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

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Question

(B) For a Carnot engine,the efficiency $\eta$ is given by $\eta = 1 - \frac{T_{sink}}{T_{source}} = \frac{Q_{absorbed} - Q_{rejected}}{Q_{absorbed}}$.

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$Q/3$

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(B) For a Carnot engine,the efficiency $\eta$ is given by $\eta = 1 - \frac{T_{sink}}{T_{source}} = \frac{Q_{absorbed} - Q_{rejected}}{Q_{absorbed}}$.Given,$T_{source} = T$,$T_{sink} = T/3$,and $Q_{absorbed} = Q$.Substituting these values into the efficiency formula:$\eta = 1 - \frac{T/3}{T} = 1 - \frac{1}{3} = \frac{2}{3}$.Now,equate this to the heat ratio:$\frac{2}{3} = \frac{Q - Q_{rejected}}{Q}$.$\frac{2}{3} = 1 - \frac{Q_{rejected}}{Q}$.$\frac{Q_{rejected}}{Q} = 1 - \frac{2}{3} = \frac{1}{3}$.Therefore,$Q_{rejected} = \frac{Q}{3}$.

$A$ 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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