Using the Carnot efficiency formula: $\eta = 1 - \frac{T_2}{T_1} = 1 - \frac{300}{600} = 1 - 0.5 = 0.5 = 50\%$
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JEE Advanced 2024 — Physics Thermodynamics
hard
numerical_value
2024
Official previous-year question
Verified 30 May 2026.
Question
A Carnot engine operates between a source at $T_1 = 600$ K and a sink at $T_2 = 300$ K. Find its efficiency (in percentage).
Answer
Exact answer: 50
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