A heat engine has an efficiency of frac{1}{6} | vedclass

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(B) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$,where $T_1$ is the source temperature and $T_2$ is the sink temperature

Vedclass · Accepted Answer

$99$

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(B) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$,where $T_1$ is the source temperature and $T_2$ is the sink temperature in Kelvin.In the first case,$\eta_1 = \frac{1}{6} = 1 - \frac{T_2}{T_1}$,which implies $\frac{T_2}{T_1} = \frac{5}{6}$ (Equation $i$).In the second case,the sink temperature is reduced by $62^{\circ}C$ (which is equivalent to $62 K$ in temperature difference),so the new sink temperature is $T_2' = T_2 - 62$. The new efficiency is $\eta_2 = 2 	imes \frac{1}{6} = \frac{1}{3}$.Thus,$\frac{1}{3} = 1 - \frac{T_2 - 62}{T_1}$,which implies $\frac{T_2 - 62}{T_1} = \frac{2}{3}$ (Equation $ii$).Subtracting Equation $ii$ from Equation $i$: $\frac{T_2}{T_1} - \frac{T_2 - 62}{T_1} = \frac{5}{6} - \frac{2}{3}$.$\frac{62}{T_1} = \frac{5-4}{6} = \frac{1}{6}$.Therefore,$T_1 = 62 	imes 6 = 372 K$.To convert Kelvin to Celsius: $T(^{\circ}C) = 372 - 273 = 99^{\circ}C$.

$A$ heat engine has an efficiency of $\frac{1}{6}$. When the temperature of the sink is reduced by $62^{\circ}C$,its efficiency gets doubled. The temperature of the source is $.....^{\circ}C$.

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