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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