Question

Consider two Carnot engines of efficiencies $\eta_1$ and $\eta_2$. The first engine absorbs heat $Q_1$ from a heat reservoir A and releases heat $Q_2$ to a heat reservoir B. The second engine takes heat $Q_2$ from B and releases heat $Q_3$ to a heat reservoir C. If $Q_1 > Q_2 > Q_3$, what is the net efficiency of this combination of the two Carnot engines?

• A. $\eta_1 + \eta_2 - \eta_1\eta_2$
• B. $\eta_1 + \eta_2 + \eta_1\eta_2$
• C. $\eta_1\eta_2$
• D. $\eta_1 + \eta_2$

Hint:

For any number of engines operating in series, the overall fraction of heat remaining is the product of the individual remaining fractions:
\[ (1 - \eta_{net}) = (1 - \eta_1)(1 - \eta_2)\dots(1 - \eta_n) \] Expanding this product for two stages immediately yields the correct net efficiency.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Efficiency is defined as \( 1 - Q_{\text{out}}/Q_{\text{in}} \). When engines work in series, the "fraction of heat remaining" multiplies.

Step 2: Detailed Explanation:
1. Engine 1: \( \eta_{1} = 1 - \frac{Q_{2}}{Q_{1}} \implies \frac{Q_{2}}{Q_{1}} = 1 - \eta_{1} \).
2. Engine 2: \( \eta_{2} = 1 - \frac{Q_{3}}{Q_{2}} \implies \frac{Q_{3}}{Q_{2}} = 1 - \eta_{2} \).
3. Net efficiency (\( \eta \)):
\[ \eta = 1 - \frac{Q_{3}}{Q_{1}} \]
Substitute \( \frac{Q_{3}}{Q_{1}} = \frac{Q_{3}}{Q_{2}} \times \frac{Q_{2}}{Q_{1}} \):
\[ \eta = 1 - (1 - \eta_{2})(1 - \eta_{1}) \]
\[ \eta = 1 - (1 - \eta_{1} - \eta_{2} + \eta_{1}\eta_{2}) \]
\[ \eta = \eta_{1} + \eta_{2} - \eta_{1}\eta_{2} \]

Step 3: Final Answer:
The net efficiency is \( \eta_{1} + \eta_{2} - \eta_{1}\eta_{2} \).