Looking for classes? Ksquare Career Institute, Bengaluru →
The ratio $\dfrac{W}{Q}$ for a Carnot engine is $\dfrac{1}{6}$. Now the temp. of sink is reduced by $62^\circ C$, then this ratio becomes twice, therefore the initial temp. of the sink and source are respectively :
A
$33^\circ C$, $67^\circ C$
B
$37^\circ C$, $99^\circ C$
C
$67^\circ C$, $33^\circ C$
D
97 K, 37 K
Detailed Solution
$\dfrac{W}{Q} = \eta = 1 - \dfrac{T_L}{T_H} = \dfrac{1}{6} \Rightarrow \dfrac{T_L}{T_H} = \dfrac{5}{6}$
When the sink temperature is reduced by 62 (a change of 62 °C = 62 K), the efficiency becomes $\dfrac{2}{6}$:
$1 - \dfrac{T_L - 62}{T_H} = \dfrac{2}{6} \Rightarrow \dfrac{T_L - 62}{T_H} = \dfrac{2}{3}$
$\Rightarrow 3T_L - 186 = 2T_H$. Substituting $T_L = \dfrac{5}{6}T_H$: $\dfrac{5}{2}T_H - 186 = 2T_H$
$\dfrac{1}{2}T_H = 186 \Rightarrow T_H = 372$ K $= 99^\circ C$
$T_L = \dfrac{5}{6} \times 372 = 310$ K $= 37^\circ C$
So the initial temperatures of the sink and source are $37^\circ C$ and $99^\circ C$.
When the sink temperature is reduced by 62 (a change of 62 °C = 62 K), the efficiency becomes $\dfrac{2}{6}$:
$1 - \dfrac{T_L - 62}{T_H} = \dfrac{2}{6} \Rightarrow \dfrac{T_L - 62}{T_H} = \dfrac{2}{3}$
$\Rightarrow 3T_L - 186 = 2T_H$. Substituting $T_L = \dfrac{5}{6}T_H$: $\dfrac{5}{2}T_H - 186 = 2T_H$
$\dfrac{1}{2}T_H = 186 \Rightarrow T_H = 372$ K $= 99^\circ C$
$T_L = \dfrac{5}{6} \times 372 = 310$ K $= 37^\circ C$
So the initial temperatures of the sink and source are $37^\circ C$ and $99^\circ C$.