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CGP EDU Academic Team
Published on: September 12, 2026
The temperature of reservoir of Carnot's engine operating with an efficiency of 70% is 1000K. The temperature of its sink is
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A
$\eta = 1 - \frac{\mathrm{T}_2}{\mathrm{T}_1} \Rightarrow \frac{70}{100} = 1 - \frac{\mathrm{T}_2}{1000} \Rightarrow \mathrm{T}_2 = 300 \mathrm{K}$
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