Question

A Carnot engine Converts one sixth of the heat input into work. When the temperature of the sink is reduced by 62° C the efficiency of the engine is doubled. The temperature of the source and sink are 

(A) 80°C, 37°C 

(B) 95°C, 28°C 

(C) 90°C, 37°C 

(D) 99°C, 37°C

Answer 1

 Correct option: (D) 99°C, 37°C

Explanation:

(1/6) of heat is converted to work

i.e.  (W/Q) = (1/6)

i.e. n = (W/Q) = (1/6)

As n = 1 – (T₁ / T₂) hence (T₂ / T₁) = [1 – (1/6)] = 5/6………(1)

When temp of sink i.e. T₂ is reduced by 62°C, η gets doubled.

i.e.  n' = [1 – (T₂' / T₁)]

∴   2n = 1 – [(T₂ – 62) / T₁]

2n = 1 – (T₂ / T₁) + (62 / T₁)

From (1), 2 × (1/6) = 1 – (5/6) + (62 / T₁)

(1/3) = (1/6) + (62 / T₁)

∴   (62 / T₁) = (1/3) – (1/6)

∴ (62 / T₁) = (1/6)

∴   T₁ = 62 × 6

∴  T₁ = 372 k = 99°C

As (T₂ / T₁) = (5/6) hence

∴  T₂ = (5/6) × 372K

∴  T₂ = 310 K

∴   T₂ = 37°C

∴ Temp of source & sink = 99°C & 37°C