A Carnot engine with its cold body at 17,^{ci | vedclass

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(B) 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

$60$

Vedclass · Answer

(B) 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 in Kelvin.Given $T_2 = 17 + 273 = 290 \, K$ and $\eta = 0.5$.$0.5 = 1 - \frac{290}{T_1} \implies \frac{290}{T_1} = 0.5 \implies T_1 = 580 \, K$.When the source temperature is increased by $145 \, ^{\circ}C$,the new temperature $T_1' = 580 + 145 = 725 \, K$.The new efficiency $\eta' = 1 - \frac{T_2}{T_1'} = 1 - \frac{290}{725} = 1 - 0.4 = 0.6$.Thus,the new efficiency is $60\%$.

$A$ Carnot engine with its cold body at $17\,^{\circ}C$ has $50\%$ efficiency. If the temperature of its hot body is now increased by $145\,^{\circ}C$,the efficiency becomes...... $\%$

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