A Carnot engine with efficiency eta operates | vedclass

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(C) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$.Taking the differential, we get $d\eta = \frac{T_2}{T_1^2} dT_1 - \frac

Vedclass · Accepted Answer

-$2$

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(C) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$.Taking the differential, we get $d\eta = \frac{T_2}{T_1^2} dT_1 - \frac{1}{T_1} dT_2$.For a small change in $T_1$ (keeping $T_2$ constant), $\Delta \eta_1 = \frac{T_2}{T_1^2} \Delta T_1 = \frac{T_2}{T_1} \left( \frac{\Delta T_1}{T_1} ight) = (1 - \eta) \frac{\Delta T_1}{T_1}$.Given $\frac{\Delta T_1}{T_1} = 0.4 \%$, so $\Delta \eta_1 = (1 - \eta) 	imes 0.004$.For a small change in $T_2$ (keeping $T_1$ constant), $\Delta \eta_2 = -\frac{1}{T_1} \Delta T_2 = -\frac{T_2}{T_1} \left( \frac{\Delta T_2}{T_2} ight) = -(1 - \eta) \frac{\Delta T_2}{T_2}$.Given $\frac{\Delta T_2}{T_2} = 0.2 \%$, so $\Delta \eta_2 = -(1 - \eta) 	imes 0.002$.Taking the ratio, $\frac{\Delta \eta_1}{\Delta \eta_2} = \frac{(1 - \eta) 	imes 0.004}{-(1 - \eta) 	imes 0.002} = -2$.

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