Q.

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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a

275 K

b

325 K

c

250 K

d

380 K

answer is C.

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Detailed Solution

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 K
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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?