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A reversible engine converts one-sixth of the heat input into work. When the temperature of the sink is reduced by 62^(@)C, the efficiency of the engine is doubled. The temperature of the source and sink are |
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Answer» `99^(@)C, 27^(@)C` `eta_(2) = 1 - ((T_(2) - 62))/(T_(1)) = 2eta_(1) = (1)/(3)` or `2T_(1) - 3T_(2) = -186` …(II) Solving (i) and (ii), we get `T_(1) = 372 K = 99^(@)C` `T_(2) = (5)/(6)T_(1) = (5)/(6) xx 372 K = 310 K = 37^(@)C` |
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