An ideal gas heat engine operates in Carnot cycle between $${227^ \circ }C$$  and $${127^ \circ }C.$$  It absorbs $$6 \times {10^4}cal$$   of heat at higher temperature. Amount of heat converted to work is

Solution :
According to the Carnot cycle in heat engine $$\frac{{{Q_2}}}{{{Q_1}}} = \frac{{{T_2}}}{{{T_1}}}$$
Given, heat absorbed, $${Q_1} = 6 \times {10^4}cal,$$
Temperature of source, $${T_1} = 227 + 273 = 500\,K$$
Temperature of sink, $${T_2} = 127 + 273 = 400\,K$$
$$\therefore \frac{{{Q_2}}}{{6 \times {{10}^4}}} = \frac{{400}}{{500}}$$
⇒ Heat rejected, $${Q_2} = \frac{4}{5} \times 6 \times {10^4} = 4.8 \times {10^4}cal$$
Now, heat converted to work
$$\eqalign{ & W = {Q_1} - {Q_2} = 6.0 \times {10^4} - 4.8 \times {10^4} \cr & = 1.2 \times {10^4}cal \cr} $$