JEE Main 2022 · 30 Jul · Shift 2 · Physics Q7

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PhysicsSingle correct+4 / -1Question 7 of 90

A Carnot engine is operating between the two reservoirs of temperature 227 °C and 127 °C respectively. If the engine performs $1.04\times10^5$ J work per cycle. Then, the heat absorbed from hot reservoir per cycle will be :

2.08\times10^4

1.04\times10^5

5.20\times10^5

2.36\times10^5

Answer and worked solution

Official answer: C

Efficiency $1-\frac{400}{500}=0.2$ , so $Q_H=\frac{1.04\times10^5}{0.2}=5.2\times10^5$ J.

(C)

Topics:Carnot cycle

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JEE Main 2022 · 30 Jul · Shift 2, Physics question 7