The efficiency of a Carnot heat engine is 0.5 | vedclass

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Question

(A) The efficiency $\eta$ of a Carnot heat engine is given by the formula: $\eta = 1 - \frac{T_{2}}{T_{1}}$.Given that $\eta = 0.5$, we have $0.5 = 1

Vedclass · Accepted Answer

$2 \,T_{1}, 2 \,T_{2}$

Vedclass · Answer

(A) The efficiency $\eta$ of a Carnot heat engine is given by the formula: $\eta = 1 - \frac{T_{2}}{T_{1}}$.Given that $\eta = 0.5$, we have $0.5 = 1 - \frac{T_{2}}{T_{1}}$, which implies $\frac{T_{2}}{T_{1}} = 0.5$, or $T_{1} = 2T_{2}$.For another Carnot engine to have the same efficiency of $0.5$, the ratio of the sink temperature to the source temperature must remain the same, i.e., $\frac{T_{2}'}{T_{1}'} = 0.5$.If we multiply both the source and sink temperatures by the same factor $k$, the new efficiency is $\eta' = 1 - \frac{k T_{2}}{k T_{1}} = 1 - \frac{T_{2}}{T_{1}} = 0.5$.Therefore, if the source temperature is $2T_{1}$ and the sink temperature is $2T_{2}$, the efficiency remains $0.5$.

The efficiency of a Carnot heat engine is $0.5$ when the temperature of the source is $T_{1}$ and that of the sink is $T_{2}$. The efficiency of another Carnot heat engine is also $0.5$. The temperatures of the source and sink of the second engine are respectively:

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