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

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(C) The efficiency $\eta$ 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 temp

Vedclass · Accepted Answer

$342$

Vedclass · Answer

(C) The efficiency $\eta$ 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.Case $1$: $\eta_1 = \frac{1}{6} = 1 - \frac{T_2}{T_1} \implies \frac{T_2}{T_1} = 1 - \frac{1}{6} = \frac{5}{6} \implies T_2 = \frac{5}{6} T_1$.Case $2$: $\eta_2 = \frac{1}{3} = 1 - \frac{T_2 - 57}{T_1} \implies \frac{T_2 - 57}{T_1} = 1 - \frac{1}{3} = \frac{2}{3} \implies T_2 - 57 = \frac{2}{3} T_1$.Substitute $T_2 = \frac{5}{6} T_1$ into the second equation:$\frac{5}{6} T_1 - 57 = \frac{2}{3} T_1$$\frac{5}{6} T_1 - \frac{4}{6} T_1 = 57$$\frac{1}{6} T_1 = 57$$T_1 = 57 	imes 6 = 342 \ K$.Thus,the temperature of the source is $342 \ K$.

$A$ Carnot engine has an efficiency of $\frac{1}{6}$. When the temperature of the sink is lowered by $57 \ K$,its efficiency becomes $\frac{1}{3}$. The temperature of the source is: (in $K$)

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