Question

A Carnot engine takes \( 3 \times 10^6 \) cal of heat from a reservoir at \( 627^\circ C \), and gives it to a sink at \( 27^\circ C \). The work done by the engine is:

• A. \( 4.2 \times 10^6 \) J
• B. \( 8.4 \times 10^6 \) J
• C. \( 16.8 \times 10^6 \) J
• D. \( 0 \)

Hint:

For a Carnot engine:
- Efficiency is determined by \( T_H \) and \( T_C \).
- More work is extracted when \( T_H \) is significantly higher than \( T_C \).
- Convert Celsius to Kelvin before calculations.

Answer 1

The Correct Option is B

Step 1: The efficiency of a Carnot engine is given by:
\[ \eta = 1 - \frac{T_C}{T_H} \] where \( T_H = 627^\circ C + 273 = 900 K \) and \( T_C = 27^\circ C + 273 = 300 K \).
Step 2: Substituting values:
\[ \eta = 1 - \frac{300}{900} = 1 - \frac{1}{3} = \frac{2}{3} \]
Step 3: The work done is given by:
\[ W = \eta Q_H = \frac{2}{3} \times (3 \times 10^6 \times 4.2) \]
\[ W = \frac{2}{3} \times 12.6 \times 10^6 = 8.4 \times 10^6 { J} \]