Question

Let ${\eta }_1$ is the efficiency of an engine at $T_1=447^{\circ }C$ and $T_2=147^{\circ }C$ while ${\eta }_2$ is the efficiency at $T_1=947^{\circ }C$ and $T_2=47^{\circ }C$. The ratio $\frac{{\eta }_1}{{\eta }_2}$ will be :

• A. $0.41$
• B. $0.56$
• C. $0.73$
• D. $0.70$

Answer 1

Answer: (B)

Efficiency $\eta =1-\frac{T_L}{T_H}$
${\eta }_1=1-\frac{147+273}{447+273}=1-\frac{420}{720}$
${\eta }_1=\frac{300}{720}$
${\eta }_2=1-\frac{47+273}{947+273}=1-\frac{320}{1220}$
${\eta }_2=\frac{900}{1220}$
$\frac{{\eta }_1}{{\eta }_2}=\frac{300}{720}\times \frac{1220}{900}=\frac{122}{72\times 3}$
$\frac{{\eta }_1}{{\eta }_2}=0.56$