A Carnot engine operating between temperature | vedclass

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Question

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

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$360\,K$ and $300\,K$

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(A) The efficiency $\eta$ of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$.Given $\eta = \frac{1}{6}$,we have $\frac{1}{6} = 1 - \frac{T_2}{T_1}$,which implies $\frac{T_2}{T_1} = \frac{5}{6}$ or $T_2 = \frac{5}{6}T_1$. (Equation $1$)When $T_2$ is lowered by $60\,K$,the new temperature is $(T_2 - 60)$. The new efficiency is $\frac{1}{3}$.So,$\frac{1}{3} = 1 - \frac{T_2 - 60}{T_1}$.This simplifies to $\frac{T_2 - 60}{T_1} = \frac{2}{3}$. (Equation $2$)Substitute $T_2 = \frac{5}{6}T_1$ from Equation $1$ into Equation $2$:$\frac{\frac{5}{6}T_1 - 60}{T_1} = \frac{2}{3}$$\frac{5}{6} - \frac{60}{T_1} = \frac{2}{3}$$\frac{60}{T_1} = \frac{5}{6} - \frac{2}{3} = \frac{5-4}{6} = \frac{1}{6}$$T_1 = 60 	imes 6 = 360\,K$.Now,find $T_2$ using $T_2 = \frac{5}{6}T_1 = \frac{5}{6} 	imes 360 = 300\,K$.Thus,$T_1 = 360\,K$ and $T_2 = 300\,K$.

$A$ Carnot engine operating between temperatures $T_1$ and $T_2$ has efficiency $\frac{1}{6}$. When $T_2$ is lowered by $60\,K$,its efficiency increases to $\frac{1}{3}$. Then $T_1$ and $T_2$ are respectively:

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