The relation between efficiency (η) of Carnot engine and coefficient of performance (η₁) of refrigerator is (a) η = (1)/(1+η₁) (b) η = (1)/(1-η₁) (c) η = (η₁)/(1-η₁) (d) η = (1+η₁)/(η₁)

Answer: (a). η = 1 - (T₂)/(T₁) = (T₁ - T₂)/(T₁) and η₁ = (T₂)/(T₁ - T₂) 1 + η₁ = (T₁)/(T₁ - T₂) = 1/η So η = (1)/(1 + η₁); the answer is (a).

The relation between efficiency (η) of Carnot engine and coefficient… | MHT CET 2025 (5 May, Shift 2)

Q8MHT CET 2025 · 5 MAY · SHIFT 2Easy

Heat Engines, Refrigerators & Second Law

The relation between efficiency $(\eta)$ of Carnot engine and coefficient of performance $(\eta_1)$ of refrigerator is

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Answer: (a)

$\eta = 1 - \frac{T_2}{T_1} = \frac{T_1 - T_2}{T_1}$ and $\eta_1 = \frac{T_2}{T_1 - T_2}$
$1 + \eta_1 = \frac{T_1}{T_1 - T_2} = \frac{1}{\eta}$
So $\eta = \frac{1}{1 + \eta_1}$ ; the answer is (a).

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