Carnot Cycle | MCQs

Explanation

Given:

Initial efficiency (η₁) = 50% = 0.5

Initial cold body temperature (T₁) = 17°C = 290 K

Initial hot body temperature (T₂) = unknown

We know that the efficiency of a Carnot engine is:

η = 1 - (T₁ / T₂)

0.5 = 1 - (290 K / T₂)

T₂ = 580 K

Now, the hot body temperature is increased by 145°C:

New hot body temperature (T₂') = 580 K + 145°C = 725 K

New efficiency (η₂) is:

η₂ = 1 - (T₁ / T₂')

= 1 - (290 K / 725 K)

≈ 0.6

New efficiency (η₂) is approximately 60%.

Explanation

The Carnot cycle is best represented on a Pressure-Volume (P-V) graph.

It consists of two isothermal and two adiabatic processes, clearly shown in the P-V diagram.