
t
807 m
1.50 kg 230 J kg °C 150°C
V
mc
9.80 m s2
m gh
m cwater
52.9 yr
The change in temperature of the rod is
T
Q
mc
1.00 10 4 J
0.350 kg 900 J kg°C
31.7°C
and the change in the length is
L
L0
T
24
10
6
°C
1
20.0 cm 31.7°C
Page 11.1
1.52
10
2
cm
0.152 mm
Chapter 11
11.5
T
Tf
Q
mc
25°C
750 cal
75 g 0.168 cal g °C
60°C
so
Tf
11.6
25°C
4.186 J
1 cal
(b)
The work done lifting her weight mg up one stair of height h is. W1
Nmgh Q or
climbing N stairs is W Nmgh , and we have W
Q
mgh
2.3
106 J
Q
(c)
106 J
2.3
55 kg 9.80 m s2
2.8
0.15 m
mgh Thus, the total work done in
10 4 stairs
If only 25% of the energy from the donut goes into mechanical energy, we have
0.25 Q
mgh
N
0.25
1
m v 2f
2
Q
mgh
0.25 2.8
1
75 kg
2
2
10 3 stairs
7.0
103 J
(b)
P
(c)
If the mechanical energy is 25% of the energy gained from converting food energy, then Wnet
0.25
Wnet
KE
4.54 103 J
5.0 s
Wnet
t
P
0.25 ( Q)
Q
t
v 02
10 4 stairs
4.5
(a)
and
11.8
103 cal
1 Cal
540 Cal
85°C
(a)
N
11.7
60°C
P
0.25
9.1
102 J s
11.0 m s
0
4.54
103 J
910 W
t , so the food energy conversion rate is
910 J s
0.25
1 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 dis
posed of through the evaporation of sweat.
(a)
The instantaneous power is
(b)
ma , and the kinematics equation v
From Newton’s second law, Fnet
taneous power expression given above may be written as
P
Fv
Q
ma 0
P
Fv , where F is the applied force and
at or P
ma2 t
Page 11.2
is the instantaneous velocity.
v0
at with v0
0 , the instant
Chapter 11
0
0
v
t
11.0 m s
5.00 s
(c)
a
v
t
(d)
P
ma2 t
(e)
Maximum instantaneous power occurs when t
2.20 m s2
2
75.0 kg 2.20 m s2
Pmax
363 J s2
5.00 s
363 kg m 2 s4
t
1.82
tmax
t
363 W s
t
5.00 s , so
103 J s
If this corresponds to 25.0% of the rate of using food energy, that rate must be
Pmax
Q
t
11.9
1
2
1
2
KEi
T
1.74 Cal s
1
2
1
4
mvi2
Q
mc
mvi2 . Thus, the change in temperature of the bullet is
1 m v2
i
4
300 m s
m clead
4 128 J kg °C
2
176 °C
The internal energy added to the system equals the gravitational potential energy given up by the 2 falling blocks,
PEg
2mb gh . Thus,
or Q
T
11.11
0.250
103 J s 1 Cal
0.250
4 186 J
The mechanical energy transformed into internal energy of the bullet is
Q
11.10
1.82
2 1.50 kg 9.80 m s2
2 mb gh
mw cw
Q
mw cw
3.00 m
0.105°C
0.200 kg 4 186 J kg °C
The quantity of energy transferred from the water-cup combination in a time interval of 1 minute is
Q
mc
mc
water
0.800 kg
T
cup
4 186
J
0.200 kg
kg °C
900
J
kg °C
1.5°C
5.3
103 J
The rate of energy transfer is
P
11.12
(a)
Q
t
5.3
103 J
60 s
88
J
s
88 W
The mechanical energy converted into internal energy of the block is Q
change in temperature of the block will be
Page 11.3
0.85 (KEi )
0.85 ( 21 mvi2 ) . The
Chapter 11
T
(b)
11.13
Q
mcCu
0.85 21 m vi2
0.85 3.0 m s
m cCu
2 387 J kg °C
2
9.9
10
3
°C
The remaining energy is absorbed by the horizontal surface on which the block slides.
L
From
L0
T , the required increase in temperature is found, using Table 10.1, as
L
steel L0
T
3.0
11
10
6
10
°C
3
1 yd
m
1
3.281 ft
1 m
3.0 ft
13 yd
23°C
The mass of the rail is
m
w
g
70 lb yd
13 yd
9.80 m s2
4.448 N
1 lb
so the required thermal energy (assuming that csteel
Q
11.14
(a)
mcsteel
T
4.1
102 kg
4.1
ciron ) is
102 kg 448 J kg °C 23°C
From the relation between compressive stress and strain, F A
4.2
Y
106 J
L L0 , 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 L0)
( T ) , where
is the coefficient of linear
expansion. Thus, the stress becomes F A
(b)
Y
T
.
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 specific heat of concrete. Using the result from part (a), the absorbed thermal
energy required to produce compressive stress F A is
Q
(c)
F A
or Q
Y
mc
Y
F
A
The mass of the given concrete slab is
m
(d)
mc
V
2.40
103 kg m3
4.00
10
2
m 1.00 m 1.00 m
96.0 kg
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
Page 11.4
Chapter 11
mc
Y
Qmax
(e)
96.0 kg 880 J kg °C
2.1
max
1010
Pa 12
10
6
1
°C
2.00
107 Pa
6.7
106 J
The change in temperature of the slab as it absorbs the thermal energy computed above is
6.7 106 J
96.0 kg 880 J kg °C
Q
mc
T
(f)
F
A
79 C
The rate the slab absorbs solar energy is
Pabsorbed
0.5Psolar
0.5 1.00
103 W
5
102 J s
so the time required to absorb the thermal energy computed in (d) above is
11.15
6.7 106 J
1h
5 102 J s 3 600 s
Q
t
Pabsorbed
4h
When thermal equilibrium is reached, the water and aluminum will have a common temperature of Tf
Assuming that the water-aluminum system is thermally isolated from the environment, Qcold
mw cw T f
mw
11.16
Ti, w
mAl cAl T f
mAl cAl T f
cw T f
65.0°C .
Qhot , so
Ti,Al , or
Ti,Al
1.85 kg 900 J kg °C 65.0°C
4 186 J kg °C 65.0°C
Ti, w
150°C
25.0°C
0.845 kg
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,
Nmpellet c
T
mc
lead
T
water
or the number of pellets required is
N
mw cw
T
mpellet clead
w
T
lead
0.500 kg 4 186 J kg °C 25.0°C
1.00
11.17
10-3
kg 128 J kg °C 200°C
20.0°C
25.0°C
467
The total energy absorbed by the cup, stirrer, and water equals the energy given up by the silver sample. Thus,
Page 11.5
Chapter 11
mc cAl
ms cCu
mw cw
T
mc
w
T
Ag
Solving for the mass of the cup gives
mc
1
cAl
mAg cAg
T
Ag
T
ms cCu
mw cw ,
w
or
mc
11.18
1
900
400 g 234
87
32
32
27
40 g 387
225 g 4 186
80 g
The mass of water is
mw
1.00 g cm3 100 cm3
wVw
100 g
0.100 kg
For each bullet, the energy absorbed by the bullet equals the energy given up by the water, so
mbcb T 20°C
mwcw 90°C T . Solving for the final temperature gives
T
mw cw 90°C
For the silver bullet, mb
Tsilver
mb cb
5.0
10
3
10
3
234 J kg °C , giving
5.0
0.100 4 186
5.0
.
kg and cb
0.100 4 186 90°C
For the copper bullet, mb
Tcopper
mb cb 20°C
mw cw
5.0
kg and cb
0.100 4 186 90°C
0.100 4 186
5.0
5.0
10
3
234 20°C
10
3
234
89.8°C
387 J kg °C , which yields
10
3
387 20°C
10
3
387
89.7°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,
mCu cCu
T
Cu
munk cunk
T
unk
mc cAl
Solving for the specific heat of the unknown material gives
Page 11.6
mwcw
T
w
Chapter 11
cunk
cunk
mc cAl
m w cw
T
munk
1
mCu cCu
w
T
Cu
, or
unk
100 g 900 J kg °C
70 g 80°C
50 g 387 J kg °C 60°C
11.20
T
250 g 4 186 J kg °C
10°C
103 J kg °C
1.8
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
T
mFe cFe
w
T
Fe
or mw
1.00 kg cFe
cw
T
T
Fe
w
giving
mw
11.21
75°F
1°C
1°C
9
5
°F
9
5
°F
1.7 kg
cCu
cAl
T
387
900
85°C
25°C
T
Cu
mCu
Al
25°C
5.0°C
200 g
2.6
102 g
0.26 kg
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 mdrumscFe T or
T
11.23
4 186 J kg °C 100°F
100°F
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
mCu cCu T Cu , or
by the aluminum, giving mAl cAl T Al
mAl
11.22
1.00 kg 448 J kg °C 500°F
(a)
1
2
mcar vi2
mdrums cFe
1
2
1 500 kg 30 m s
2
32 kg 448 J kg °C
47°C
Assuming that the tin-lead-water mixture is thermally isolated from the environment, we have
Page 11.7
Chapter 11
Qcold
and since and mSn
Tf
mPb
mmetal
Ti, w
mSn cSn T f
0.400 kg , and Ti,Sn
mw cwTi, w
mmetal cSn
cPb Thot
mw cw
mmetal cSn
cPb
Ti,Pb
mPb cPb T f
Thot
60.0°C this yields
128 J kg °C 60.0°C
1.00 kg 4 186 J kg °C
0.400 kg 227 J kg °C
128 J kg °C
21.3°C
If an alloy containing a mass mSn of tin and a mass mPb of lead undergoes a rise in temperature
thermal energy absorbed would be Q QSn QPb , or
mPb calloy
T
mSn cSn
T
mPb cPb
If the alloy is a half-and-half mixture, so mSn
calloy
227 J kg °C
T
giving
128 J kg °C
mSn cSn
mSn
calloy
mPb , this reduces to calloy
(cSn
T , the
mPb cPb
mPb
cPb ) 2 and yields
178 J kg °C
2
For a substance forming monatomic molecules, the number of atoms in a mass equal to the molecular weight
0.400 kg 400 g of
of that material is Avogadro’s number, N A . Thus, the number of tin atoms in mSn
tin with a molecular weight of MSn 118.7 g mol is
N Sn
mSn
MSn
NA
400 g
118.7 g mol
6.02
10 23 mol
1
2.03
10 24
mPb
M Pb
NA
400 g
207.2 g mol
6.02
10 23 mol -1
1.16
10 24
10 24
10 24
1.75
and, for the lead,
N Pb
(d)
Ti,Pb
0.400 kg 227 J kg °C
mSn
(c)
Ti,Sn
1.00 kg 4 186 J kg °C 20.0°C
yielding T f
(b)
mw cw T f
Qhot or
We have
NSn
N Pb
2.03
1.16
Page 11.8
Chapter 11
and observe that
cSn
cPb
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
Qcold ,
Assuming that the unknown-water-calorimeter system is thermally isolated from the environment, Qhot
Ti, x ) mwcw (Tf Ti, w ) mAl cAl (Tf Ti,Al ) and, since Ti, w Ti,Al Tcold
25.0°C , we
or mx cx (Tf
have cx
(mwcw
mAl cAl )(Tf
Tcold ) mx (Ti, x
Tf )
or
0.285 kg 4 186 J kg °C
cx
0.125 kg 95.0°C
yielding cx
11.25
0.150 kg 900 J kg °C
1.18
32.0
25.0 °C
32.0°C
103 J kg °C .
Remember that energy must be supplied to melt the ice before its temperature will begin to rise. Then, assuming a
Qhot , or
thermally isolated system, Qcold
mice L f
mice cwater T f
0°C
mwcwater T f
25°C
and
Tf
yielding T f
11.26
mw cwater 25°C
mice
mice L f
825 g
4 186 J kg °C 25°C
mw cwater
75 g
825 g
16°C .
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 f
50 g cwater 100°C
0°C
5.0 g Lv
Page 11.9
75 g
3.33
4 186 J kg °C
105 J kg
Chapter 11
Thus,
Q
0.050 kg
10 5
3.33
J
kg
0.050 kg
which gives Q
11.27
104 J
4.9
4 186
J
kg °C
100°C
5.0
3
10
kg
0°C
2 .26
106
J
kg
49 kJ .
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
mice L f
mice cw T
0°C
mwcw 80°C
T
This yields
T
mw cw 80°C
mice
mice L f
m w cw
so
T
11.28
1.0 kg 4 186 J kg °C 80°C
0.100 kg 3.33
10 5 J kg
1.1 kg 4 186 J kg °C
65 °C
The energy required is the following sum of terms:
Q
energy to reach melting point
energy to melt
energy to reach boiling point
energy to vaporize
energy to reach 110°C
Mathematically,
Q
m cice 0°C
10°C
Lf
cw 100°C
Page 11.10
0°C
Lv
csteam 110°C
100°C
Chapter 11
This yields
Q
40
10
3
kg
2 090
4 186
J
kg °C
J
kg °C
10°C
100°C
10 5
3.33
106
2.26
J
kg
J
kg
2 010
J
kg °C
10°C
or
Q
11.29
0.12 MJ
Assuming all work done against friction is used to melt snow, the energy balance equation is f
Since f
s
k
(a)
s
msnow L f .
mskier g , the distance traveled is
1.0 kg 3.33
msnow L f
mskier g
k
11.30
105 J
1.2
105 J kg
2.3
0.20 75 kg 9.80 m s2
103 m
2.3 km
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
Qice
1.00
2 090
Qmelt
1.00
Qwater
1.00
QAl
Qalc
L J kg
Tf °C
Ti °C
Expression
10.0
mice cice 0
0
10.0°C
105 0
0
mice L f
4 186
T
0
mice cwater T
0.500
900
T
20.0
mAl cAl T
20.0°C
6.00
2 430
T
30.0
malc calc T
30.0°C
3.33
Page 11.11
0
Chapter 11
(c)
mice cice 10.0°C
(d)
T
mice L f
mAl cAl 20.0°C
mice cwater T
0
malc calc 30.0°C
mice cwater
mAl cAl T
20.0°C
mice cice 10.0°C
mAl cAl
malc calc T
30.0°C
0
Lf
malc calc
Substituting in numeric values from the table in (b) above gives
0.500 900 20.0
T
1.00 4 186
and yields T
11.31
6.00 2 430 30.0
1.00
0.500 900
2 090 10.0
3.33
6.00 2 430
4.81°C .
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
mice cice 0°C
78°C
Lf
cw T
0°C
mw cw
mcal cCu
Qhot yields
T
25°C
or
mw cw
T
mcal cCu
mw
With mw
0.560 kg, mcal
cCu
25°C
mice cice 78°C
mice cw
mcal cCu
0.080 g, mice
387 J kg °C , cice
Lf
0.040 g, cw
4 186 J kg °C ,
2 090 J kg °C , and L f
3.33
105 J kg
this gives
T
or T
11.32
0.560 4 186
0.080 387
0.560
25°C
0.040
0.040 4 186
2 090 78°C
3.33
10 5
0.080 387
16 °C and the assumption that all ice melts is seen to be valid.
At a rate of 400 kcal h , the excess internal energy that must be eliminated in a half-hour run is
Q
400
105
103
cal
h
4.186 J
1 cal
0.500 h
8.37
10 5 J
The mass of water that will be evaporated by this amount of excess energy is
Page 11.12
Chapter 11
8.37 105 J
2.5 106 J kg
Q
Lv
mevaporated
0.33 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
400 kcal h 0.500 h
mproduced
22 g
9.0 kcal gram of fat
22
10
3
kg
so the fraction of water needs provided by burning fat is
f
11.33
mproduced
22
mevaporated
10 3 kg
0.33 kg
The mass of 2.0 liters of water is mw
0.066 or 6.6%
103 kg m3
V
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
Qboil
mwcw
mAl cAl
T
or
Qboil
2.0 kg
4 186
J
kg
0.25 kg
900
J
kg
100°C
20°C
6.9
10 5 J
The time required for the 14 000 Btu h burner to produce this much energy is
tboil
Qboil
14 000 Btu h
6.9 105 J
1 Btu
14 000 Btu h 1.054 10 3 J
4.7
10
2
h
2.8 min
Once the boiling temperature is reached, the additional energy required to evaporate all of the water is
Qevaporate
mw Lv
2.0 kg 2.26
106 J kg
4.5
106 J
and the time required for the burner to produce this energy is
tboil
11.34
Qevaporate
14 000 Btu h
4.5 106 J
1 Btu
14 000 Btu h 1.054 10 3 J
0.31 h
18 min
In 1 hour, the energy dissipated by the runner is
E
P t
Ninety percent, or Q
300 J s 3 600 s
0.900 1.08
106 J
1.08
106 J
9.72
105 J , of this is used to evaporate bodily fluids. The mass
of fluid evaporated is
Page 11.13
Chapter 11
m
Q
Lv
9.72 10 5 J
2.41 106 J kg
0.403 kg
Assuming the fluid is primarily water, the volume of fluid evaporated in 1 hour is
V
11.35
m
0.403 kg
1 000 kg m 3
4.03
10
4
m3
10 6 cm 3
1 m3
403 cm 3
The energy required to melt 50 g of ice is
Q1
mice L f
0.050 kg 333 kJ kg
16.7 kJ
The energy needed to warm 50 g of melted ice from 0°C to 100°C is
Q2
(a)
mice cw
T
0.050 kg 4.186 kJ kg °C 100°C
20.9 kJ
If 10 g of steam is used, the energy it will give up as it condenses is
Q3
ms Lv
0.010 kg 2 260 kJ kg
22 .6 kJ
Since Q3 Q1 , all of the ice will melt. However, Q3
100°C. From conservation of energy, we find
mice L f
cw T
0°C
msteam Lv
Q1
cw 100°C
Q2 , so the final temperature is less than
T
or
T
msteam Lv
mice
cw 100°C
mice L f
msteam cw
giving
T
(b)
10 g
2.26
106
4 186 100
50 g
10 g
If only 1.0 g of steam is used, then Q3
up as it cools from 100°C to 0°C is
Q4
Since Q3
ms cw
T
1.0
10
3
50 g 3.33
10 5
4 186
ms Lv
40 °C
2.26 kJ . The energy 1.0 g of condensed steam can give
kg 4.186 kJ kg °C 100°C
0.419 kJ
Q4 is less than Q1 , not all of the 50 g of ice will melt, so the final temperature will
be 0 °C . The mass of ice which melts as the steam condenses and the condensate cools to 0°C is
m
Q3
Q4
Lf
2.26
0.419 kJ
333 kJ kg
8.0
Page 11.14
10
3
kg
8.0 g
Chapter 11
11.36
First, we use the ideal gas law (with V 0.600 L
the quantity of water vapor in each exhaled breath:
PV
nRT
m
nMwater
103 Pa
3.20
PV
RT
n
0.600
10
0.600
8.31 J mol K
3
10
m3 and T
3
m3
37.0°C
7.45
310 K
10
4
310 K ) to determine
mol
or
7.45
4
10
mol
18.0
g
1 kg
103 g
mol
1.34
10
5
kg
The energy required to vaporize this much water, and hence the energy carried from the body with each breath is
Q
mLv
1.34
5
10
kg 2.26
106 J kg
30.3 J
The rate of losing energy by exhaling humid air is then
P
11.37
Q
respiration rate
30.3
J
breath
22.0
breaths
min
1 min
60 s
11.1 W
(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 final 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
KEi
mbullet clead 0°C
3.00
10
3
kg
2.40
30.0°C
102 m s
1
mbullet vi2
2
mbullet clead 30.0°C
2
128 J kg °C 30.0°C
2
97.9 J
The mass of ice that melts when this quantity of thermal energy is absorbed is
m
11.38
(a)
Q
Lf
water
97.9 J
3.33 105 J kg
2.94
10
4
kg
103 g
1 kg
0.294 g
The rate of energy transfer by conduction through a material of area A, thickness L, with thermal conductivity
k A Th Tc L . For the given windowpane, this is
k, and temperatures Th Tc on opposite sides is P
Page 11.15
Chapter 11
P
(b)
P
t
6.8
103 J s 8.64
The thermal conductivity of concrete is k
P
11.40
(a)
25°C
1.0 m 2.0 m
0.62
0°C
10
2
103 J s
6.8
m
6.8
10 3 W
The total energy lost per day is
E
11.39
J
s m °C
0.84
kA
Th
Tc
1.3
L
104 s
108 J
5.9
1.3 J s m °C , so the energy transfer rate through the slab is
J
s m °C
20°C
5.0 m 2
12
10
2
m
1.1
10 3 J s
1.1
10 3 W
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:
1.0 10 3 m
0.020 W m K
Rskin
0.50 10 2 m
0.20 W m K
Rfat
5.0
2.5
10
10
2
2
m2 K W
m2 K W
and
3.0 10 2 m
0.50 W m K
Rtissue
6.0
10
2
m2 K W
so the total R value of the three layers taken together is
3
Rtotal
(b)
P
5.0
2.5
6.0
10
2
m2 K
W
14
10
2
m2 K
W
0.14
The rate of energy transfer by conduction through these three layers with a surface area of A
37 0 °C 37°C 37 K is
temperature difference of T
P
11.41
i 1
Ri
kA
T
L
A
T
Rtotal
, with k
2.0 m 2
0.14
0.200
m2
37 K
K W
5.3
cal
102 cm
cm °C s
1m
Page 11.16
10 2 W
4.186 J
1 cal
83.7
J
s m °C
m2 K
W
2.0 m2 and
Chapter 11
Thus, the energy transfer rate is
P
83.7
4.02
11.42
J
8.00 m 50.0 m
s m °C
J
108
402 MW
s
200 C 20.0°C
1.50 10 2 m
The total surface area of the house is
A
Aside walls
Aend walls
Agables
Aroof
where
Aside walls
2
5.00 m
10.0 m
100 m2
Aend walls
2
5.00 m
8.00 m
80.0 m2
Agables
1
2
2
Aroof
base
altitude
2 10.0 m
1
2
2
8.00 m
4.00 m tan 37.0°
24.1 m2
100 m2
4.00 m cos 37.0°
Thus,
100 m2
A
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
E
P
E
1.5
kA
t
T
4.8
t
L
10
4
kW m °C 304 m2
25.0°C
0.210 m
or
106 kJ
1.5
109 J
The volume of gas that must be burned to replace this energy is
V
11.43
R
R
Ri
0.17
E
heat of combustion
Routside
air film
0.87
Rshingles
1.32
9 300 kcal m 3
Rsheathing
3 3.70
109 J
1.5
0.45
Rcellulose
0.17
4 186 J kcal
Rdry wall
ft 2 °F
Btu h
Page 11.17
39 m 3
Rinside
air film
14
ft 2 °F
Btu h
86 400 s
Chapter 11
11.44
The rate of energy transfer through a compound slab is
A
P
(a)
T
R
, where R
For the thermopane, R
Li ki
Rpane
Rtrapped air
Rpane
2Rpane
Rtrapped air ..
Thus,
R
0.50 10 2 m
0.84 W m °C
2
1.0 10 2 m
0.0234 W m °C
0.44
m 2 °C
W
and
1.0 m 2
P
(b)
0.44
m2
23°C
52 W
°C W
For the 1.0 cm thick pane of glass:
1.0 10 2 m
0.84 W m °C
R
1.2
10
2
m 2 °C
W
so
P
11.45
1.0 m 2
1.2
10
2
23°C
m 2 °C W
1.9
1.9 kW , 37 times greater
103 W
When the temperature of the junction stabilizes, the energy transfer rate must be the same for each of the rods, or
PCu PAl 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,
PCu
kCu A
T
LCu
kAl A
T
LAl
LAl
kAl
kCu
LCu
238 W m °C
397 W m °C
PAl ,
which gives
11.46
15 cm
The energy transfer rate is
Page 11.18
9.0 cm
Chapter 11
Thus,
P
k
11.47
10 5 J kg
5.0 kg 3.33
mice L f
Q
t
P
t
58 W
8.0 h 3 600 s 1 h
k A ( T L) gives the thermal conductivity as
58 W 2.0
P L
A
T
0.80 m2
2
10
25°C
m
7.2
5.0°C
10
2
W m °C
473 K and that of the surroundings is T0
The absolute temperature of the sphere is T
black-body radiator, the emissivity is e 1 The net power radiated by the sphere is
Pnet
Ae T 4
5.67
295 K For a perfect
T04
10
8
m2
W
K4
4
0.060 m
2
4
473 K
295 K
4
or
Pnet
11.48
102 W
1.1
0.11 kW
Since 97.0% of the incident energy is reflected, 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
Assuming the sail radiates equally from both sides (so A
2
4.20
2 1.00 km
2
107 W
2.00
106 m2 ), the rate at which it
will radiate energy to a 0 K environment when it has absolute temperature T is
Prad
Ae T 4
0
5.669 6
At the equilibrium temperature, where
3.40
11.49
10
3
W
K4
T4
10
8
Prad
4.20
W
m2 K 4
2.00
106 m 2
0.03
T4
3.40
10
3
W
K4
T4
Pabsorbed , we then have
10 7 W or
The absolute temperatures of the two stars are TX
diated powers is
T
4.20 107 W
3.40 10 3 W K 4
6 000 K and TY
Page 11.19
14
333 K
12 000 K Thus, the ratio of their ra-
Chapter 11
PY
PX
11.50
AeTY4
AeTX4
TY
TX
Pnet
T04
2
Pnet
The net power radiated is
T
4
4
16
Ae T 4
T04 , so the temperature of the radiator is
1
4
Ae
If the temperature of the surroundings is T0
T
295 K
11.51
1
4
5.67
10
1.8
8
W m2 K 4
2.5
10
5
m2
0.90
103 °C
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
11.52
295 K ,
25 W
4
103 K
2.1
22°C
kA
T
L
1.2
10 4 W
397
W
m °C
0.10 m
102°C 100°C
2 .0 10 3 m
2
12 kW
The mass of the water in the heater is
m
103
V
kg
m3
50.0 gal
3.786 L
1 gal
1 m3
103 L
189 kg
The energy required to raise the temperature of the water from 20.0°C to 60.0°C is
Q
mc
T
189 kg 4 186 J kg 60.0°C
20.0°C
The time required for the water heater to transfer this energy is
t
Q
P
3.17 10 7 J
1h
4 800 J s
3 600 s
1.83 h
Page 11.20
3.17
107 J
Chapter 11
11.53
The energy conservation equation is
mPb c Pb 98°C
12°C
mice L f
mice
mw cw
mcup cCu
12°C
0°C
This gives
J
kg °C
m Pb 128
86°C
10 5 J kg
0.040 kg 3.33
0.24 kg 4 186 J kg °C
or m Pb
11.54
12°C
2.3 kg .
The energy needed is
Q
mc
T
103
The power input is
t
11.55
0.100 kg 357 J kg °C
Q
P
V c
kg
m3
T
1.00 m 3
P
4 186 J kg 40.0°C
550 W m2
6.00 m2
1.67 108 J
1h
3
3.30 10 J s 3 600 s
3.30
1.67
10 8 J
103 J s , so the time required is
14.1 h
The conservation of energy equation is
mw cw
mcup cglass
T
27°C
mCu cCu 90°C
T
This gives
T
mCu cCu 90°C
mw cw
mw cw
mcup cglass
mcup cglass
27°C
mCu cCu
or
T
11.56
(a)
0.200 387 90°C
0.400 4 186
0.400 4 186
0.300 837
0.300 837
0.200 387
27°C
29 °C
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.
Page 11.21
Chapter 11
(b)
The mass of nitrogen vaporized in a 4.0 h period is
m
11.57
P
Q
Lf
t
25 J s 4.0 h 3 600 s h
Lf
1.8 kg
105 J kg
2.01
Assuming the aluminum-water-calorimeter system is thermally isolated from the environment, Qcold
mAl cAl T f
Since Tf
cAl
Ti,Al
mw cw T f
66.3°C and Ti,cal
mw cw
mcal ccal
mAl T f
Ti, w
Ti, w
Ti, w
mcal ccal T f
Qhot , or
Ti,cal
70.0°C , this gives
Tf
Ti,Al
or
0.400 kg
J
4 186
0.040 kg
kg °C
cAl
0.200 kg 66.3
630
J
70.0
kg °C
66.3 °C
8.00
27.0 °C
102
J
kg °C
The variation between this result and the value from Table 11.1 is
%
variation
accepted value
100%
800
900 J kg °C
900 J kg °C
100%
11.1%
which is within the 15% tolerance.
11.58
(a)
37°C 273 310 K and surroundings at temperature
With a body temperature of T
T0 24°C 273 297 K , the rate of energy transfer by radiation is
Prad
(b)
Ae T 4
T04
5.669 6
10
8
m2
W
K4
2.0 m 2
0.97
310 K
4
297 K
4
1.6
2.7
102 W
10 2 W
The rate of energy transfer by evaporation of sweat is
Psweat
Q
t
mLv ,sweat
0.40 kg 2.43
t
103 kJ kg 103 J kJ
3 600 s
Page 11.22
Chapter 11
(c)
The rate of energy transfer by evaporation from the lungs is
Plungs
(d)
38
kJ
h
103 J
1 kJ
1h
3 600 s
11 W
The excess thermal energy that must be dissipated is
Pexcess
0.80Pmetabolic
10 3
0.80 2.50
kJ
h
103 J
1 kJ
1h
3 600 s
10 2 W
5.6
so the rate energy must be transferred by conduction and convection is
Pc & c
11.59
Prad
Psweat
Plungs
5.6
1.6
2.7
.11
From
Q
t
P
m
Lv
t
k A (T
T
0.500
kg
min
1 min
60 s
2.26
106
J
kg
P L
100°C
kA
1.88
104 W 0.500
238 W m °C
10
2
0.120 m
m
2
The energy added to the air in one hour is
Ptotal t
10 200 W
3 600 s
106 J
7.20
and the mass of air in the room is
m
1.3 kg m3
V
6.0 m 15.0 m 3.0 m
The change in temperature is
T
giving T
Q
mc
T0
In the steady state,
1.88
100°C) L , the temperature of the bottom surface is
100°C
Q
11.61
102 W
1.2
The rate at which energy must be added to the water is
P
11.60
Pexcess
7.2
3.5
T
PAu
20°C
106 J
10 2 kg 837 J kg °C
25°C
45°C .
PAg , or
Page 11.23
25°C
3.5
102 kg
109 °C
10 4 W
102 W
Chapter 11
kAu A
80.0°C
L
T
kAg A
T
30.0°C
L
This gives
kAu 80.0°C
T
11.62
(a)
kAu
314 80.0°C
kAg
314
427 30.0°C
427
51.2 °C
The rate work is done against friction is
P
(b)
kAg 30.0°C
f
50 N 40 m s
v
103 J s
2.0
2.0 kW
In a time interval of 10 s, the energy added to the 10-kg of iron is
Q
P t
103 J s 10 s
2.0
2.0
104 J
and the change in temperature is
T
11.63
(a)
2.0 10 4 J
10 kg 448 J kg °C
Q
mc
4.5°C
The energy required to raise the temperature of the brakes to the melting point at 660°C is
Q
mc
T
6.0 kg 900 J kg °C 660°C
20°C
3.46
4.69
105 J
106 J
The internal energy added to the brakes on each stop is
Q1
KE
1
mcar v i2
2
1
1 500 kg 25 m s
2
2
The number of stops before reaching the melting point is
N
(b)
11.64
Q
Q1
3.46
4.69
106 J
10 5 J
7 stops
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.
When liquids 1 and 2 are mixed, the conservation of energy equation is
mc1 17°C
10°C
mc2 20°C
17°C , or c2
When liquids 2 and 3 are mixed, energy conservation yields
Page 11.24
7
c
3 1
Chapter 11
mc3 30°C
28°C
20°C , or c3
mc2 28°C
Then, mixing liquids 1 and 3 will give mc1 T
10°C
28
c1
3
4 c2
mc3 30°C
T
or
c1 10°C
T
11.65
(a)
c1
V
where
c
1
28°C
28 3
Q added to the volume V of liquid that flows through the calorimeter in time
( V )c ( T ) .Thus, the rate of adding energy is
t is
V
t
T
From the result of part (a), the specific heat is
Q
T
t
V
40 J s
t
0.72 g
103 g
1 kg
J
g °C
2.7
(a)
28 3 30°C
t is the flow rate through the calorimeter.
c
11.66
10°C
c3
The internal energy
Q ( m)c ( T )
Q
t
(b)
c3 30°C
2.7
The surface area of the stove is Astove
Astove
2
0.200 m
2
2
5
9
(70.0°F
5.8°C 3.5 cm 3 s
103 J kg °C
Aends
Acylindrical
32.0°F)
5
9
2
r2
2 rh , or
side
0.200 m 0.500 m
The temperature of the stove is Ts
room is Tr
cm3
0.880 m2
(400°F
32.0°F)
204°C
477 K while that of the air in the
21.1°C
294 K If the emissivity of the stove is e
power radiated to the room is
P
Astove e Ts4
5.67
10
8
Tr4
W m2 K 4
0.880 m2
or
Page 11.25
0.920
477 K
4
294 K
4
0.920 , the net
Chapter 11
P
(b)
103 W
2.03
The total surface area of the walls and ceiling of the room is
A
4 Awall
Aceiling
4
8.00 ft 25.0 ft
25.0 ft
2
103 ft 2
1.43
If the temperature of the room is constant, the power lost by conduction through the walls and ceiling must
A Th Tc
Ri , the
equal the power radiated by the stove. Thus, from thermal conduction equation, P
net R value needed in the walls and ceiling is
A Th
Ri
103 ft 2
1.43
Tc
P
70.0°F
103
2.03
32.0°F
Js
1 054 J
1 Btu
1h
3 600 s
or
7.84 ft 2 °F h Btu
Ri
11.67
A volume of 1.0 L of water has a mass of m
103 kg m3 1.0
V
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 T
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 m
0.50 600 W m2
2
59 W
4
so the time required to evaporate the water is
t
11.68
(a)
Q
P
2.59 106 J
59 J s
4.4
From the thermal conductivity equation,
10 4 s
P
1h
3 600 s
k A Th
12 h
Tc
L , the total energy lost by conduction
through the insulation during the 24-h period will be
Q
P1 12.0 h
P2 12.0 h
kA
L
37.0°C
or
Page 11.26
23.0°C
37.0°C
16.0°C
12.0 h
Chapter 11
0.012 0 J s m°C 0.490 m 2
Q
0.095 0 m
14.0°C
21.0°C 12.0 h
3 600 s
1h
9.36
10 4 J
The mass of molten wax which will give off this much energy as it solidifies (all at 37°C) is
(b)
9.36 10 4 J
205 103 J kg
Q
Lf
m
0.457 kg
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 final temperature of 50.0°C is
Qm
mcsteam
m
T
2.01
1
mLv
mcliquid
T
water
103 J kg °C 130°C
2
100°C
2.26
10 6 J kg
4 186 J kg °C 100°C
106
m 2.53
50.0°C
J kg
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
Qneeded
mw cw
2.76
mg cg
T
0.200 kg 4 186 J kg °C
0.100 kg 837 J kg °C
30.0°C
10 4 J
Equating the energy available from the steam to the energy required gives
m 2.53
11.70
106 J kg
2.76
104 J or
m
2.76 10 4 J
2.53 106 J kg
0.010 9 kg
10.9 g
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
L37°C
v
cwater
L37°C
v
2.5
T
L100°C
v
4 186 J kg °C 100°C
or
106 J kg
Page 11.27
37°C
2.26
106 J kg
Chapter 11
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
1 000 W m2
t
0.60 m2
3 600 s
2.2
106 J
The quantity of water this much energy could evaporate from your body is
m
2.2 106 J
2.5 106 J kg
Q
L37°C
v
0.9 kg
The volume of this quantity of water is
V
m
0.9 kg
kg m3
10
103
3
m3
1L
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 first 50 minutes, the energy input is used converting m kilograms of ice at 0°C into liquid water at 0°C.
mL f
m 3.33 105 J kg , so the constant power input must be
The energy required is Q1
P
Q1
t1
m 3.33
105 J kg
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
P
Q2
t 2
m
10 kg c
t
T
m
10 kg 4 186 J kg °C 2.0°C
10 min
2
Since the power input is the same in the two periods, we have
m 3.33
105 J kg
m
10 kg 4 186 J kg °C 2.0°C
50 min
which simplifies to 8.0 m
m
11.72
(a)
10 kg
7.0
10 min
m
10 kg , or
1.4 kg
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.
Page 11.28
Chapter 11
(b)
The total energy that must be removed is
Q
Qcool water
Qfreeze
to 0°C
Qcool ice
at 0°C
to
mw cw 0°C
8.00°C
Ti
mw L f
mw cice T f
0°C
or
Q
75.0
3.25
11.73
(a)
10
3
kg
10 4 J
4 186
J
20.0°C
kg °C
3.33
10 5
J
kg
2 090
J
kg °C
32.5 kJ
In steady state, the energy transfer rate is the same for each of the rods, or
PAl
PFe . Thus, kAl A
100°C
L
T
kFe A
T
0°C
L
giving
T
(b)
If L
kAl
kAl
kFe
15 cm and A
Q
PAl t
3.6
10 4 J
100°C
238
238 79.5
100°C
75.0°C
5.0 cm2 , the energy conducted in 30 min is
238
W
m °C
5.0
10
4
m2
36 kJ
Page 11.29
100°C 75.0°C
0.15 m
1 800 s
8.00°C