A Carnot engine absorbs an amount Q of heat f | vedclass

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(B) For a Carnot engine,the efficiency $\eta$ is given by the ratio of temperatures of the sink $(T_2)$ and the source $(T_1)$:$\eta = 1 - \frac{T_2}{

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$Q / 3$

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(B) For a Carnot engine,the efficiency $\eta$ is given by the ratio of temperatures of the sink $(T_2)$ and the source $(T_1)$:$\eta = 1 - \frac{T_2}{T_1}$Given,$T_1 = T$ and $T_2 = T/3$.Therefore,$\eta = 1 - \frac{T/3}{T} = 1 - \frac{1}{3} = \frac{2}{3}$.Also,the efficiency is defined as $\eta = 1 - \frac{Q_2}{Q_1}$,where $Q_1$ is the heat absorbed and $Q_2$ is the heat rejected.Equating the two expressions for efficiency:$\frac{2}{3} = 1 - \frac{Q_2}{Q}$$\frac{Q_2}{Q} = 1 - \frac{2}{3} = \frac{1}{3}$$Q_2 = \frac{Q}{3}$.Thus,the amount of heat rejected is $Q/3$.

$A$ 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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