Question

A Carnot engine is operated between a hot and a cold reservoir at a temperature difference of 623.15 K. If the efficiency of this engine is 25%, the temperature of the cold reservoir is
1. 1050∘C
2. 2033∘C
3. 1240∘C
4. 640∘C

Answer 1

Correct Answer - Option 1 : 1050∘C

The correct answer is option 1) i.e. 1050​∘C

CONCEPT:

• Carnot engine: Carnot engine is a heat engine that operates on a Carnot cycle.
  • A heat engine is a closed-cycle device that takes heat from a hot reservoir, does useful work, and rejects the remaining heat to a cold reservoir.  
  • In each cycle, the engine absorbs a heat Q1 from the hot reservoir at a temperature T1, converts it into useful work W and the remaining heat Q2 is rejected to the cold reservoir at a temperature T2.
  • The efficiency η of a Carnot engine is given as

\(⇒ η = 1 - \frac{T_2}{T_1}\)

CALCULATION:

Given that:

Temperature difference, ΔT = T₁ - T₂ = 623.15 K = 623.15 - 273.15 = 350^∘C

Efficiency, η = 25% = 0.25

We know, \( η = 1 - \frac{T_2}{T_1}\)

\(⇒ 0.25 = 1 - \frac{T_2}{T_1} = \frac{T_1 - T_2}{T_1}\)

\(⇒ 0.25 = \frac{\Delta T}{T_1} = \frac{350}{T_1}\)

⇒ T₁ = 1400∘C

⇒ ΔT = T1 - T2 ⇒ T₂ = T₁ - ΔT

• The temperature of the cold reservoir, 

⇒ T2 = 1400 - 350 = 1050∘C