The efficiency of an ideal Carnot engine work | 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 temperature

Vedclass · Accepted Answer

$60$

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.Given $\eta = 1/3$,we have $\frac{1}{3} = 1 - \frac{T_2}{T_1}$,which implies $\frac{T_2}{T_1} = 1 - \frac{1}{3} = \frac{2}{3}$.When the sink temperature is reduced by $40 \%$,the new sink temperature $T_2^{\prime} = T_2 - 0.40 T_2 = 0.6 T_2$.The new efficiency $\eta^{\prime}$ is given by $\eta^{\prime} = 1 - \frac{T_2^{\prime}}{T_1} = 1 - \frac{0.6 T_2}{T_1}$.Substituting $\frac{T_2}{T_1} = \frac{2}{3}$,we get $\eta^{\prime} = 1 - 0.6 	imes \frac{2}{3} = 1 - 0.4 = 0.6$.Converting to percentage,$\eta^{\prime} = 0.6 	imes 100 \% = 60 \%$.

The efficiency of an ideal Carnot engine working between temperatures $T_1$ and $T_2$ is $1/3$. If the temperature of the sink is reduced by $40 \%$,then its efficiency will be: (in $\%$)

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