Efficiency of a Carnot engine is 50% when temperature of outlet is 500 K. In order to increase efficiency up to 60% keeping the inlet temperature same, what is temperature of outlet?

$\eta =\frac{T_1-T_2}{T_1}$

$$\begin{gathered} \text{ When efficiency is }50\mathrm{\%}\text{, outlet temperature, }{\mathrm{T}}_2=500\mathrm{K} \\ \mathrm{We} \mathrm{know}, \\ \mathrm{\eta }=1-\frac{{\mathrm{T}}_2}{{\mathrm{T}}_1} \\ \Rightarrow \frac{50}{100}=1-\frac{500}{{\mathrm{T}}_1} \\ \Rightarrow {\mathrm{T}}_1=\frac{500\times 100}{50}=1000\mathrm{K} \\ \text{ To increase the efficiency upto }60\mathrm{\%}\text{, with }{\mathrm{T}}_1=1000\mathrm{K}\text{, then } \\ \mathrm{\eta }=1-\frac{{\mathrm{T}}_2}{{\mathrm{T}}_1} \\ \Rightarrow \frac{60}{100}=1-\frac{{\mathrm{T}}_2}{1000} \\ ⟹{\mathrm{T}}_2=\frac{1000\times 40}{100}=400\mathrm{K} \end{gathered}$$