A Carnot engine having efficiency 60 % receiv | vedclass

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(C) 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 temperatur

Vedclass · Accepted Answer

$1200$

Vedclass · Answer

(C) 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: $\eta_1 = 0.60$, $T_1 = 600 	ext{ K}$.$0.60 = 1 - \frac{T_2}{600} \implies \frac{T_2}{600} = 0.40 \implies T_2 = 240 	ext{ K}$.For the second case: $\eta_2 = 0.80$, $T_2 = 240 	ext{ K}$.$0.80 = 1 - \frac{240}{T_1'} \implies \frac{240}{T_1'} = 0.20 \implies T_1' = \frac{240}{0.20} = 1200 	ext{ K}$.Thus, the new source temperature required is $1200 	ext{ K}$.

$A$ Carnot engine having efficiency $60 \%$ receives heat from a source at a temperature $600 \text{ K}$. For the same sink temperature, to increase its efficiency to $80 \%$, the temperature of the source is (in $\text{ K}$)

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