Physics Solution Manual for 1100 and 2101

# According to equation 161 then the wavelength must be

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Unformatted text preview: t. We will apply this expression to each engine and utilize the fact that in each case the same work is done. We expect to find that engine 2 requires less input heat to do the same amount of work, because it has the greater efficiency. SOLUTION Applying Equation 15.11 to each engine gives W = e1 QH1 14244 4 3 and W = e2 QH2 14 244 4 3 Engine 1 Engine 2 Since the work is the same for each engine, we have e1 QH1 = e2 QH2 or e QH2 = 1 QH1 e 2 It follows, then, that e 0.18 QH2 = 1 QH1 = ( 5500 J ) = 3800 J e 0.26 2 which is less than the input heat for engine 1, as expected. 87. SSM REASONING AND SOLUTION a. Since the energy that becomes unavailable for doing work is zero for the process, we have from Equation 15.19, Wunavailable = T0 ∆Suniverse = 0 . Therefore, ∆Suniverse = 0 and according to the discussion in Section 15.11, the process is reversible . Chapter 15 Problems 821 b. Since the process is reversible, we have (see Section 15.11) ∆S universe = ∆Ssystem + ∆Ssurroundings = 0 Therefore, ∆Ssurroundings = −∆Ssystem = –125 J/K 88. REASONING AND SOLUTION We know that QC = QH − W = 14 200 J − 800 J = 13 400 J Therefore, Q 13 400 J TC = TH C = ( 301 K ) = 284 K Q 14 200 J H 89. REASONING The coefficient of performance of an air conditioner is QC / W , according to Equation 15.16, where QC is the magnitude of the heat removed from the house and W is the magnitude of the work required for the removal. In addition, we know that the first law of thermodynamics (energy conservation) applies, so that W = QH − QC , according to Equation 15.12. In this equation QH is the magnitude of the heat discarded outside. While we have no direct information about QC and QH , we do know that the air conditioner is a Carnot device. This means that Equation 15.14 applies: QC / QH = TC / TH . Thus, the given temperatures will allow us to calculate the coefficient of performance. SOLUTION Using Equation 15.16 for the definition of the coefficient of performance and Equation 15.12 for the fact that W = QH − QC , we have Coefficient of performance = QC W = QC QH − QC = QC / QH 1 − QC / QH Equation 15.14 applies, so that QC / QH = TC / TH . With this substitution, we find 822 THERMODYNAMICS Coefficient of performance = = 90. REASONING QC / QH 1 − QC / QH TC TH − TC = = TC / TH 1 − TC / TH 297 K = 21 ( 311 K ) − ( 297 K ) The efficiency eCarnot of a Carnot engine is eCarnot = 1 − TC TH (Equation 15.15), where TC and TH are, respectively, the Kelvin temperatures of the cold and hot reservoirs. After the changes are made to the temperatures, this same equation still applies, except that the variables must be labeled to denote the new values. We will use a “prime” for this purpose. From the original efficiency and the information given about the changes made to the temperatures, we will be able to obtain the new temperature ratio and, hence, the new efficiency. SOLUTION After the reservoir temperatures are changed, the engine has an efficiency that, according to Equation 15.15, is T′ ′ eCarnot = 1 − C ′ TH where the “prime” denotes the new engine. Using unprimed symbols to denote the original ′ ′ engine, we know that TC = 2TC and TH = 4TH . With these substitutions, the efficiency of the new engine becomes T′ 2T 1T ′ (1) eCarnot = 1 − C = 1 − C = 1 − C 2T ′ 4TH TH H To obtain the original ratio TC / TH , we use Equation 15.15: eCarnot = 1 − TC TH TC or TH = 1 − eCarnot Substituting this original temperature ratio into Equation (1) gives 1T ′ eCarnot = 1 − C 2T H 1 = 1 − 2 (1 − eCarnot ) = 1 2 (1 + eCarnot ) = 1 (1 + 0.40 ) = 0.70 2 Chapter 15 Problems 823 91. SSM REASONING AND SOLUTION a. Starting at point A, the work done during the first (vertical) straight-line segment is W1 = P1∆V1 = P1(0 m3) = 0 J For the second (horizontal) straight-line segment, the work is W2 = P2∆V2 = 10(1.0 × 104 Pa)6(2.0 × 10–3 m3) = 1200 J For the third (vertical) straight-line segment the work is W3 = P3∆V3 = P3(0 m3) = 0 J For the fourth (horizontal) straight-line segment the work is W4 = P4∆V4 = 15(1.0 × 104 Pa)6(2.0 × 10–3 m3) = 1800 J The total work done is W = W1 + W2 + W3 + W4 = +3.0 × 103 J b. Since the total work is positive, work is done by the system . 92. REASONING AND SOLUTION a. The work is the area under the path ACB. There are 48 "squares" under the path, so that ( )( ) W = − 48 2.0 × 104 Pa 2.0 × 10–3 m3 = −1900 J The minus sign is included because the gas is compressed, so that work is done on it. Since there is no temperature change between A and B (the line AB is an isotherm) and the gas is ideal, ∆U = 0, so Q = ∆U + W = W = −1900 J b. The negative answer for W means that heat flows out of the gas. 824 THERMODYNAMICS 93. SSM REASONING AND SOLUTION The change in volume is ∆V = – sA, where s is the distance through which the piston drops and A is the piston area. The minus sign is included because the volume decreases. Thus, s= – ∆V A The ideal gas law states that ∆V = nR∆T/P. But...
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