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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 temperature of the cold reservoir and $T_1$ is the t

Vedclass · Accepted Answer

$380$

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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 temperature of the cold reservoir and $T_1$ is the temperature of the hot reservoir.Given $T_2 = 7 + 273 = 280 \, K$ and initial efficiency $\eta_1 = 0.5$.$0.5 = 1 - \frac{280}{T_1} \implies \frac{280}{T_1} = 0.5 \implies T_1 = 560 \, K$.For the new efficiency $\eta_2 = 0.7$ with the same cold reservoir temperature $T_2 = 280 \, K$:$0.7 = 1 - \frac{280}{T_1'} \implies \frac{280}{T_1'} = 0.3 \implies T_1' = \frac{280}{0.3} = 933.33 \, K$.The increase in temperature of the high temperature reservoir is $\Delta T = T_1' - T_1 = 933.33 - 560 = 373.33 \, K$.Rounding to the nearest provided option,the increase is approximately $380 \, K$.

$A$ Carnot engine whose low temperature reservoir is at $7\,^{\circ}C$ has an efficiency of $50\%$. It is desired to increase the efficiency to $70\%$. By how many degrees should the temperature of the high temperature reservoir be increased? (in $K$)

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