Question

In a Carnot engine, if the absolute temperature of the source is \( 25% \) more than the absolute temperature of the sink, then the efficiency of the engine is

• A. \( 25% \)
• B. \( 50% \)
• C. \( 20% \)
• D. \( 40% \)

Hint:

The Carnot engine is the most efficient heat engine operating between two given temperatures. Its efficiency depends only on the absolute temperatures of the hot source (\( T_s \)) and the cold sink (\( T_k \)). Remember the formula \( \eta = 1 - \frac{T_k}{T_s} \). Always use absolute temperatures (Kelvin) for these calculations. If temperatures are given in Celsius, convert them to Kelvin by adding 273.15. Be careful with percentage increases; "25% more than \( X \)" means \( X + 0.25X = 1.25X \).

Answer 1

The Correct Option is C

Step 1: Define the temperatures of the source and sink.
Let \( T_s \) be the absolute temperature of the source (hot reservoir).
Let \( T_k \) be the absolute temperature of the sink (cold reservoir).
According to the problem statement, the absolute temperature of the source is \( 25% \) more than the absolute temperature of the sink.
This can be written as:
\( T_s = T_k + 0.25 T_k \)
\( T_s = 1.25 T_k \)
Step 2: State the formula for the efficiency of a Carnot engine.
The efficiency (\( \eta \)) of a Carnot engine is given by:
\( \eta = 1 - \frac{T_k}{T_s} \)
where \( T_k \) and \( T_s \) are absolute temperatures.
Step 3: Substitute the relationship between \( T_s \) and \( T_k \) into the efficiency formula.
Substitute \( T_s = 1.25 T_k \) into the efficiency formula:
\( \eta = 1 - \frac{T_k}{1.25 T_k} \)
\( \eta = 1 - \frac{1}{1.25} \)
Step 4: Calculate the efficiency.
Convert \( 1.25 \) to a fraction: \( 1.25 = \frac{5}{4} \).
\( \eta = 1 - \frac{1}{\frac{5}{4}} \)
\( \eta = 1 - \frac{4}{5} \)
\( \eta = \frac{5-4}{5} = \frac{1}{5} \)
Convert the efficiency to a percentage:
\( \eta = \frac{1}{5} \times 100% = 20% \).