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