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A Carnot engine whose low temperature reservoir is at 7oC has an efficiency of 50%. It is desired to increase the efficiency to 70%. By how many degrees should the temperature of the high temperature reservoir be increased

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a

840 K

b

280 K

c

560 K

d

380 K

answer is D.

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Detailed Solution

Initially η=T1−T2T1⇒0.5=T1−(273+7)T1⇒ 12=T1−280T1⇒T1=560KFinallyη1′=T1′−T2T1′⇒0.7=T1′−(273+7)T1′⇒T1′=933K∴ Increase in temperature =933−560=373K≈380K

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