Question

A Carnot engine operates between two temperatures, \( T_L = 100\,\text{K} \) and \( T_H = 150\,\text{K} \). Each cycle of the engine lasts for 0.5 seconds during which the power delivered is \( 500\,\text{J/s} \). Let \( Q_H \) be the heat absorbed by the engine and \( Q_L \) be the heat lost. Identify the correct statement(s).

• A. \( Q_H = 750\,\text{J} \)
• B. \( \frac{Q_H}{Q_L} = \frac{3}{2} \)
• C. The change in entropy of the engine and the hot bath in a cycle is \( 5\,\text{J/K} \)
• D. The change in entropy of the engine in 0.5 seconds is zero

Hint:

In a Carnot engine, efficiency depends only on the reservoir temperatures, and the total entropy change of the system per cycle is always zero.

Answer 1

The Correct Option is A, C, D

Step 1: Determine work done per cycle.
Given power \( P = 500\,\text{J/s} \) and each cycle takes \( 0.5\,\text{s} \): \[ W = P \times t = 500 \times 0.5 = 250\,\text{J}. \] Step 2: Efficiency of Carnot engine.
\[ \eta = 1 - \frac{T_L}{T_H} = 1 - \frac{100}{150} = \frac{1}{3}. \] Step 3: Calculate heat absorbed from the hot reservoir.
\[ \eta = \frac{W}{Q_H} \Rightarrow Q_H = \frac{W}{\eta} = \frac{250}{1/3} = 750\,\text{J}. \] Step 4: Calculate \( Q_L \).
\[ Q_L = Q_H - W = 750 - 250 = 500\,\text{J}. \] \[ \frac{Q_H}{Q_L} = \frac{750}{500} = \frac{3}{2}. \] Step 5: Entropy change in a Carnot cycle.
For a reversible Carnot cycle, \[ \Delta S_{\text{system}} = 0, \] because the entropy gained and lost per cycle are equal in magnitude and opposite in sign. Step 6: Final Answer.
Correct options are (A) and (D).