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(D) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_L}{T_H}$,where $T_L$ is the temperature of the low-temperature reservoir and $T_

Vedclass · Accepted Answer

$380$

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(D) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_L}{T_H}$,where $T_L$ is the temperature of the low-temperature reservoir and $T_H$ is the temperature of the high-temperature reservoir in Kelvin.Given $T_L = 7^{\circ} C = 7 + 273 = 280 \ K$.For the first case,$\eta_1 = 50 \% = 0.5$.$0.5 = 1 - \frac{280}{T_{H1}} \implies \frac{280}{T_{H1}} = 0.5 \implies T_{H1} = 560 \ K$.For the second case,$\eta_2 = 70 \% = 0.7$.$0.7 = 1 - \frac{280}{T_{H2}} \implies \frac{280}{T_{H2}} = 0.3 \implies T_{H2} = \frac{280}{0.3} = \frac{2800}{3} \approx 933.33 \ K$.The increase in temperature is $\Delta T = T_{H2} - T_{H1} = \frac{2800}{3} - 560 = \frac{2800 - 1680}{3} = \frac{1120}{3} \approx 373.33 \ K$.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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