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 $$ W= \eta\times Q$$ $$=0.25\times 10^8 Cal$$
$$ =0.25\times 10^8\times4.81J$$
$$ =1.05\times10^8J$$