A Carnot engine operating between temperature | vedclass

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(D) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$,where $T_2$ is the sink temperature and $T_1$ is the source temperature

Vedclass · Accepted Answer

$380$

Vedclass · Answer

(D) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$,where $T_2$ is the sink temperature and $T_1$ is the source temperature.Given $T_2 = 7^{\circ}C = 280 \ K$ and $\eta_1 = 0.5$.$0.5 = 1 - \frac{280}{T_1} \Rightarrow \frac{280}{T_1} = 0.5 \Rightarrow T_1 = 560 \ K$.Now,for $\eta_2 = 0.7$ with the same sink temperature $T_2 = 280 \ K$:$0.7 = 1 - \frac{280}{T_1'} \Rightarrow \frac{280}{T_1'} = 0.3 \Rightarrow T_1' = \frac{280}{0.3} = 933.33 \ K$.The increase in source temperature is $\Delta T = T_1' - T_1 = 933.33 - 560 = 373.33 \ K$.Rounding to the nearest given option,the increase is approximately $380 \ K$.

$A$ Carnot engine operating between temperatures $7^{\circ}C$ and a higher temperature has an efficiency of $50\%$. To increase its efficiency to $70\%$,by how much must the temperature of the source be increased in $K$?

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