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Q.

A carnot engine has an efficiency of 40%. Keeping the source temperature unchanged, what should be the percentage change in the temperature of the sink in order to increase the efficiency by 10%?

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a

4.44% increase

b

3.64%, decrease

c

6.66% decrease

d

5.18% increase

answer is C.

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Detailed Solution

1−T2T1=η⇒T2=T1(1−η)Initially, T2=T1(1−0.4)=0.6 T1Finally, T21=T1[1−(0.4+10100×0.4)]=0.56 T 1∴ T2−T21T2=0.6T1−0.56T10.6T=6.66%So temperature of the sink should be decreased by 6.66%

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A carnot engine has an efficiency of 40%. Keeping the source temperature unchanged, what should be the percentage change in the temperature of the sink in order to increase the efficiency by 10%?