JEE Main 2022 · 24 Jun · Shift 1 · Physics Q47

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PhysicsSingle correct+4 / -1Question 47 of 90

A Carnot engine whose heat sinks at 27°C, has an efficiency of 25%. By how many degrees should the temperature of the source be changed to increase the efficiency by 100% of the original efficiency ?

Increases by 18°C

Increases by 200°C

Increases by 120°C

Increases by 73°C

Answer and worked solution

Official answer: B

Original source. $0.25=1-\dfrac{300}{T_1}$ , so $T_1=400$ K.

Doubled efficiency. $0.5=1-\dfrac{300}{T_1'}$ , so $T_1'=600$ K.

Change. The source temperature increases by 200 K, which is 200°C.

(B)

Topics:Carnot cycle

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JEE Main 2022 · 24 Jun · Shift 1, Physics question 47