A Carnot engine has the same efficiency between (i) \(100 \mathrm{~K}\) and \(600 \mathrm{~K}\) and (ii) \(\mathrm{T} \mathrm{K}\) and \(960 \mathrm{~K}\). The temperature \(T\) in kelvin of the sink is

\(
\begin{array}{ll}
& \text { Efficiency of a carnot engine is } \eta=1-\frac{T_C}{T_H} \\
& \text { For case (i), } \\
& T_C=100 \mathrm{~K} \text { and } T_H=600 \mathrm{~K} \\
\therefore \quad & \eta_1=1-\frac{100}{600}=\frac{5}{6} \\
& \text { For case (ii), } \\
& \mathrm{T}_{\mathrm{C}}=\mathrm{T} \mathrm{K} \text { and } \mathrm{T}_{\mathrm{H}}=960 \mathrm{~K} \\
\therefore \quad & \eta_2=1-\frac{\mathrm{T}}{960} \\
& \text { Given } \eta_1=\eta_2 \\
\therefore \quad & \frac{5}{6}=1-\frac{\mathrm{T}}{960} \\
& \frac{5}{6}=\frac{960-\mathrm{T}}{960}
\end{array}
\)
Solving for \(\mathrm{T}\), we get \(\mathrm{T}=160 \mathrm{~K}\)