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Question

(B) The efficiency $\eta$ of a Carnot engine is given by $\eta = \frac{W}{Q} = 1 - \frac{T_2}{T_1}$,where $W$ is work done,$Q$ is input heat,$T_1$ is

Vedclass · Accepted Answer

$600 \ K, 450 \ K$

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(B) The efficiency $\eta$ of a Carnot engine is given by $\eta = \frac{W}{Q} = 1 - \frac{T_2}{T_1}$,where $W$ is work done,$Q$ is input heat,$T_1$ is the source temperature,and $T_2$ is the sink temperature.Initially,$\eta = \frac{1}{4}$. Therefore,$1 - \frac{T_2}{T_1} = \frac{1}{4} \implies \frac{T_2}{T_1} = \frac{3}{4} \implies T_2 = \frac{3}{4}T_1$ (Equation $i$).When the sink temperature is reduced by $50 \ K$,the new sink temperature is $T_2' = T_2 - 50$. The new efficiency is $\eta' = 33 \frac{1}{3} \% = \frac{1}{3}$.Thus,$1 - \frac{T_2 - 50}{T_1} = \frac{1}{3} \implies 1 - \frac{T_2}{T_1} + \frac{50}{T_1} = \frac{1}{3}$ (Equation $ii$).Substituting $\frac{T_2}{T_1} = \frac{3}{4}$ from Equation $i$ into Equation $ii$:$1 - \frac{3}{4} + \frac{50}{T_1} = \frac{1}{3} \implies \frac{1}{4} + \frac{50}{T_1} = \frac{1}{3}$.$\frac{50}{T_1} = \frac{1}{3} - \frac{1}{4} = \frac{4 - 3}{12} = \frac{1}{12}$.$T_1 = 50 	imes 12 = 600 \ K$.Now,$T_2 = \frac{3}{4} 	imes 600 = 450 \ K$.Therefore,the initial temperatures are $600 \ K$ and $450 \ K$.

$A$ reversible Carnot heat engine converts $\frac{1}{4}$ of its input heat into work. When the temperature of the sink is reduced by $50 \ K$,its efficiency becomes $33 \frac{1}{3} \%$. The initial temperatures of the source and the sink respectively are

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