CHAPTER ELEVEN SOLUTIONS
CHAPTER ELEVEN SOLUTIONS
11.1 From Q = mgh=PE, we find 1 food cal =
1000 cals= 4186 Joules
Q= 50 (9.8)(10) 1cal /4186J =1.17 Cal
Q
11.4 From Q = mcΔT, we find: ΔT = mc =
1200 J
(0.05 kg)(387 J/kg °C) = 62 ° C.
Thus, the final temperature is T = 25°C +
62°C = 87 °C
11.7 Consider a 1 kg mass of water:
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CHAPTER ELEVEN SOLUTIONS
Q = ΔPE = mgh = (1.0 kg)(9.8 m/s2)(50
m) = 490 J
Q
Also, ΔT = mc =
490 J
=
0.117
°
C.
(1.0 kg)(4.19 X 103 J/kg °C)
So, Tf = 10.1 C°.
11.9 Let us find the heat extracted from the
system in one minute.
Q = mcupccupΔT + mwatercwaterΔT, or
Q = [(0.2 kg)(900 J/kg °C) + (0.8
kg)(4186 J/kg °C)](1.5°C) = 5293 J.
If this much heat is removed each minute,
the rate of removal of heat is
Q 5293 J
P=
= 60 s = 88.2 J/s = 88 W.
Δt
11.11 The heat needed to raise the temperature
of the water to 25 °C is:
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CHAPTER ELEVEN SOLUTIONS
Qneeded = (0.5 kg)(4186 J/kg °C)(5 °C) =
1.05 X 104 J.
The heat received from each pellet is:
Qpellet = (10-3 kg)(128 J/kg °C)(175 °C)
= 22.4 J/pellet
Thus, the number of pellets needed is:
Qneeded
1.05 X 104 J
n= Q
= 22.4 J/pellet = 470
pellet
pellets.
11.12 Our heat loss = heat gain
equation becomes: mironcironΔTiron
= mwatercwΔTw
or, (0.4 kg)(448 J/kg °C)(500 °C - T) =
(20 kg)(4186 J/kg °C)(T - 22 °C).
From which, we find: T = 23 °C.
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CHAPTER ELEVEN SOLUTIONS
11.13 (heat gain = heat loss) becomes:
mwatercwΔTw = mgoldcgΔTg.
Thus, mwater(4186 J/kg °C)(25 °C) = (3
kg)(129 J/kg °C)(50 °C), or
mwater = 0.185 kg = 190 g.
11.14 (heat gain = heat loss) becomes:
mcupccΔTc + mwcwΔTw +
mstirrercsΔTs = msilvercsilΔTsil , or
mcup(900 J/kg °C)(5°C) + (0.225
kg)(4186 J/kg °C)(5°C)
+ (0.04 kg)(387 J/kg °C)(5 °C) =
(0.4 kg)(234 J/kg °C)(55 °C).
We find: mcup = 80 X 10-3 kg = 80 g.
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CHAPTER ELEVEN SOLUTIONS
11.15 ΔQsystem = 0, since the system
does not exchange heat with the
environment.
ΔQsystem = (0.2 kg)(4.19 x 103
J/kg°C)(Tf - 10°C) + (0.3 kg)(900
J/kg°C)(Tf - 10°C)
+ (0.1 kg)(4.19 x 103
J/kg°C)(Tf - 100°C) = 0
Which gives Tf = 35°C.
11.18 First we will assume that both the aluminum and
water gain heat while the copper loses the heat.
mwcwΔT + macaΔT = mcccΔT
(0.250 kg)(4186 J/kg °C)(T-20) + (0.4 kg)(900
J/kg °C)](Τ−26°C) = 0.100 (386 J/kg C)(100-T)
If this much heat is removed T = 23.6 degrees C.
This means our assumption that the aluminum
gained the heat is incorrect! What if we assumed
water gained the heat (while the other 2 lost it)
(0.250 kg)(4186 J/kg °C)(T-20) + (0.4 kg)(900
J/kg °C)](26°C-T) = 0.100 (386 J/kg C)(100-T)
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CHAPTER ELEVEN SOLUTIONS
= 23.6 degrees C as well!
11.19 heat loss = heat gain
mcuccuΔTcu = malcalΔTal +
mwcwΔTw but ΔTw = 0
(0.200 kg)(387 J/kg °C)(85 °C - 25
°C) = mal(900 J/kg °C)(25 °C - 5 °C)
mal = 0.26 kg = 260 g
11.20 Q = (heat to melt) + (heat to warm
melted ice to 100 °C) + (heat to
vaporize 5 g)
Q = (5 X 10-2 kg)(3.33 X 105 J/kg) +
(5 X 10-2 kg)(4186 J/kg °C)(100 °C)
+ (5 X 10-3 kg)(2.26 X 106
J/kg) = 4.9 X 104 J
11.21 heat gain = heat loss
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CHAPTER ELEVEN SOLUTIONS
(heat to melt ice) + (Heat to warm
melted ice to Tf) = (loss of heat by 1
kg of water)
(0.100 kg)(3.34 X 105 J/kg) + (0.100
kg)(4186 J/kg °C)(Tf)
= (1 kg)(4186 J/kg
°C)(80 - Tf)
Tf = 66 °C
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CHAPTER ELEVEN SOLUTIONS
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