Question

Efficiency of Carnot engine is always:

• A. Greater than 1
• B. Less than 1
• C. Equal to 1
• D. Infinite

Hint:

An efficiency of $1$ ($100\%$) would violate the Kelvin-Planck statement of the Second Law of Thermodynamics.
Therefore, all real and ideal heat engine efficiencies are strictly less than $1$.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The question asks about the upper limit of the thermal efficiency of a Carnot heat engine.

Step 2: Key Formula or Approach:
The thermal efficiency of a Carnot heat engine is given by:
\[ \eta_c = 1 - \frac{T_C}{T_H} \]
where:
$T_C$ is the temperature of the cold reservoir (in Kelvin).
$T_H$ is the temperature of the hot reservoir (in Kelvin).

Step 3: Detailed Explanation:

• According to the Second Law of Thermodynamics, it is impossible for any heat engine to convert $100\%$ of absorbed heat into useful work. Some heat must always be rejected to a colder reservoir.

• Mathematically, for the efficiency $\eta_c$ to be equal to $1$ ($100\%$):
- Either the cold reservoir temperature must be absolute zero ($T_C = 0\text{ K}$).
- Or the hot reservoir temperature must be infinite ($T_H \rightarrow \infty$).

• According to the Third Law of Thermodynamics, absolute zero ($0\text{ K}$) cannot be reached in a finite number of steps. Infinite temperatures are also physically impossible.

• Therefore, the term $\frac{T_C}{T_H}$ is always strictly greater than $0$, which means $\eta_c$ must always be strictly less than $1$.

Step 4: Final Answer:
The efficiency of a Carnot engine is always less than 1.