⚛️ Physics · Thermodynamics · NEET & JEE

A Carnot engine rejects heat to a sink at 300 K and runs at 40% efficiency. To lift its efficiency to 50% with the same sink temperature, the source temperature must be raised by:

Answer: 100 K.

A 100 K
B 50 K
C 200 K
D 150 K

Correct answer: A. 100 K

Explanation: T<sub>h</sub> = T<sub>c</sub>/(1−η): for 40%, 300/0.6 = 500 K; for 50%, 300/0.5 = 600 K. Rise = 100 K.

The Carnot cycle alternates two isothermal steps (heat absorbed at T_H, released at T_C) with two adiabatic steps (no heat exchange); the enclosed loop area equals the net work output, and η = 1−T_C/T_H is the maximum possible efficiency between those two temperatures.

Concept context

Laws of thermodynamics, heat engines, entropy, and gas processes.

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A Carnot engine rejects heat to a sink at 300 K and runs at 40% efficiency. To lift its efficiency to 50% with the same sink temperature, the source temperature must be raised by:

Concept context

More Thermodynamics questions