A Carnot’s engine, with its cold body at ${17}^{0}C$ has 50% efficiency. If the temperature of its hot body is now increased by${145}^{0}C$ , the efficiency becomes

$\begin{array}{l}\mathrm{\eta }=1-\frac{{\mathrm{T}}_2}{{\mathrm{T}}_1}=0.5\\ \therefore {\mathrm{T}}_1=2{\mathrm{T}}_2=2(17+273)=580\mathrm{K}={307}^0\mathrm{C}\end{array}$

Now the temperature of hot body is increased by ${145}^{0}C$

$\begin{array}{l}\therefore {{\mathrm{T}}_1}^{\text{'}}={307}^0+{145}^0={452}^0\mathrm{C}=725\mathrm{K}\\ {\mathrm{T}}_2=17+273=290\mathrm{K}\\ \therefore \mathrm{\eta }\text{'}=\left(1-\frac{290}{725}\right)\times 100=60\mathrm{\%}\end{array}$