In a Carnot engine, the temperature of reservoir is $${927^ \circ }C$$  and that of sink is $${27^ \circ }C.$$  If the work done by the engine when it transfers heat from reservoir to sink is $$12.6 \times {10^6}J,$$   the quantity of heat absorbed by the engine from the reservoir is

Solution :
As we know $$\eta = \frac{W}{{{Q_1}}} = 1 - \frac{{{T_2}}}{{{T_1}}}$$
$$\eqalign{ & \Rightarrow \eta = 1 - \frac{{300K}}{{1200K}} = \frac{3}{4} \cr & \frac{3}{4} = \frac{W}{{{Q_1}}} \Rightarrow {Q_1} = W \times \frac{4}{3} \Rightarrow {Q_1} = 12.6 \times {10^6} \times \frac{4}{3} \cr & {Q_1} = 16.8 \times {10^6}J. \cr} $$