A reversible engine has an efficiency of $\frac{1}{4}$. If the temperature of the sink is reduced by $58^{\circ} \mathrm{C}$, its efficiency becomes double. Calculate the temperature of the sink :

(C) $180.4^{\circ} \mathrm{C}$

JEE Main · 2021 · 31 Aug 2021 · Shift 1 · Heat & Thermodynamics

A reversible engine has an efficiency of

\frac{1}{4}

. If the temperature of the sink is reduced by

5 8^{\circ} C

, its efficiency becomes double. Calculate the temperature of the sink :

(A) $28 0^{\circ} C$
(B) $38 2^{\circ} C$
(C) $180. 4^{\circ} C$
(D) $17 4^{\circ} C$ ✓ Correct

Answer: (D)

Solution

For a Carnot engine

η = 1 - \frac{T _{C}}{T _{H}}

. Initially

\frac{1}{4} = 1 - \frac{T _{C}}{T _{H}}

, so

\frac{T _{C}}{T _{H}} = \frac{3}{4}

. After the sink falls by

58

\frac{1}{2} = 1 - \frac{T _{C} - 58}{T _{H}}

, so

\frac{T _{C} - 58}{T _{H}} = \frac{1}{2}

. Dividing the two relations,

\frac{T _{C} - 58}{T _{C}} = \frac{2}{3}

, hence

3 (T_{C} - 58) = 2 T_{C}

and

T_{C} = 174

. The sink temperature is

174

JEE Main · 2021 · 31 Aug 2021 · Shift 1 · Single correct option

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