A Carnot engine operating between temperature | vedclass

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(B) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$.Given $\eta_1 = \frac{1}{6}$,we have $\frac{1}{6} = 1 - \frac{T_2}{T_1}

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$372\, K$ and $310\, K$

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(B) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$.Given $\eta_1 = \frac{1}{6}$,we have $\frac{1}{6} = 1 - \frac{T_2}{T_1}$,which implies $\frac{T_2}{T_1} = \frac{5}{6} \implies T_2 = \frac{5}{6}T_1$ $...(i)$When $T_2$ is lowered by $62\, K$,the new temperature is $(T_2 - 62)$. The new efficiency is $\eta_2 = \frac{1}{3}$.So,$\frac{1}{3} = 1 - \frac{T_2 - 62}{T_1} \implies \frac{T_2 - 62}{T_1} = \frac{2}{3}$ $...(ii)$Substitute $T_2 = \frac{5}{6}T_1$ into equation $(ii)$:$\frac{\frac{5}{6}T_1 - 62}{T_1} = \frac{2}{3}$$\frac{5}{6} - \frac{62}{T_1} = \frac{2}{3}$$\frac{62}{T_1} = \frac{5}{6} - \frac{2}{3} = \frac{5-4}{6} = \frac{1}{6}$$T_1 = 62 	imes 6 = 372\, K$Now,find $T_2$ using equation $(i)$:$T_2 = \frac{5}{6} 	imes 372 = 5 	imes 62 = 310\, K$Thus,$T_1 = 372\, K$ and $T_2 = 310\, K$.

$A$ Carnot engine operating between temperatures $T_1$ and $T_2$ has an efficiency of $\frac{1}{6}$. When $T_2$ is lowered by $62\, K$,its efficiency increases to $\frac{1}{3}$. Then $T_1$ and $T_2$ are,respectively:

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