Let η 1 is the efficiency of an engine at T 1 = 447 ° C and T 2 = 147 ° C while η 2 is…

The efficiency of an engine is given by $\eta =1-\frac{T_L}{T_H}$. Here, $T_L$ and $T_H$ are in Kelvin scale.

$\Rightarrow {\eta }_1=1-\frac{147+273}{447+273}=1-\frac{420}{720}$

$\Rightarrow {\eta }_1=\frac{300}{720}$

And ${\eta }_2=1-\frac{47+273}{947+273}=1-\frac{320}{1220}$

$\Rightarrow {\eta }_2=\frac{900}{1220}$

$\Rightarrow \frac{{\eta }_1}{{\eta }_2}=\frac{300}{720}\times \frac{1220}{900}=\frac{122}{72\times 3}$

$\Rightarrow \frac{{\eta }_1}{{\eta }_2}=0.56$