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Unformatted text preview: Prof. Adrian Lee Physics 7B Fall 07 Final Exam Solutions Problem 1 a The efficiency of the engine is a simple computation from the definition: e r = W Q in = 600 J 1600 J = 3 8 . (1) (“r” stands for “real” here.) The Carnot engine efficiency depends on the heat reservoir temperatures: e C = T H T L T H = 850 K 400 K 850 K = 9 17 ≈ . 53 . (2) (“C” stands for “Carnot” here.) Notice that the Carnot efficiency is bigger than the real efficiency. b The total entropy change of the universe consists of the entropy change of the engine and the entropy change of the environment. The engine itself is running through a cycle, so its entropy doesn’t change because entropy is s state function. The environment, however, is not running through a cycle. Each of the heat reservoirs exchanges heat with the engine, but even though the reservoirs are losing/gaining heat they remain at the same temperature because they’re so big. So we can use the fact that Δ S = Q/T when temperature is constant. The high temperature reservoir loses heat, and so loses entropy, while the low temperature reservoir gains heat (and entropy). In equations, we have Δ S universe = Δ S engine + Δ S env = Δ S env (3) = Δ S H + Δ S L = Q in T H + Q out T L (4) = Q in T H + Q in W T L = 1600 J 850 K + 1600 J 600 J 400 K (5) ≈ . 62 J/K . (6) 1 c For a Carnot engine, we can use the same procedure as before except we will also use the relation W = Q in e C which relates the work to the heat input. Note that the Q and W appearing here are not necessarily the same as those above. Δ S universe = Δ S engine + Δ S env = Δ S env (7) = Δ S H + Δ S L = Q in T H + Q out T L (8) = Q in T H + Q in W T L = Q in T H + Q in Q in e C T L (9) = Q in 1 T H + 1 e C T L = Q in 1 T H + 1 T L 1 T L + 1 T H (10) = 0 . (11) d The difference in work done, assuming they have the same energy input Q in , is equal to Q in ( e C e r ). We can also go back to part b and use the formulas we got for the entropy change in the universe for a cycle of the real engine. Δ S universe = Q in T H + Q in W T L (12) = Q in T H + Q in Q in e r T L (13) = Q in 1 T H + 1 e r T L . (14) Now we have T L Δ S universe = Q in T L T H + 1 e r = Q in ( e C e r ) = Δ W . (15) Problem 2 Throughout this problem we will normalize our potentials so that the negative side of the applied potential difference is at V = 0....
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 Fall '08
 Packard
 Physics, Heat, Magnetic Field, Electric charge, Qin Qin

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