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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)` |
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Answer» Correct Answer - C `eta =1-(T_(2))/(T_(1))` `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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