Question

If the temperature of a hot reservoir is 600K and the cold reservoir is 300K, the efficiency of the Carnot engine is:

• A. 25%
• B. 75%
• C. 50%
• D. 100%

Hint:

Always verify that the given temperatures are in Kelvin (K) before using the Carnot efficiency formula.
If they are provided in Celsius, you must convert them by adding 273.15.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
A Carnot engine is an idealized heat engine that operates on the Carnot cycle.
Its efficiency is solely determined by the temperatures of the hot and cold reservoirs.
Step 2: Key Formula or Approach:
The efficiency \( \eta \) of a Carnot engine is given by the formula:
\[ \eta = 1 - \frac{T_C}{T_H} \]
To express the efficiency as a percentage, we multiply by 100%:
\[ \eta (%) = \left( 1 - \frac{T_C}{T_H} \right) \times 100% \]
Step 3: Detailed Explanation:
Given the temperature of the hot reservoir \( T_H = 600\text{ K} \).
Given the temperature of the cold reservoir \( T_C = 300\text{ K} \).
Substituting these values into the efficiency formula:
\[ \eta = 1 - \frac{300}{600} \]
\[ \eta = 1 - \frac{1}{2} = \frac{1}{2} \]
Converting this to a percentage:
\[ \eta (%) = \frac{1}{2} \times 100% = 50% \]
Step 4: Final Answer:
The efficiency of the Carnot engine is 50%.