1. 200^° C 2. 27^° C 3. 15^° C 4. 100^° C

Answer: 2. 27^° C Carnot efficiency: η=1-T₂/T₁ with temperatures in kelvin. T₁=327+273=600 K, η=0.5⇒ T₂=0.5×600=300 K = 27^° C. The answer is 27^° C.

A Carnot engine has an efficiency of 50% when its source is at a temperature 327^° C. The temperature of the sink is :

1. \(200^\circ\) C

2. \(27^\circ\) C

3. \(15^\circ\) C

4. \(100^\circ\) C

Doubtly Asked 30 Oct 2026 NEET UG 2023 neet Save WhatsApp Report

Best answer

Answer: 2. \(27^\circ\) C

Carnot efficiency: \(\eta=1-\dfrac{T_2}{T_1}\) with temperatures in kelvin.

\(T_1=327+273=600\) K, \(\eta=0.5\Rightarrow T_2=0.5\times600=300\) K = \(27^\circ\)C.

The answer is \(27^\circ\)C.

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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