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A Carnot engine works between `120^(@)C` and `30^(@)C`. Calculate the efficiency. If the power produced by the engine is `400W`, calculate the heat abosorbed from the source. |
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Answer» Efficiency, `eta =(T_(2)-T_(1))/(T_(2))` Here `T_(2) = 273 +120 = 393K` `T_(1) = 273 +30 = 303 K` `:. eta = (393-303)/(393) = 0.229 = 22.9%` Again, `eta = (Q_(2)-Q_(1))/(Q_(2)) = (W)/(Q_(2))` Thus, heat absorbed form the source. `Q_(2) = (W)/(eta) = (400)/(0.229) = 1747W` |
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