JEE Main · 2023 · 01 Feb 2023 · Shift 2 · Heat & Thermodynamics

A Carnot engine operating between two reservoirs has efficiency

\frac{1}{3}

. When the temperature of cold reservoir raised by x, its efficiency decreases to

\frac{1}{6}

. The value of x, if the temperature of hot reservoir is

9 9^{\circ} C

, will be:

(A)16.5 K
(B)33 K
(C)66 K
(D)62 K✓ Correct

Answer: (D)

Solution

For a Carnot engine the efficiency is

η = 1 - \frac{T _{C}}{T _{H}}

, with

T_{H} = 99^{\circ} C = 372

K. The initial efficiency

\frac{1}{3}

gives

\frac{T _{C}}{372} = \frac{2}{3}

, so

T_{C} = 248

K. Raising the cold reservoir to

T_{C} + x

lowers the efficiency to

\frac{1}{6}

, so

\frac{T _{C} + x}{372} = \frac{5}{6}

, giving

T_{C} + x = 310

K. Hence

x = 310 - 248 = 62

JEE Main · 2023 · 01 Feb 2023 · Shift 2 · Single correct option

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