Question

A heat engine extracts heat (\(Q_H\)) from a thermal reservoir at a temperature of 1000 K and rejects heat (\(Q_L\)) to a thermal reservoir at a temperature of 100 K, while producing work (\(W\)). Which one of the combinations of [\(Q_H\), \(Q_L\) and \(W\)] given is allowed?

• A. \(Q_H = 2000\) J, \(Q_L = 500\) J, \(W = 1000\) J
• B. \(Q_H = 2000\) J, \(Q_L = 750\) J, \(W = 1250\) J
• C. \(Q_H = 6000\) J, \(Q_L = 500\) J, \(W = 5500\) J
• D. \(Q_H = 6000\) J, \(Q_L = 600\) J, \(W = 5500\) J

Hint:

When checking heat engine problems, always perform two checks in order: 1. Does it obey the First Law? \(W = Q_H - Q_L\). This is a simple energy balance. 2. If yes, does it obey the Second Law? Is \(\eta_{\text{actual}} \le \eta_{\text{Carnot}}\)? An engine cannot be more efficient than a reversible Carnot engine.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
For a heat engine to be thermodynamically possible, it must satisfy both the First Law and the Second Law of Thermodynamics.
1. First Law (Conservation of Energy): The work produced must be equal to the net heat supplied. \(W = Q_H - Q_L\).
2. Second Law (Clausius Inequality): The efficiency of any real heat engine must be less than or equal to the efficiency of a Carnot engine operating between the same two temperatures. \(\eta_{\text{actual}} \le \eta_{\text{Carnot}}\).
Step 2: Key Formula or Approach:
- First Law: \(W = Q_H - Q_L\)
- Actual Efficiency: \(\eta_{\text{actual}} = \frac{W}{Q_H}\)
- Carnot Efficiency: \(\eta_{\text{Carnot}} = 1 - \frac{T_L}{T_H}\)
- Condition for operation: \(\eta_{\text{actual}} \le \eta_{\text{Carnot}}\)
Step 3: Detailed Calculation:
First, calculate the maximum possible (Carnot) efficiency: \[ \eta_{\text{Carnot}} = 1 - \frac{T_L}{T_H} = 1 - \frac{100 \text{ K}}{1000 \text{ K}} = 1 - 0.1 = 0.9 \text{ or } 90% \] Now, let's check each option:
(A) \(Q_H = 2000\), \(Q_L = 500\), \(W = 1000\):
- First Law check: \(Q_H - Q_L = 2000 - 500 = 1500\) J. Given \(W = 1000\) J. Since \(W \neq Q_H - Q_L\), this violates the First Law. Not allowed.
(B) \(Q_H = 2000\), \(Q_L = 750\), \(W = 1250\):
- First Law check: \(Q_H - Q_L = 2000 - 750 = 1250\) J. Given \(W = 1250\) J. First Law is satisfied.
- Second Law check: \(\eta_{\text{actual}} = \frac{W}{Q_H} = \frac{1250}{2000} = 0.625\).
- Compare efficiencies: \(0.625<0.9\). Since \(\eta_{\text{actual}}<\eta_{\text{Carnot}}\), the Second Law is satisfied. Allowed.
(C) \(Q_H = 6000\), \(Q_L = 500\), \(W = 5500\):
- First Law check: \(Q_H - Q_L = 6000 - 500 = 5500\) J. Given \(W = 5500\) J. First Law is satisfied.
- Second Law check: \(\eta_{\text{actual}} = \frac{W}{Q_H} = \frac{5500}{6000} \approx 0.9167\).
- Compare efficiencies: \(0.9167>0.9\). Since \(\eta_{\text{actual}}>\eta_{\text{Carnot}}\), this violates the Second Law. Not allowed.
(D) \(Q_H = 6000\), \(Q_L = 600\), \(W = 5500\):
- First Law check: \(Q_H - Q_L = 6000 - 600 = 5400\) J. Given \(W = 5500\) J. Since \(W \neq Q_H - Q_L\), this violates the First Law. Not allowed.
Step 4: Final Answer:
Only the combination in option (B) is allowed as it satisfies both the First and Second Laws of Thermodynamics.
Step 5: Why This is Correct:
Option (B) is the only one where the work output equals the net heat input (\(1250 = 2000 - 750\)) and the resulting efficiency (62.5%) is less than the maximum possible Carnot efficiency (90%).