A reversible Carnot heat engine converts $\frac{1}{4}$ th of its input heat into work. When the temperature of the sink is reduced by $50 \mathrm{~K}$, its efficiency becomes $33 \frac{1}{3} \%$. The initial temperatures of the source and the sink respectively are

• A

  600 K, 550 K
• Correct answer
  B

  600 K, 450 K
• C

  300 K, 150 K
• D

  450 K, 350 K

Solution

Given work, $W=\frac{Q}{4}$ where, $Q=$ input heat. $$ \text { Efficiency, } \eta=\frac{W}{Q}=\frac{Q / 4}{Q} \Rightarrow \eta=\frac{1}{4} $$ Also, $\quad \eta=1-\frac{T_2}{T_1}$ where, $T_2=$ temperature of sink and $\quad T_1=$ temperature of source. $$ 1-\frac{T_2}{T_1}=\frac{1}{4} $$ $\Rightarrow$ When temperature of sink is reduced by $50 \mathrm{~K}$, efficiency becomes $\frac{100}{3} \%$ $$ \eta=\frac{100}{3} \times \frac{1}{100}=\frac{1}{3} $$ So, $\quad 1-\frac{T_2-50}{T_1}=\frac{1}{3}$
Subtracting Eq (ii) from Eq. (i), we get $$ \begin{aligned} & \frac{50}{T_1}=\frac{1}{3}-\frac{1}{4}=\frac{1}{12} \\ & T_1=600 \mathrm{~K} \\ & T_2=\frac{3}{4} \times 600=450 \mathrm{~K} \end{aligned} $$

Hence, the correct answer is option: 'B'.