An engine has an efficiency of 1/6. When the | vedclass

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(D) 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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(D) 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.Given $\eta_1 = 1/6$,so $1/6 = 1 - \frac{T_2}{T_1}$,which implies $\frac{T_2}{T_1} = 5/6$ or $T_2 = \frac{5}{6}T_1$ $...(i)$.When the sink temperature is reduced by $62^{\circ}C$ (which is equivalent to a change of $62 \ K$),the new efficiency $\eta_2 = 2 	imes \eta_1 = 2 	imes (1/6) = 1/3$.The new sink temperature is $T_2' = T_2 - 62$.Using the efficiency formula: $1/3 = 1 - \frac{T_2 - 62}{T_1}$.Rearranging gives $\frac{T_2 - 62}{T_1} = 1 - 1/3 = 2/3$.Substituting $T_2 = \frac{5}{6}T_1$ from equation $(i)$:$\frac{\frac{5}{6}T_1 - 62}{T_1} = 2/3$.$\frac{5}{6} - \frac{62}{T_1} = 2/3$.$\frac{62}{T_1} = \frac{5}{6} - \frac{4}{6} = 1/6$.$T_1 = 62 	imes 6 = 372 \ K$.To convert to Celsius: $T(^{\circ}C) = 372 - 273 = 99^{\circ}C$.

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

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