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Unformatted text preview: t = QH
t − Power = 2.4 × 109 J/s The rejected heat is carried away by the flowing water and, according to Equation 12.4,
QC = cm ∆T . Therefore,
QC
t = cm ∆T
t or QC
t m = c ∆T
t Solving the last equation for ∆t, we have
∆T = QC
tc(m / t ) = QC / t
c (m / t ) = 2.4 × 109 J/s ( 4186 J/ ( kg ⋅ C° ) 1.0 ×105 kg/s ) = 5.7 C° 804 THERMODYNAMICS 63. SSM WWW REASONING AND SOLUTION Equation 15.14 holds for a Carnot air
conditioner as well as a Carnot engine. Therefore, solving Equation 15.14 for QC , we
have
T
QC = QH C
T
H 299 K 5
5 = 6.12 × 10 J = 5.86 × 10 J 312 K ( ) 64. REASONING The magnitude QC of the heat removed from the inside compartment (cold
reservoir) of the refrigerator is found from the first law of thermodynamics:
QC = QH − W (Equation 15.12), where QH is the magnitude of the heat deposited in the
kitchen (hot reservoir) and W is the magnitude of the work done. For a Carnot refrigerator,
QC TC
the magnitudes of the heat removed and deposited are related by
(Equation
=
QH TH
15.14), where TC and TH are, respectively, the Kelvin temperatures of the inside
compartment and the kitchen.
SOLUTION Solving QC
QH = TC
TH (Equation 15.14) for QH yields T QH = QC H TC (1) Substituting Equation (1) into QC = QH − W (Equation 15.12), we obtain
T
QC = QC H TC − W or QC = W TH T − 1
C Therefore, the magnitude of the heat that can be removed from the inside compartment is
QC = W TH T − 1
C = 2500 J
= 3.1× 104 J
299 K
−1
277 K Chapter 15 Problems 65. REASONING 805 The coefficient of performance COP is defined as COP = QC / W (Equation 15.16), where QC is the magnitude of the heat removed from the cold reservoir
and W is the magnitude of the work done on the refrigerator. The work is related to the
magnitude QH of the heat deposited into the hot reservoir and QC by the conservation of
energy, W = QH − QC . Thus, the coefficient of performance can be written as (after some
algebraic manipulations)
COP = 1
QH
QC −1 The maximum coefficient of performance occurs when the refrigerator is a Carnot
refrigerator. For a Carnot refrigerator, the ratio QH / QC is equal to the ratio TH / TC of the
Kelvin temperatures of the hot and cold reservoirs, QH / QC = TH / TC (Equation 15.14).
SOLUTION Substituting QH / QC = TH / TC into the expression above for the COP gives COP = 1
TH
−1
TC = 1
= 13
296 K
−1
275 K 66. REASONING According to the first law of thermodynamics (see Equation 15.12), the
magnitude W of the work required is given by W = QH − QC , where QH is the
magnitude of the heat deposited in the hot reservoir and QC is the magnitude of the heat
removed from the cold reservoir. Since the air conditioners are Carnot devices, we know
that the ratio QC / QH is equal to the ratio TC / TH of the Kelvin temperatures of the cold
and hot reservoirs: QC / QH = TC / TH (Equation 15.14).
SOLUTION Combining Equations 15.12 and 15.14, we have
Q T W = QH − QC = H − 1 QC = H − 1 QC
T Q C C Applying this result to each air conditioner gives 806 THERMODYNAMICS Unit A 309.0 K W =
− 1 ( 4330 J ) = 220 J 294.0 K Unit B 309.0 K W =
− 1 ( 4330 J ) = 120 J 301.0 K We can find the heat deposited outside directly from Equation 15.14 by solving it for QH . QC
QH = TC
TH T QH = H QC
T C or Applying this result to each air conditioner gives
Unit A 309.0 K QH = ( 4330 J ) = 4550 J 294.0 K Unit B 309.0 K QH = ( 4330 J ) = 4450 J 301.0 K 67. REASONING Since the refrigerator is a Carnot device, we know that QC
QH = TC
TH (Equation 15.14). We have values for TH (the temperature of the kitchen) and QC (the
magnitude of the heat removed from the food). Thus, we can use this expression to
determine TC (the temperature inside the refrigerator), provided that a value can be obtained
for QH (the magnitude of the heat that the refrigerator deposits into the kitchen). Energy
conservation dictates that QH = W + QC (Equation 15.12), where W is the magnitude of
the work that the appliance uses and is known.
SOLUTION From Equation 15.14, it follows that QC
QH = TC
TH or Q TC = TH C Q H Substituting QH = W + QC (Equation 15.12) into this result for TC gives Chapter 15 Problems Q QC
TC = TH C = TH QH W + QC 807 2561 J = ( 301 K ) = 275 K 241 J + 2561 J 68. REASONING AND SOLUTION
a. The work is
W = QH − QC = 3140 J − 2090 J = 1050 J
b. The coefficient of performance (COP) is COP = QH
W = 3140 J
= 2.99
1050 J 69. REASONING AND SOLUTION Suppose this device were a Carnot engine instead of a
heat pump. We know that its efficiency e would be
e = 1 − (TC/TH) = 1 − [(265 K)/(298 K)] = 0.111 The efficiency, however, is also given by e= W
QH Since the heat pump’s coefficient of performance COP is COP = QH / W , we have that
COP = QH
W = 1
1
=
=
e
0.111 9.03 70. REASONING The mass m of ice that the ice maker can produce in one day depends upon
the magnitude QC of the heat it can extract from water as the water cools from Ti = 15.0 °C
to Tf = 0 °C and then freezes at 0 °C. To cool a mass m of water from Ti to Tf requires
extracting an amount of heat Q1 = cm ∆T (Equation 12.4) from it, where c is the specific
he...
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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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