The efficiency of the Carnot engine is $50 \%$ and the temperature of the sink is $500 \mathrm{~K}$. If the…

The efficiency of the Carnot engine is $50 \%$ and the temperature of the sink is $500 \mathrm{~K}$. If the temperature of the source is kept constant and its efficiency raised to $60 \%$, then the required temperature of the sink will be:
  1. $100 \mathrm{~K}$
  2. $600 \mathrm{~K}$
  3. $400 \mathrm{~K}$
  4. $500 \mathrm{~K}$

Solution

$\begin{aligned} & \% \eta=\left(1-\frac{T_2}{T_1}\right) \times 100 \\ & \therefore \text { For } 50 \%, \frac{500}{T_1}=1-\frac{50}{11} \\ & \Rightarrow T_1=1000 \mathrm{~K} \\ & \text { aim for } 60 \%, \frac{60}{100}=1-\frac{T_2}{1000} \\ & \Rightarrow T_2=400 \mathrm{~K} \end{aligned}$ ^

Asked in: NEET 2002

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