Question

An ideal gas heat engine operates in a Carnot cycle between $227^{\circ} C $ and $127^{\circ} C. $ It absorbs $6\, kcal$ at the higher temperature. The amount of heat (in kcal) converted into work is equal to

• A. 4.8
• B. 3.5
• C. 1.6
• D. 1.2

Answer 1

The Correct Option is D

To solve this problem, we need to understand the working principle of a Carnot cycle, which is a theoretical thermodynamic cycle that provides insight into the efficiency of heat engines.

The Carnot efficiency \eta of an engine operating between two temperatures is given by the formula:

\eta = 1 - \frac{T_{2}}{T_{1}}

where:

• T_{1} is the absolute temperature (in Kelvin) of the hot reservoir.
• T_{2} is the absolute temperature (in Kelvin) of the cold reservoir.

Given temperatures:

• Hot reservoir temperature: 227^\circ C = 227 + 273 = 500\, K
• Cold reservoir temperature: 127^\circ C = 127 + 273 = 400\, K

Substituting these values into the efficiency formula, we get:

\eta = 1 - \frac{400}{500} = 1 - 0.8 = 0.2

This means the efficiency of the engine is 0.2 or 20%.

The engine absorbs 6\, kcal of heat at the higher temperature. The work done by the engine is the amount of heat absorbed multiplied by the efficiency:

\text{Work} = \eta \times \text{Heat Absorbed} = 0.2 \times 6\, \text{kcal} = 1.2\, \text{kcal}

Thus, the amount of heat converted into work is 1.2 kcal, which corresponds to the correct answer.