Question

If a heat engine is filled at temperature \( 27^\circ C \) and heat of 100 kcal is taken from the source at temperature \( 677^\circ C \), work done (in J) is

• A. \( 0.28 \times 10^6 \)
• B. \( 2.8 \times 10^6 \)
• C. \( 28 \times 10^6 \)
• D. \( 0.028 \times 10^6 \)

Hint:

When solving problems involving heat engines, use the Carnot efficiency formula to calculate the work done based on the temperature of the source and sink.

Answer 1

The Correct Option is A

For the heat engine, we can use the efficiency of a Carnot engine to calculate the work done. The efficiency \( \eta \) of a Carnot engine is given by: \[ \eta = 1 - \frac{T_2}{T_1} \] Where: - \( T_1 = 677^\circ C = 950K \), - \( T_2 = 27^\circ C = 300K \). Thus, the efficiency becomes: \[ \eta = 1 - \frac{300}{950} = \frac{13}{19} \] Now, the work done \( W \) is related to the heat absorbed from the source \( Q_1 \) by: \[ W = \eta Q_1 \] Given that \( Q_1 = 100 \, \text{kcal} = 418400 \, \text{J} \), we substitute the values into the equation: \[ W = \frac{13}{19} \times 418400 \approx 0.28 \times 10^6 \, \text{J} \] Thus, the correct answer is (A).