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Unformatted text preview: at capacity of water (see Table 12.2 in the text), and ∆T = 15 C° is the difference between
the higher initial temperature Ti and the lower final temperature Tf. For the freezing process,
the heat loss is given by Q2 = mLf (Equation 12.5), where Lf is the latent heat of fusion of
water (see Table 12.3 in the text). Therefore, the magnitude QC of the total heat to be
extracted from the water each day is 808 THERMODYNAMICS QC = cm ∆T
123 + Heat removed
to cool the water (1) mLf
{
Heat removed
to freeze the water The ice maker operates in the same way as a refrigerator or an air conditioner, so the ice
Q
maker’s coefficient of performance (COP) is given by COP = C (Equation 15.16), where
W
W is the magnitude of the work done by the ice maker in extracting the heat of magnitude
QC from the water. Solving Equation 15.16 for QC, we find that QC = W ( COP ) (2) The wattage P of the ice maker is the rate at which it does work. This rate is given by
W
P=
(Equation 6.10a). Therefore, the magnitude W of the total amount of work the ice
t
maker performs in one day is
W = Pt (3) SOLUTION Solving Equation (1) for the desired mass m of ice, we find that
m ( c ∆T + Lf ) = QC or m= QC
c ∆T + Lf (4) Substituting Equation (2) into Equation (4), then, yields
m= QC
c ∆T + Lf = W ( COP )
c ∆T + Lf (5) Lastly, Substituting Equation (3) into Equation (5), we obtain
m= W COP
Pt ( COP )
=
c ∆T + Lf c ∆T + Lf The elapsed time of t = 1 day is equivalent to t = 8.64×104 s (see the inside of the front
cover of the text). Therefore, the maximum amount of ice that the ice maker can produce in
one day is Chapter 15 Problems (
)( 809 ) ( 225 W ) 8.64 ×104 s ( 3.60 )
Pt ( COP )
m=
=
= 176 kg
c ∆T + Lf 4186 J/ kg ⋅ Co 15.0 Co + 33.5 × 104 J/kg ( ) 71. SSM REASONING The conservation of energy applies to the air conditioner, so that
QH = W + QC , where QH is the amount of heat put into the room by the unit, QC is the amount of heat removed from the room by the unit, and W is the amount of work needed to
operate the unit. Therefore, a net heat of QH − QC = W is added to and heats up the room.
To find the temperature rise of the room, we will use the COP to determine W and then use
the given molar specific heat capacity.
SOLUTION Let COP denote the coefficient of performance. By definition (Equation
15.16), COP = QC / W , so that W= QC
COP = 7.6 ×104 J
= 3.8 ×104 J
2.0 The temperature rise in the room can be found as follows: QH − QC = W = CV n ∆T .
Solving for ∆T gives
∆T = W
CV n = W ( 5 R) n
2 = 3.8 ×104 J
= 0.48 K
5 8.31 J/ mol ⋅ K 3800 mol
(
) (
)
2 72. REASONING AND SOLUTION The magnitude of the heat removed from the ice QC is
4 5 QC = mLf = (2.0 kg)(33.5 × 10 J/kg) = 6.7 × 10 J The magnitude of the heat leaving the refrigerator QH is therefore,
5 5 QH = QC (TH/TC) = (6.7 × 10 J)(300 K)/(258 K) = 7.8 × 10 J The magnitude of the work done by the refrigerator is therefore,
5 W = QH − QC = 1.1 × 10 J 810 THERMODYNAMICS 6 At $0.10 per kWh (or $0.10 per 3.6 × 10 J), the cost is ( ) $0.10
1.1 × 105 J = $3.0 × 10−3 = 0.30 cents
6
3.6 × 10 J 73. SSM WWW REASONING Let the coefficient of performance be represented by the
symbol COP. Then according to Equation 15.16, COP = QC / W . From the statement of
energy conservation for a Carnot refrigerator (Equation 15.12), W = QH − QC . Combining
Equations 15.16 and 15.12 leads to
COP = QC
QH − QC = QC / QC ( QH − QC ) / QC = 1 ( QH / QC ) − 1 Replacing the ratio of the heats with the ratio of the Kelvin temperatures, according to
Equation 15.14, leads to
1
COP =
(1)
TH / TC − 1
The heat QC that must be removed from the refrigerator when the water is cooled can be
calculated using Equation 12.4, QC = cm∆T ; therefore, W= QC
COP = cm ∆T
COP (2) SOLUTION
a. Substituting values into Equation (1) gives
COP = 1
=
TH
−1
TC 1
=
( 20.0 + 273.15 ) K − 1
( 6.0 + 273.15) K 2.0 ×101 b. Substituting values into Equation (2) gives W= cm ∆T 4186 J/ ( kg ⋅ C° ) ( 5.00 kg ) ( 20.0 °C − 6.0 °C ) =
= 1.5 ×104 J
1
COP
2.0 ×10 Chapter 15 Problems 811 74. REASONING The efficiency of the Carnot engine is, according to Equation 15.15, e = 1− TC
TH = 1− 842 K 1
=
1684 K 2 The magnitude of the work delivered by the engine is, according to Equation 15.11,
W = e QH = 1
2 QH The heat pump removes an amount of heat QH from the cold reservoir. Thus, the amount
of heat Q ′ delivered to the hot reservoir of the heat pump is
Q′ = QH + W = QH + 1
2 QH = 3
2 QH Therefore, Q′ / QH = 3 / 2 . According to Equation 15.14, Q′ / QH = T ′ / TC , so T ′ / TC = 3 / 2 .
SOLUTION Solving for T ′ gives
T ′ = TC = (842 K) = 1.26 ×103 K
3
2 3
2 75. REASONING The total entropy change ∆Suniverse of the universe is the sum of the entropy
changes of the hot and cold reservoir. For each reservoir, the entropy change is given by
Q
∆S = (Equation 15.18). As indicated by the label “R,” this equation applies only to T R
reversible processes. For the two irreversible engines, therefore, we apply this equation to
an imaginary process that removes the given heat from the hot reservoir reversibly and
rejects the g...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.
 Spring '13
 CHASTAIN
 Physics, The Lottery

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