Efficiency of a heat engine whose sink is at temperature of $300 \mathrm{~K}$ is $40 \%$. To increase the…

Efficiency of a heat engine whose sink is at temperature of $300 \mathrm{~K}$ is $40 \%$. To increase the efficiency to $60 \%$, keeping the sink temperature constant, the source temperature must be increased by
  1. $750 \mathrm{~K}$
  2. $500 \mathrm{~K}$
  3. $250 \mathrm{~K}$
  4. $1000 \mathrm{~K}$

Solution

$ \begin{array}{ll} \frac{T_2}{T_1}=1-\eta=1-\frac{40}{100}=\frac{3}{5} \\ \Rightarrow & T_1=\frac{5}{3} T_2 \\ \Rightarrow & T_1=\frac{5}{3} \times 300=500 \mathrm{~K} \end{array} $ New efficiency $\eta^{\prime}=60 \%$ $ \begin{aligned} & \frac{T_2}{T_1^{\prime}}=1-\eta^{\prime}=1-\frac{60}{100}=\frac{2}{5} \\ \Rightarrow \quad & T_1^{\prime}=\frac{5}{2} \times 300=750 \mathrm{~K} \end{aligned} $

Asked in: AP EAMCET 2013

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