An ideal gas heat engine operates in a Carnot cycle between $${227^ \circ }C$$  and $${127^ \circ }C.$$  It absorbs $$6\,kcal$$  at the higher temperature. The amount of heat (in $$kcal$$ ) converted into work is equal to

Solution :
The efficiency of heat engine is $$\eta = 1 - \frac{{{T_2}}}{{{T_1}}}$$
or $$\frac{W}{{{Q_1}}} = 1 - \frac{{{T_2}}}{{{T_1}}}$$
$${{T_2}} =$$  Temperature of sink
$${{T_1}} =$$  Temperature of source
$$W =$$  Work done
Given, $${{Q_1}} =$$  heat absorbed from the source $$=6\,kcal$$
$${T_1} = 227 + 273 = 500\,K$$
and $${T_2} = 127 + 273 = 400\,K$$
Hence, $$\frac{W}{6} = 1 - \frac{{400}}{{500}}$$
or $$\frac{W}{6} = \frac{{100}}{{500}}\,\,{\text{or}}\,\,W = 1.2\,kcal$$
Thus, amount of heat converted into work is $$1.2\,kcal.$$