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An ideal gas heat engine operates in carnot cycle between $227^{\circ} \mathrm{C}$ and $127^{\circ} \mathrm{C}$. It absorbs $6 \times 10^4$ cal of heat at higher temperature. Amount of heat converted into work, is:
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Verified Answer
The correct answer is:
$1.2 \times 10^4$ cals
For a carnot engine
$$
\text { efficiency } \begin{aligned}
e & =1-\frac{T_2}{T_1} \\
& =1-\frac{127+273}{227+273} \\
& =1-\frac{400}{500}=\frac{1}{5}
\end{aligned}
$$
Now, $\quad e=\frac{\text { Work Output }}{\text { Heat Output }}$
$$
=\frac{\mathrm{W}}{6 \times 10^4}
$$
$$
\begin{aligned}
\mathrm{W} & =e \times 6 \times 10^4 \\
& =\frac{1}{5} \times 6 \times 10^4 \\
& =1.2 \times 10^4 \mathrm{cal} .
\end{aligned}
$$
$$
\text { efficiency } \begin{aligned}
e & =1-\frac{T_2}{T_1} \\
& =1-\frac{127+273}{227+273} \\
& =1-\frac{400}{500}=\frac{1}{5}
\end{aligned}
$$
Now, $\quad e=\frac{\text { Work Output }}{\text { Heat Output }}$
$$
=\frac{\mathrm{W}}{6 \times 10^4}
$$
$$
\begin{aligned}
\mathrm{W} & =e \times 6 \times 10^4 \\
& =\frac{1}{5} \times 6 \times 10^4 \\
& =1.2 \times 10^4 \mathrm{cal} .
\end{aligned}
$$
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