A Carnot engine whose sink is at 300 K has an efficiency of 40%. By how much should the temperature of the source be increased so that its efficiency is increased 50% of original efficiency?

The efficiency of Carnot engine is defined as the ratio of work done to the heat supplied, i.e.,η = Work doneHeat supplied = WQ1 = Q1-Q2Q1    = 1- Q2Q1 = 1-T2T1Here, T1 is the temperature of source and T2 is the temperature of the sinkAs given, η = 40% = 40100 = 0.4and T2 = 300 KSo, 0.4 = 1-300T1T1 = 3001-0.4 = 3000.6 = 500 KLet temperature of the source be increased by x K, then efficiency becomesη' = 40% + 50% of η     = 40100+50100×0.4 = 0.4 + 0.5×0.4 = 0.6Hence, 0.6 = 1 -300500+x          300500 + x = 0.4         500 + x = 3000.4 = 750         x = 750 - 500 = 250 K