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A Carnot engine whose sink is at 300 K has an efficiency of 40 %. By how much should the temperature of source be increased so as to increase its efficiency by 50% of original efficiency? |
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Answer» `380 K` or, `T_(2)/T_(1) = 1 - eta = 1 - 40/100 = 3/5` ` :. T_(1) = 5/3 xx T_(2) = 5/3 xx 300 = 500 K`. Increase in efficiency = `50% " of " 40% = 20%` New efficiency, `eta. = 40% + 20% = 60%` ` :. T_(2)/T_(1) = 1 - 60/100 = 2/5` ` rArr T_(1). = 5/2 xx T_(2) = 5/2 xx 300 = 750 K` ` :. ` Increse in temperature of SOURCE ` = T_(1). - T_(1)` ` = 750 - 500 = 250 K` |
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