Question

In a Carnot engine, when $T_{2} = 0^{o}C$ and $T_{1} = 200^{o}C,$ its efficiency is $\eta_{1}$and when $T_{1} = 0^{o}C$ and $T_{2} = - 200^{o}C$, Its efficiency is $\eta_{2}$, then what is $\eta_{1}/\eta_{2}$

• A. 0.577
• B. 0.733
• C. 0.638
• D. Can not be calculated

Answer 1

Answer: (A)

0.577

Answer 2

$\eta = 1 - \frac{T_{2}}{T_{1}} = \frac{T_{1} - T_{2}}{T_{1}}$ $\Rightarrow \eta_{1} = \frac{(473 - 273)}{473} = \frac{200}{473}$

and $\eta_{2} = \frac{273 - 73}{273} = \frac{200}{273}$

So required ratio $\frac{\eta_{1}}{\eta_{2}} = \frac{273}{473} = 0.577$