A Carnot heat engine has an efficiency of 50 | vedclass

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Question

(A) 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

$1500$

Vedclass · Answer

(A) 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 $(T_2 = 500 \ K)$.For $\eta_1 = 50\% = 0.5$:$0.5 = 1 - \frac{500}{T_1} \Rightarrow \frac{500}{T_1} = 0.5 \Rightarrow T_1 = 1000 \ K$.For $\eta_2 = 80\% = 0.8$,let the new source temperature be $T_1' = T_1 + \Delta T$:$0.8 = 1 - \frac{500}{T_1 + \Delta T} \Rightarrow \frac{500}{T_1 + \Delta T} = 0.2$.$T_1 + \Delta T = \frac{500}{0.2} = 2500 \ K$.Since $T_1 = 1000 \ K$,the increment $\Delta T = 2500 \ K - 1000 \ K = 1500 \ K$.

$A$ Carnot heat engine has an efficiency of $50 \%$. The temperature of the sink is maintained at $500 \ K$. To increase the efficiency up to $80 \%$,the required increment in the source temperature is: (in $K$)

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