ch20-p071 - Δ S = (260 J)(1/300 K – 1/400 K) = 0.217...

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71. The change in entropy in transferring a certain amount of heat Q from a heat reservoir at T 1 to another one at T 2 is Δ S = Δ S 1 + Δ S 2 = Q (1/ T 2 1/ T 1 ). (a) Δ S = (260 J)(1/100 K – 1/400 K) = 1.95 J/K. (b) Δ S = (260 J)(1/200 K – 1/400 K) = 0.650 J/K. (c)
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Unformatted text preview: Δ S = (260 J)(1/300 K – 1/400 K) = 0.217 J/K. (d) Δ S = (260 J)(1/360 K – 1/400 K) = 0.072 J/K. (e) We see that as the temperature difference between the two reservoirs decreases, so does the change in entropy....
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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