Question:
A heat engine operates between a cold reservoir at temperature $\mathrm{T}_{2}=400 \mathrm{~K}$ and a hot reservoir at temperature $\mathrm{T}_{1}$. It takes $300 \mathrm{~J}$ of heat from the hot reservoir and delivers $240 \mathrm{~J}$ of heat to the cold reservoir in a cycle. The minimum temperature of the hot reservoir has to be_____________K.
Solution:
$Q_{\text {in }}=300 \mathrm{~J} ; \mathrm{Q}_{\text {out }}=240 \mathrm{~J}$
Work done $=Q_{\text {in }}-Q_{\text {out }}=300-240=60 \mathrm{~J}$
Efficiency $=\frac{\mathrm{W}}{\mathrm{Q}_{\mathrm{in}}}=\frac{60}{300}=\frac{1}{5}$
efficiency $=1-\frac{T_{2}}{T_{1}}$
$\frac{1}{5}=1-\frac{400}{\mathrm{~T}_{1}} \Rightarrow \frac{400}{\mathrm{~T}_{1}}=\frac{4}{5}$
$\mathrm{T}_{1}=500 \mathrm{k}$