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

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(C) For a Carnot engine,the efficiency $\eta$ is given by $\eta = 1 - \frac{T_L}{T_H}$,where $T_L$ is the sink temperature and $T_H$ is the source tem

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$900$

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(C) For a Carnot engine,the efficiency $\eta$ is given by $\eta = 1 - \frac{T_L}{T_H}$,where $T_L$ is the sink temperature and $T_H$ is the source temperature.Given $\eta_1 = 40 \% = 0.4$ and $T_{H1} = 600 \ K$:$0.4 = 1 - \frac{T_L}{600}$$\frac{T_L}{600} = 1 - 0.4 = 0.6$$T_L = 0.6 	imes 600 = 360 \ K$.Now,for the desired efficiency $\eta_2 = 60 \% = 0.6$ with the same sink temperature $T_L = 360 \ K$:$0.6 = 1 - \frac{360}{T_{H2}}$$\frac{360}{T_{H2}} = 1 - 0.6 = 0.4$$T_{H2} = \frac{360}{0.4} = 900 \ K$.

$A$ Carnot engine,whose efficiency is $40 \%$,takes heat from a source maintained at a temperature of $600 \ K$. If it is desired to have an efficiency of $60 \%$,then the intake temperature for the same exhaust (sink) temperature should be: (in $K$)

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