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An ideal gas heat engine operates in a Carnot cycle between $227^{\circ} \mathrm{C}$ and $127^{\circ} \mathrm{C}$. It absorbs $6 \mathrm{kcal}$ at the higher temperature. The amount of heat (in kcal) converted into work is equal to:
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Verified Answer
The correct answer is:
1.2
$$
\begin{aligned}
& \text { Efficiency of carnot engine }=\frac{W}{Q_1} \\
& =1-\frac{T_2}{T_1} \\
& \frac{W}{6}=1-\frac{400}{500} \\
& \Rightarrow \quad W=\frac{6}{5}=1.2 \mathrm{Kcal}
\end{aligned}
$$
\begin{aligned}
& \text { Efficiency of carnot engine }=\frac{W}{Q_1} \\
& =1-\frac{T_2}{T_1} \\
& \frac{W}{6}=1-\frac{400}{500} \\
& \Rightarrow \quad W=\frac{6}{5}=1.2 \mathrm{Kcal}
\end{aligned}
$$
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