In Carnot engine, efficiency is 40% at hot reservoir temperature T. For efficiency 50%, what will be the temperature of hot reservoir?

Question

MARKS Content Team · Accepted Answer

Correct answer: (C) 6T/5. The efficiency η=(work done)/(heat input)=W/Q η=1-T_2/T_1 Where, T_2= temperature of sink And T_1= temperature of hot reservoir 40/100=1-T_2/T_1 ⇒ T_2/T_1=0.6 ⇒ T_2=0.6T_1 Now, 50/100=1-T_2/(T_(1′⇒T2)T1′)=0.5 0.6T_1/(T1′)=0.5 ⇒ T1′=0.6/0.5T_1 ⇒ T1′=6/5T_1 ⇒ T1′=6/5T ∵ T_1=T

In Carnot engine, efficiency is $40 %$ at hot reservoir temperature T. For efficiency $50 %,$ what will be the temperature of hot reservoir?

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