A Carnot engine has an efficiency of $1 / 6$. When the temperature of the sink is reduced by $62^{\circ} \mathrm{C}$, its efficiency is doubled. The temperatures of the source and the sink are, respectively

(B) $99^{\circ} \mathrm{C}, 37^{\circ} \mathrm{C}$

(A) $62^{\circ} \mathrm{C}, 124^{\circ} \mathrm{C}$

(C) $37^{\circ} \mathrm{C}, 99^{\circ} \mathrm{C}$

(D) $124^{\circ} \mathrm{C}, 62^{\circ} \mathrm{C}$

JEE Main · 2019 · 12 Apr 2019 · Shift 2 · Heat & Thermodynamics

A Carnot engine has an efficiency of

1/6

. When the temperature of the sink is reduced by

6 2^{\circ} C

, its efficiency is doubled. The temperatures of the source and the sink are, respectively

Answer: (B)

For a Carnot engine

η = 1 - \frac{T _{2}}{T _{1}}

. With

η = \frac{1}{6}

\frac{T _{2}}{T _{1}} = \frac{5}{6}

. Dropping the sink by

6 2^{\circ}

C doubles the efficiency to

\frac{1}{3}

, so

1 - \frac{T _{2} - 62}{T _{1}} = \frac{1}{3}

, that is

\frac{T _{2} - 62}{T _{1}} = \frac{2}{3}

. Subtracting the two efficiency relations,

\frac{T _{2}}{T _{1}} - \frac{T _{2} - 62}{T _{1}} = \frac{62}{T _{1}} = \frac{5}{6} - \frac{2}{3} = \frac{1}{6}

, so

T_{1} = 372

= 9 9^{\circ}

C and

T_{2} = \frac{5}{6} \times 372 = 310

= 3 7^{\circ}

C. The source is at

9 9^{\circ}

C and the sink at

3 7^{\circ}

JEE Main · 2019 · 12 Apr 2019 · Shift 2 · Single correct option