A Carnot engine whose sink is at 300K 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?
A
325 K
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B
250 K
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C
380 K
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D
275 K
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Solution
The correct option is B 250 K Efficiency of the carnot engine, η=T1−T2T1
where T1=Source temperature T2 = sink temperature 0.40=T1−300T1 T1=500 K
Since η is increased by 50% , the new efficiency will be 60% 0.60=T1−300T1 T1=750 K
Increase in Temprature=750−500=250 K