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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 $6 \times 10^4 \mathrm{cal}$ of heat at higher temperature. Amount of heat converted to work is
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The correct answer is:
$1.2 \times 10^4 \mathrm{cal}$
The heat converted to work is the amount of heat that remains after going through sink.
From the relation $\frac{Q_2}{Q_1}=\frac{T_2}{T_1}$
Given,
$\begin{aligned} Q_1 & =6 \times 10^4 \mathrm{cal}, \\ T_1 & =227+273=500 \mathrm{~K} \\ T_2 & =127+273=400 \mathrm{~K}\end{aligned}$
$\therefore \frac{Q_2}{6 \times 10^4}=\frac{400}{500}$
$\Rightarrow Q_2=\frac{4}{5} \times 6 \times 10^4=4.8 \times 10^4 \mathrm{cal}$
Now, heat converted to work
$\begin{aligned}& =Q_1-Q_2 \\& =6.0 \times 10^4-4.8 \times 10^4 \\
& =1.2 \times 10^4 \mathrm{cal}\end{aligned}$
From the relation $\frac{Q_2}{Q_1}=\frac{T_2}{T_1}$
Given,
$\begin{aligned} Q_1 & =6 \times 10^4 \mathrm{cal}, \\ T_1 & =227+273=500 \mathrm{~K} \\ T_2 & =127+273=400 \mathrm{~K}\end{aligned}$
$\therefore \frac{Q_2}{6 \times 10^4}=\frac{400}{500}$
$\Rightarrow Q_2=\frac{4}{5} \times 6 \times 10^4=4.8 \times 10^4 \mathrm{cal}$
Now, heat converted to work
$\begin{aligned}& =Q_1-Q_2 \\& =6.0 \times 10^4-4.8 \times 10^4 \\
& =1.2 \times 10^4 \mathrm{cal}\end{aligned}$
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