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A Carnot engine works as a refrigerator in between `250K` and `300K`. If it acquires `750` calories form heat source at low temperature, then what is the heat generated at higher temperature (in calories)? |
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Answer» Correct Answer - `900` Calories `eta = (Q_(2))/(Q_(1)-Q_(2)) = (T_(2))/(T_(1)-T_(2)) rArr (750)/(Q_(1) - 750) = (250)/(300-250)` `Q_(1) = 900` Calories |
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