An ideal heat engine working between temperat | vedclass

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(B) The efficiency of an ideal heat engine (Carnot engine) is given by the formula: $\eta = 1 - \frac{T_2}{T_1} = \frac{T_1 - T_2}{T_1}$.In the first

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$\eta$

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(B) The efficiency of an ideal heat engine (Carnot engine) is given by the formula: $\eta = 1 - \frac{T_2}{T_1} = \frac{T_1 - T_2}{T_1}$.In the first case,the efficiency is $\eta = \frac{T_1 - T_2}{T_1}$.In the second case,the new source temperature is $T_1' = 2T_1$ and the new sink temperature is $T_2' = 2T_2$.The new efficiency $\eta'$ is given by: $\eta' = \frac{T_1' - T_2'}{T_1'} = \frac{2T_1 - 2T_2}{2T_1}$.Simplifying the expression: $\eta' = \frac{2(T_1 - T_2)}{2T_1} = \frac{T_1 - T_2}{T_1} = \eta$.Therefore,the efficiency remains unchanged.

An ideal heat engine working between temperatures $T_1$ and $T_2$ has an efficiency $\eta$. What will be the new efficiency if both the source and sink temperatures are doubled?

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