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