A heat engine absorbs heat of \(10^5Kcal\) from a source, which is at \(127^\circ\) and rejects a part of heat to sink at \(27^\circ\). Calculate the efficiency of the engine and the work done by it.

**Solution:**

The efficiency of the engine is $$ \eta=1-\frac{T_2}{T_1}$$ $$=1-\frac{300}{400}$$

$$ =0.25$$ that is, \(25\% \)

Work done by the engine $$ W= \eta\times Q$$ $$=0.25\times 10^8 Cal$$

$$ =0.25\times 10^8\times4.81J$$

$$ =1.05\times10^8J$$

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