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Question

A Carnot engine operates with sources at $$127^{\circ}C$$ and sink at $$27^{\circ}C$$. If the source supplies $$40\ kJ$$ of heat energy, the work done by the engine is


A
30 kJ
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B
10 kJ
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C
4 kJ
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D
1 kJ
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Solution

The correct option is A $$10\ kJ$$
Efficiency, $$\eta = 1 - \dfrac {T_{2}}{T_{1}}$$
$$\therefore \eta = 1 - \dfrac {(273 + 27)}{(273 + 127)} = 1 - \dfrac {300}{400} = 1 - \dfrac {3}{4} = \dfrac {1}{4}$$
Also $$\eta = \dfrac {\text {Work done by engine}}{\text {Heat supplied by source}} = \dfrac {W}{40\ kJ}$$
$$\therefore W = 40\eta = 40\times \dfrac {1}{4} = 10\ kJ$$.

Physics

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