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

When efficiency is 50 %, outlet temperature is ${\mathrm{T}}_2=500\mathrm{K}$

$\eta =1-\frac{{\mathrm{T}}_2}{{\mathrm{T}}_1};\frac{50}{100}=1-\frac{500}{{\mathrm{T}}_1};\frac{1}{2}=1-\frac{500}{{\mathrm{T}}_1};\frac{500}{{\mathrm{T}}_1}=1-\frac{1}{2}=\frac{1}{2}$

${\mathrm{T}}_1=1000\mathrm{K}$ 

To increase the efficiency up to 60 % with ${\mathrm{T}}_1=1000\mathrm{K}$,

$\eta =1-\frac{{\mathrm{T}}_2}{{\mathrm{T}}_1};\frac{60}{100}=1-\frac{{\mathrm{T}}_2}{1000};\frac{{\mathrm{T}}_2}{1000}=1-\frac{60}{100};$

$\frac{{\mathrm{T}}_2}{1000}=\frac{100-60}{100};\frac{{\mathrm{T}}_2}{1000}=\frac{40}{100};{\mathrm{T}}_2=400\mathrm{K}$