Question

A heat engine operates between a hot source at \(T_{\text{H}} = 900\text{ K}\) and a cold sink at the ambient temperature \(T_{0} = T_{\text{C}} = 300\text{ K}\). During one cycle the engine receives \(Q_{\text{H}} = 12,000\text{ kJ}\) from the hot source and rejects \(Q_{\text{C}} = 8,000\text{ kJ}\) to the cold sink. Which of the following statements is correct?

• A. The entropy generation per cycle is \(S_{\text{gen}} = 4.00\text{ kJ/K}\).
• B. The process is reversible because the work equals the Carnot work for these temperatures.
• C. The entropy generation per cycle is \(S_{\text{gen}} = 13.333\text{ kJ/K}\).
• D. The entropy generation is zero.

Hint:

An entropy generation value of zero ($S_{\text{gen}} = 0$) indicates a fully reversible process.
If $S_{\text{gen}} > 0$, the process is irreversible.
A negative value ($S_{\text{gen}} < 0$) violates the Second Law of Thermodynamics and is physically impossible.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The question asks us to analyze a heat engine cycle operating between two temperatures to determine if the process is reversible or irreversible, and to calculate the entropy generated during one cycle.

Step 2: Key Formula or Approach:
The entropy change of the universe, which is equal to the entropy generation (\(S_{\text{gen}}\)), is calculated by summing the entropy changes of the system and its surroundings:
\[ S_{\text{gen}} = \Delta S_{\text{system}} + \Delta S_{\text{surrounds}} \]
For a complete engine cycle, the change in entropy of the engine itself is zero (\(\Delta S_{\text{engine}} = 0\)).
The entropy change of the reservoirs is:
\[ S_{\text{gen}} = -\frac{Q_{\text{H}}}{T_{\text{H}}} + \frac{Q_{\text{C}}}{T_{\text{C}}} \]

Step 3: Detailed Explanation:

• Identify the given parameters from the problem:
Source temperature, \(T_{\text{H}} = 900\text{ K}\).
Sink temperature, \(T_{\text{C}} = 300\text{ K}\).
Heat absorbed from source, \(Q_{\text{H}} = 12000\text{ kJ}\).
Heat rejected to sink, \(Q_{\text{C}} = 8000\text{ kJ}\).

• Calculate the entropy change of the hot reservoir:
\[ \Delta S_{\text{hot}} = -\frac{Q_{\text{H}}}{T_{\text{H}}} = -\frac{12000}{900} = -13.333\text{ kJ/K} \]

• Calculate the entropy change of the cold reservoir:
\[ \Delta S_{\text{cold}} = \frac{Q_{\text{C}}}{T_{\text{C}}} = \frac{8000}{300} = 26.667\text{ kJ/K} \]

• Calculate the total entropy generation per cycle:
\[ S_{\text{gen}} = \Delta S_{\text{hot}} + \Delta S_{\text{cold}} = -13.333 + 26.667 = 13.334\text{ kJ/K} \]

• Since \(S_{\text{gen}} > 0\), the process is irreversible, which makes Options B and D incorrect.

Step 4: Final Answer:
The entropy generation per cycle is \(13.333\text{ kJ/K}\).