In a Carnot engine, when T_2=0^ C and T_1=200^ C, its efficiency … - Vidyadip

MCQ

In a Carnot engine, when $T_2=0^{\circ} \mathrm{C}$ and $T_1=200^{\circ} \mathrm{C}$, its efficiency is $\eta_1$ and when $T_1=0{ }^{\circ} \mathrm{C}$ and $T_2=-200{ }^{\circ} \mathrm{C}$, its efficiency is $\eta_2$, then what is $\eta_1 / \eta_2$

✓
$0.577$
B
$0.733$
C
$0.638$
D
Can not be calculated

✓

Answer

Correct option: A.

$0.577$

$\eta=1-\frac{T_2}{T_1}=\frac{T_1-T_2}{T_1} \Rightarrow \eta_1=\frac{(473-273)}{473}=\frac{200}{473}$and $\eta_2=\frac{273-73}{273}=\frac{200}{273}$
So required ratio $\frac{\eta_1}{\eta_2}=\frac{273}{473}=0.577$

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