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HW11-solutions

HW11-solutions - baum(lmb2768 HW11 distler(56295 This...

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baum (lmb2768) – HW11 – distler – (56295) 1 This print-out should have 24 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points A particular engine has a power output of 6 kW and an e ffi ciency of 14%. If the engine expels 11470 J of thermal energy in each cycle, find the heat absorbed in each cycle. Correct answer: 13337 . 2 J. Explanation: Let : P = 6 kW , = 14% = 0 . 14 , and Q c = 11470 J . = W Q h = Q h - Q c Q h = 1 - Q c Q h Q c Q h = 1 - Q h = Q c 1 - = 11470 J 1 - 0 . 14 = 13337 . 2 J . 002 (part 2 of 2) 10.0 points Find the time for each cycle. Correct answer: 0 . 311202 s. Explanation: The net work done by a heat engine through a cyclic process ( Δ U = 0) is W = Q h - Q c , and from P = W t t = Q h - Q c P = 13337 . 2 J - 11470 J 6 kW 1 kW 1000 W = 0 . 311202 s . 003 10.0 points What is the entropy decrease in 1 . 31 mol of helium gas which is cooled at 1 atm from room temperature 290 K to a final temperature of 5 . 16 K? The specific heat of helium is 21 J / mol · K. Correct answer: 110 . 836 J / K. Explanation: Let : n = 1 . 31 mol , T 1 = 290 K , and T 2 = 5 . 16 K . Since in this cooling process pressure is con- stant, d Q = n c p dT , where n is the number of moles of helium. The change of entropy is Δ S = d Q T = T 2 T 1 n c p d T T = n c p log T 2 T 1 = (1 . 31 mol)(21 J / mol · K) × log 5 . 16 K 290 K = - 110 . 836 J / K , a decrease of 110 . 836 J / K . 004 10.0 points An ice cube melts on the warm sidewalk on a hot summer day. Let the entropies of the ice cube, of the pavement and of the ice cube- sidewalk system be S ice , S sw and S sys . What happens to these entropies? 1. S ice increases, S sw increases, S sys in- creases 2. S ice decreases, S sw increases, S sys de- creases 3. S ice increases, S sw decreases, S sys does not change.

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baum (lmb2768) – HW11 – distler – (56295) 2 4. S ice increases, S sw decreases, S sys in- creases correct Explanation: Heat is added to the ice cube, increasing its entropy. Heat is removed from the sidewalk, decreasing its entropy. The total entropy change, however, is positive. 005 (part 1 of 2) 10.0 points Use Stefan’s law to calculate the to- tal power radiated per unit area by a tungsten filament at a temperature of 1830 K. The Stephan-Boltzmann constant is 5 . 6696 × 10 - 8 W / m 2 / K 4 . Assume the fila- ment is an ideal radiator. Correct answer: 6 . 35853 × 10 5 W / m 2 . Explanation: Let : T = 1830 K . From Stefan’s law p P A = σ T 4 = (5 . 6696 × 10 - 8 W / m 2 / K 4 )(1830 K) 4 = 6 . 35853 × 10 5 W / m 2 .
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HW11-solutions - baum(lmb2768 HW11 distler(56295 This...

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