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$