Heat engine

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Question

A Carnot engine is operating between temperatures $27^{0} C$ and $127^{0} C$ . Energy received by the engine per second is 40 kJ. Then power out put of the engine is

Moderate

Solution

$η = 1 - \frac{273 + 27}{273 + 127} = 1 - \frac{300}{400} = \frac{1}{4}$
$∴ \frac{Out put Power}{In put power} = \frac{1}{4} \frac{P_{out}}{40 kW} = \frac{1}{4} \Rightarrow P_{out} = 10 kW$

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