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An ideal heat engine working between temperature `T_(1)` and `T_(2)` has an efficiency `eta`, the new efficiency if both the source and sink temperature are doubled, will beA. `eta/2`B. `eta`C. `2 eta`D. `3 eta` |
Answer» Correct Answer - B In first case , `eta_(1) = (T_1-T_2)/(T_1)` in second case `eta_(2) = (2T_(1)-2T_(2))/(2T_1) = (T_1-T_2)/(T_1) = eta` . |
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