A carnot engine takes \( 300 \) calories of heat from a source at \( 500 \mathrm{~K} \) and rejects \( 150 \) calories of
heat to the sink. The temperature of the sink is

Given, heat taken from source, \(Q_{1}=300\) calories; temperature of source \(T_{1}=500 \mathrm{~K} ;\) heat rejected to the
sink, \(Q_{2}=150\) calories.
Temperature of sink, \(T_{2}=?\)
Now, efficiency, \(\eta=\frac{Q_{1}-Q_{2}}{Q}\)
Also, efficiency, \(\eta=\frac{T_{1}-T_{2}}{T_{1}}\)
Therefore, \(\frac{Q_{1}-Q_{2}}{Q_{1}}=\frac{T_{1}-T_{2}}{T_{1}}\)
\(\Rightarrow \frac{300-150}{300}=\frac{500-T_{2}}{500}\)
\(\Rightarrow \frac{150}{300}=1-\frac{T_{2}}{500}\)
\(\Rightarrow \frac{1}{2}=1-\frac{T_{2}}{500}\)
\(\Rightarrow \frac{T_{2}}{500}=1-\frac{1}{2}\)
\(\Rightarrow T_{2}=500 \times \frac{1}{2}=250 \mathrm{~K}\)
Thus, temperature of sink \(=250 \mathrm{~K}\)