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A reversible heat engine converts one-sixth of heat, which it extracts from source, into work. When the temperature of the sink is reduce by 40^(@)C, its efficiency is doubled. Find the temperature of source. |
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Answer» SOLUTION :Efficiency, `eta = (W)/(Q_(1)) = (1)/(6)` Again `eta = 1 - (T_(2))/(T_(1))` or `(1)/(6) = 1- (T_(2))/(T_(2)) rArr T_(2) = (5T_(1))/(6)` ….(i) finally, efficiency `= eta = (2 xx 1)/(6) = (1)/(3)` `:. (1)/(3) = 1 - (T_(2) - 40)/(T_(1))` or `(T_(2) - 40)/(T_(1)) = (1-(1)/(3)) = (2)/(3)` or `2T_(1) = 3T_(2) - 120` Put `T_(2) = (5T_(1))/(6)` (from (i)) or `2T_(1) = (3 xx (5T_(1)))/(6) - 120` or `2.5T_(1) - 2T_(1) = 120` or `T_(1) = (120)/(0.5) = 240 K`. |
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