P21_023 - 23(a The eciency is = TH TL(235 115 K = = 0.236 =...

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23. (a) The efficiency is ε = T H T L T H = (235 115) K (235 + 273) K =0 . 236 = 23 . 6% . We note that a temperature diference has the same value on the Kelvin and Celsius scales. Since the temperatures in the equation must be in Kelvins, the temperature in the denominator is converted
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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