A Carnot engine absorbs \(1000\,J\) of heat energy from a reservoir at \(127\,^oC\) and rejects \(600\,J\) of heat energy during each cycle. The efficiency of engine and temperature of sink will be

            \(Q_2=600 J\)

            \({T_1} = {127^ \circ }C = 400\,K\)

            \({T_2} = ?\)

            \(\eta  = ?\)

Efficiency of carnot engine,

\(\eta  = \frac{W}{Q_1} \times 100\% \)

\(or,\,\,\eta  = \frac{{{Q_2} - {Q_1}}}{{{Q_1}}} \times 100\% \)

\(or,\,\,\,\eta  = \frac{{1000 - 600}}{{1000}} \times 100\% \)

\(\eta  = 40\% \)

\(Now,for\,carnot\,cycle\frac{{{Q_2}}}{{{Q_1}}} = \frac{{{T_2}}}{{{T_1}}}\)

\(\frac{{600}}{{1000}} = \frac{{{T_2}}}{{400}}\)

\({T_2} = \frac{{600 \times 400}}{{1000}} = 240\,K = 240 - 273\)

\(\therefore {T_2} =  - {33^ \circ }C\)