Questions
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?
detailed solution
Correct option is C
η=1-T2T1⇒1-300T1=0.4⇒T1=500 K Now η'=0.4+0.4×50100=0.6 Therefore 0.6=1-300500+∆T⇒∆T=250 KTalk to our academic expert!
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