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An ideal heat engine has an efficiency `eta`. The cofficient of performance of the engine when driven backward will beA. `1-((1)/(eta))`B. `eta-((1)/(eta))`C. `((1)/(eta))-1`D. `((1)/(1-eta))` |
Answer» Correct Answer - C `eta = 1-(T_2)/(T_1)` and `omega = (T_2)/(t_1-T_2) = (T_2//T_1)/(1-(T_2//T_1)) = (1-eta)/(eta) = 1/(eta)-1`. |
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