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…

Question

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?

MARKS Content Team · Accepted Answer

Correct answer: 250. T_2/T_1=1-η=1-40/100=3/5 ∴ T_1=5/3T_2=5/3×300=500K Increase in efficiency =50% of 40%=20% ∴ New efficiency η^(′)=40+20=60% ∴ T_2/(T′_1)=1-η^(′)=1-60/100=2/5 T1′=5/2×300=750 K Increase in temperature of source =T1′-T_1 =750-500=250 K

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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