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Unformatted text preview: Homework 10 Solution Physics 404 - Spring 2011 Homework 10 Solutions Question 1 K+K Chapter 8, Problem 5. The best case scenario is that the plant operates as a thermodynamic heat engine on the Carnot cycle and has an efficiency given by: c = W Q h = T h- T l T h (1) Also Q l = Q h T l /T h , which follows from the fact that the process is reversible (hence dS=0). Using this we can write the work in terms of Q l instead of Q h as: W = Q l T h- T l T l (2) (pay attention to the denominator). We are given that the upper limit to the heat that can be released in one second is Q l = 1 . 5 10 9 J . Using this we can obtain the maximum amount of work per second W = 1 . 5 10 9 J 500 o C- 20 o C 293 K = 2 . 46 10 9 J (3) or P = W = 2 . 46 GW . Note that in the numerator we used temperatures in units of Celsius. This is ok, because the difference of two temperatures is the same in both units. The denominator on the other hand has to be in the absolute temperature scale....
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