The temperature of the sink of a Carnot engin | vedclass

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(C) Temperature of the sink, $T_2 = 300 \,K$. Efficiency of the engine, $\eta_1 = 0.25$.The formula for efficiency is $\eta = 1 - \frac{T_2}{T_1}$.Sub

Vedclass · Accepted Answer

$0.15$

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(C) Temperature of the sink, $T_2 = 300 \,K$. Efficiency of the engine, $\eta_1 = 0.25$.The formula for efficiency is $\eta = 1 - \frac{T_2}{T_1}$.Substituting the values: $0.25 = 1 - \frac{300}{T_1} \Rightarrow \frac{300}{T_1} = 0.75 \Rightarrow T_1 = \frac{300}{0.75} = 400 \,K$.When the source temperature is increased by $100 \,K$, the new source temperature $T_1' = 400 + 100 = 500 \,K$.The new efficiency $\eta_2 = 1 - \frac{T_2}{T_1'} = 1 - \frac{300}{500} = 1 - 0.6 = 0.40$.The increase in efficiency is $\Delta\eta = \eta_2 - \eta_1 = 0.40 - 0.25 = 0.15$.

The temperature of the sink of a Carnot engine is $300 \,K$ and the efficiency of the engine is $0.25$. If the temperature of the source of the engine is increased by $100 \,K$, the efficiency of the engine increases by:

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