Question

The efficiency of a Carnot’s engine, working between steam point and ice point, will be

• A. 26.81%
• B. 37.81%
• C. 47.81%
• D. 57.81%

Answer 1

The Correct Option is A

The given question is about the efficiency of a Carnot engine, which operates on a theoretical thermodynamic cycle. The goal is to determine the efficiency of a Carnot engine working between two temperatures: the steam point (100°C) and the ice point (0°C).

The efficiency \eta of a Carnot engine is given by the formula:

\eta = \left(1 - \frac{T_C}{T_H}\right) \times 100\%

where:

• T_C is the absolute temperature (in Kelvin) of the cold reservoir (ice point).
• T_H is the absolute temperature (in Kelvin) of the hot reservoir (steam point).

First, convert the given temperatures from Celsius to Kelvin:

• The steam point (hot reservoir): T_H = 100°C + 273.15 = 373.15 \, K
• The ice point (cold reservoir): T_C = 0°C + 273.15 = 273.15 \, K

Substitute the values into the efficiency formula:

\eta = \left(1 - \frac{273.15}{373.15}\right) \times 100\%

Calculate the efficiency:

\eta = \left(1 - 0.7314\right) \times 100\%

\eta = 0.2686 \times 100\% = 26.86\%

Given the options, the closest efficiency is 26.81%.

Thus, the correct answer is 26.81%.

This demonstrates that a Carnot engine working between the boiling and freezing points of water has a maximum possible efficiency of approximately 26.81% under ideal conditions.