Question

In a Carnot engine, when $T_2=0^{\circ }C$ and $T_1={200}^{\circ }C$, its efficiency is ${\eta }_1$ and when $T_1=0^{\circ }C$ and $T_2=-{200}^{\circ }C$, Its efficiency is ${\eta }_2$, then what is ${\eta }_1/{\eta }_2$

• A. $0.733$
• B. $0.638$
• C. cannot be calculated
• D. $0.577$

Answer 1

The correct option is D $0.577$

When $T_1={200}^{\circ }C$ and $T_2=0^{\circ }C$

$\eta =1-\frac{T_2}{T_1}=\frac{T_1-T_2}{T_1}$

$\Rightarrow {\eta }_1=\frac{(473-273)}{473}=\frac{200}{473}$

When $T_1=0^{\circ }C$ and $T_2=-{200}^{\circ }C$ ,

${\eta }_2=\frac{273-73}{273}=\frac{200}{273}$

So required ratio

$\frac{{\eta }_1}{{\eta }_2}=\frac{273}{473}=0.577$