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 ?

Question

Dr. Abubacker · Accepted Answer

[Solution from source] With the sink at 300 K, 0.4 = 1 − 300/T₁ gives T₁ = 500 K. The new efficiency is 0.4 + 0.5 × 0.4 = 0.6, so 300/T₁' = 0.4 and T₁' = 750 K. The source must be raised by 750 − 500 = 250 K; reading '50% more' as an efficiency of 90% gives an impossible value. Answer (3).

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 ?

Explanation

Reviewed by Dr. Abubacker

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