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An engine has an efficiency of $\dfrac{1}{6}$. When the temperature of sink is reduced by $62^\circ C$, its efficiency is doubled. Temperature of the source is
A
$37^\circ C$
B
$62^\circ C$
C
$99^\circ C$
D
$124^\circ C$
Detailed Solution
Efficiency of the engine: $\eta = 1 - \dfrac{T_2}{T_1}$, where $T_1$ is the temperature of the source and $T_2$ that of the sink (in kelvin).
According to the problem, $\dfrac{1}{6} = 1 - \dfrac{T_2}{T_1}$ ...(1)
From (1): $\dfrac{T_2}{T_1} = \dfrac{5}{6}$
When the temperature of the sink is reduced by $62^\circ C$ (a fall of 62 K), the efficiency is doubled: $2\left(\dfrac{1}{6}\right) = 1 - \dfrac{T_2 - 62}{T_1}$ ...(2)
$\dfrac{1}{3} = 1 - \dfrac{T_2}{T_1} + \dfrac{62}{T_1} = \dfrac{1}{6} + \dfrac{62}{T_1}$
$\dfrac{62}{T_1} = \dfrac{1}{3} - \dfrac{1}{6} = \dfrac{1}{6}$
$T_1 = 62 \times 6 = 372$ K
$T_1 = 372 - 273 = 99^\circ C$ = temperature of the source.
(The sink is at $T_2 = \dfrac{5}{6} \times 372 = 310$ K $= 37^\circ C$.)
According to the problem, $\dfrac{1}{6} = 1 - \dfrac{T_2}{T_1}$ ...(1)
From (1): $\dfrac{T_2}{T_1} = \dfrac{5}{6}$
When the temperature of the sink is reduced by $62^\circ C$ (a fall of 62 K), the efficiency is doubled: $2\left(\dfrac{1}{6}\right) = 1 - \dfrac{T_2 - 62}{T_1}$ ...(2)
$\dfrac{1}{3} = 1 - \dfrac{T_2}{T_1} + \dfrac{62}{T_1} = \dfrac{1}{6} + \dfrac{62}{T_1}$
$\dfrac{62}{T_1} = \dfrac{1}{3} - \dfrac{1}{6} = \dfrac{1}{6}$
$T_1 = 62 \times 6 = 372$ K
$T_1 = 372 - 273 = 99^\circ C$ = temperature of the source.
(The sink is at $T_2 = \dfrac{5}{6} \times 372 = 310$ K $= 37^\circ C$.)