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A Carnot engine absorbs an amount $Q$ of heat from a reservoir at an absolute temperature $T$ and rejects heat to a sink at a temperature of $T / 3$. The amount of heat rejected is
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The correct answer is:
$Q / 3$
$\boxtimes \eta=1-\frac{T_2}{T_1}=\frac{W}{Q_1}=\frac{Q_1-Q_2}{Q_1}$
where $Q_1=$ heat absorbed, $Q_2=$ heat rejected
$\begin{aligned} \Rightarrow & 1-\frac{T / 3}{T}=\frac{W}{Q_1} \Rightarrow \frac{2}{3}=\frac{W}{Q_1}=\frac{Q_1-Q_2}{Q_1} \\ \Rightarrow & \frac{2}{3}=1-\frac{Q_2}{Q_1} \Rightarrow \frac{Q_2}{Q_1}=\frac{1}{3} \Rightarrow Q_2=\frac{Q_1}{3}=\frac{Q}{3}\end{aligned}$
where $Q_1=$ heat absorbed, $Q_2=$ heat rejected
$\begin{aligned} \Rightarrow & 1-\frac{T / 3}{T}=\frac{W}{Q_1} \Rightarrow \frac{2}{3}=\frac{W}{Q_1}=\frac{Q_1-Q_2}{Q_1} \\ \Rightarrow & \frac{2}{3}=1-\frac{Q_2}{Q_1} \Rightarrow \frac{Q_2}{Q_1}=\frac{1}{3} \Rightarrow Q_2=\frac{Q_1}{3}=\frac{Q}{3}\end{aligned}$
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