For an ideal heat engine, the temperature of the source is $127^{\circ} \mathrm{C}$. In order to hav

Question

For an ideal heat engine, the temperature of the source is $127^{\circ} \mathrm{C}$. In order to have $60 \%$ efficiency the temperature of the sink should be $\qquad$ ${ }^{\circ} \mathrm{C}$. (Round off to nearest integer).\

Professor PC Thomas Classes · Answer

(113*) : Given, temperature of source, $T_{1}=127^{\circ} \mathrm{C}=127+273=400 \mathrm{~K}, \eta=0.6$\ We know that for Carnot engine, $\eta=1-\frac{T_{2}}{T_{1}}$\ where $T_{2}$ is the sink temperature.\ $0.6=1-\frac{T_{2}}{400} \Rightarrow \frac{T_{2}}{400}=0.4$\ $	herefore \quad T_{2}=160 \mathrm{~K}=160-273=-113^{\circ} \mathrm{C}$\