Question

A Carnot engine efficiency is equal to $\frac{1}{7}$. If the temperature of the sink is reduced by 65 K, the efficiency becomes $\frac{1}{4}$. Then temperature of source and the sink in the first case are respectively :

• A. 610 K and 520 K
• B. 520 K and 606 67 K
• C. 606.67 K and 520 K
• D. 520 K and 610 K

Answer 1

Answer: (C)

606.67 K and 520 K

$\because 1-\frac{T_2}{T_1}=\frac{1}{7}$(in case 1)
$\therefore \frac{T_2}{T_1}=\frac{6}{7}$ ....(i)
$1-\frac{T_2-65}{T_1}=\frac{1}{4}$ (in case 2)
$\therefore \frac{T_2-65}{T_1}=\frac{3}{4}$ .....(ii)
From Eqs. (i) and (ii) taking ratios
$\frac{T_2/T_1}{\frac{T_2-65}{T_1}}=\frac{6/7}{3/4}=\frac{8}{7}$
$\Rightarrow \frac{T_2}{T_2-65}=\frac{8}{7}$
$\Rightarrow 7T_2=8T_2-8\times 65$
$T_2=520K$
$\because \frac{T_2}{T_1}=\frac{6}{7}$
$\therefore T_1=\frac{T_2\times 7}{6}=\frac{520\times 7}{6}$
$T1=606.67K$