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?

Question

Infinity Learn · Accepted Answer

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

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?

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