Physics Solution Manual for 1100 and 2101

19 where t0 is the kelvin temperature of the coldest

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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.

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