A Carnot engine has efficiency of $50 \%$. If the temperature of sink is reduced by $40^{\circ} \mathrm{C}$, its efficiency increases by $30 \%$. The temperature of the source will be :

JEE Main · 2022 · 28 Jul 2022 · Shift 1 · Heat & Thermodynamics

A Carnot engine has efficiency of

50%

. If the temperature of sink is reduced by

4 0^{\circ} C

, its efficiency increases by

30%

. The temperature of the source will be :

(A)166.7 K
(B)255.1 K
(C)266.7 K✓ Correct
(D)367.7 K

Answer: (C)

Solution

Carnot efficiency:

η = 1 - \frac{T _{L}}{T _{H}}

. Initially

0.5 = 1 - \frac{T _{L}}{T _{H}}

, so

T_{L} = 0.5 T_{H}

. After reducing sink temperature by 40 K, new efficiency is

0.5 \times 1.3 = 0.65 = 1 - \frac{T _{L} - 40}{T _{H}}

. This gives

\frac{T _{L} - 40}{T _{H}} = 0.35

. Substituting

T_{L} = 0.5 T_{H}

\frac{0.5 T _{H} - 40}{T _{H}} = 0.35

, so

0.5 T_{H} - 40 = 0.35 T_{H}

, yielding

0.15 T_{H} = 40

, thus

T_{H} = 266.7

JEE Main · 2022 · 28 Jul 2022 · Shift 1 · Single correct option

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