The efficiency of a Carnot cycle is frac{1}{6 | 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

$325 \ K, 260 \ K$

Vedclass · Answer

(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.For the first case: $\frac{1}{6} = 1 - \frac{T_2}{T_1} \Rightarrow \frac{T_2}{T_1} = \frac{5}{6} \Rightarrow T_2 = \frac{5}{6} T_1$ ....$(i)$For the second case,the sink temperature is reduced by $65 \ K$,so the new sink temperature is $(T_2 - 65)$.$\frac{1}{3} = 1 - \frac{T_2 - 65}{T_1} \Rightarrow \frac{T_2 - 65}{T_1} = \frac{2}{3}$ ....(ii)Substituting $T_2 = \frac{5}{6} T_1$ from equation $(i)$ into equation (ii):$\frac{\frac{5}{6} T_1 - 65}{T_1} = \frac{2}{3} \Rightarrow \frac{5}{6} - \frac{65}{T_1} = \frac{2}{3}$$\frac{65}{T_1} = \frac{5}{6} - \frac{2}{3} = \frac{5-4}{6} = \frac{1}{6}$$T_1 = 65 	imes 6 = 390 \ K$Now,calculating the initial sink temperature $T_2$:$T_2 = \frac{5}{6} 	imes 390 = 325 \ K$The final sink temperature is $T_2 - 65 = 325 - 65 = 260 \ K$.Thus,the initial and final temperatures of the sink are $325 \ K$ and $260 \ K$ respectively.

The efficiency of a Carnot cycle is $\frac{1}{6}$. By lowering the temperature of the sink by $65 \ K$,it increases to $\frac{1}{3}$. The initial and final temperatures of the sink are:

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