A Carnot engine has an efficiency of 1/6. Whe | vedclass

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(B) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_{	ext{sink}}}{T_{	ext{source}}}$.Given $\eta_1 = 1/6$,we have:$\frac{1}{6} = 1

Vedclass · Accepted Answer

$99\,^{\circ}C, 37\,^{\circ}C$

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(B) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_{	ext{sink}}}{T_{	ext{source}}}$.Given $\eta_1 = 1/6$,we have:$\frac{1}{6} = 1 - \frac{T_{	ext{sink}}}{T_{	ext{source}}} \implies \frac{T_{	ext{sink}}}{T_{	ext{source}}} = \frac{5}{6} \quad ...(i)$When the sink temperature is reduced by $62\,^{\circ}C$ (or $62\,K$),the efficiency doubles,so $\eta_2 = 2 	imes (1/6) = 1/3$. Thus:$\frac{1}{3} = 1 - \frac{T_{	ext{sink}} - 62}{T_{	ext{source}}} \implies \frac{T_{	ext{sink}} - 62}{T_{	ext{source}}} = \frac{2}{3} \quad ...(ii)$Substitute $\frac{T_{	ext{sink}}}{T_{	ext{source}}} = \frac{5}{6}$ from $(i)$ into $(ii)$:$\frac{T_{	ext{sink}}}{T_{	ext{source}}} - \frac{62}{T_{	ext{source}}} = \frac{2}{3} \implies \frac{5}{6} - \frac{62}{T_{	ext{source}}} = \frac{4}{6}$$\frac{62}{T_{	ext{source}}} = \frac{1}{6} \implies T_{	ext{source}} = 372\,K$.Converting to Celsius: $T_{	ext{source}} = 372 - 273 = 99\,^{\circ}C$.From $(i)$,$T_{	ext{sink}} = \frac{5}{6} 	imes 372 = 310\,K$.Converting to Celsius: $T_{	ext{sink}} = 310 - 273 = 37\,^{\circ}C$.Therefore,the source and sink temperatures are $99\,^{\circ}C$ and $37\,^{\circ}C$ respectively.

$A$ Carnot engine has an efficiency of $1/6$. When the temperature of the sink is reduced by $62\,^{\circ}C$,its efficiency is doubled. The temperatures of the source and the sink are,respectively:

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