Question

A Carnot engine operates between 600 K and 300 K. Efficiency is:

• A. 25%
• B. 50%
• C. 75%
• D. 100%

Hint:

To maximize the efficiency of a Carnot engine, you must either increase the temperature of the source (\( T_H \)) or decrease the temperature of the sink (\( T_L \)). A 100% efficient engine is theoretically impossible as it would require a sink at absolute zero (0 K).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
A Carnot engine is a theoretical ideal engine that operates on the Carnot cycle, which is the most efficient cycle possible for a heat engine.
The efficiency of a Carnot engine is determined entirely by the temperatures of the hot reservoir (source) and the cold reservoir (sink).
It does not depend on the working substance used in the engine.
Key Formula or Approach:
The formula for the efficiency (\(\eta\)) of a Carnot engine is:
\[ \eta = 1 - \frac{T_L}{T_H} \]
To express efficiency as a percentage:
\[ \eta% = \left( 1 - \frac{T_L}{T_H} \right) \cdot 100 \]
Where:
- \(T_H\) is the absolute temperature of the hot source.
- \(T_L\) is the absolute temperature of the cold sink.
Step 2: Detailed Explanation:
Given Temperatures:
Temperature of the source (\(T_H\)) = 600 K.
Temperature of the sink (\(T_L\)) = 300 K.
Both temperatures are already in Kelvin (K), so no conversion is needed.
Substitute the values into the efficiency formula:
\[ \eta = 1 - \frac{300}{600} \]
\[ \eta = 1 - \frac{1}{2} = 0.5 \]
Convert the decimal value to a percentage:
\[ \eta% = 0.5 \cdot 100 = 50% \]
Step 3: Final Answer:
The efficiency of the engine is 50%.
This matches option (B).