A Carnot engine, whose efficiency is 40%, tak | vedclass

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Question

(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 (exhaust)

Vedclass · Accepted Answer

$900$

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 (exhaust) temperature.For the first case: $\eta_1 = 0.40$, $T_1 = 600	ext{ K}$.$0.40 = 1 - \frac{T_2}{600} \implies \frac{T_2}{600} = 0.60 \implies T_2 = 360	ext{ K}$.For the second case: $\eta_2 = 0.60$, $T_2 = 360	ext{ K}$.$0.60 = 1 - \frac{360}{T_1'} \implies \frac{360}{T_1'} = 1 - 0.60 = 0.40$.$T_1' = \frac{360}{0.40} = 900	ext{ K}$.Thus, the required intake temperature is $900	ext{ K}$.

$A$ Carnot engine, whose efficiency is $40\%$, takes heat from a source maintained at temperature $600\text{ K}$. To have an efficiency of $60\%$, the intake temperature for the same exhaust temperature should be: (in $\text{ K}$)

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