Question

A Carnot engine (E) is working between two temperatures 473K and 273K. In a new system two engines - engine \(E_1\) works between 473K to 373K and engine \(E_2\) works between 373K to 273K. If \(\eta_{12}\), \(\eta_1\) and \(\eta_2\) are the efficiencies of the engines \(E\), \(E_1\) and \(E_2\), respectively, then:

• A. \(\eta_{12} = \eta_1 \eta_2\)
• B. \(\eta_{12} = \eta_1 + \eta_2\)
• C. \(\eta_{12}<\eta_1 + \eta_2\)
• D. \(\eta_{12}>\eta_1 + \eta_2\)

Hint:

Remember, when dealing with multiple Carnot engines operating between different temperature ranges in series, their efficiencies multiply rather than add.

Answer 1

The Correct Option is C

Let's denote the temperatures as follows:

• $ T_H = 473 $ K (highest temperature)
• $ T_M = 373 $ K (intermediate temperature)
• $ T_C = 273 $ K (lowest temperature)

The Carnot engine (E) operates between $ T_H $ and $ T_C $, so its efficiency is:

$$ \eta_{12} = 1 - \frac{T_C}{T_H} = 1 - \frac{273}{473} = \frac{473 - 273}{473} = \frac{200}{473} $$

Engine $ E_1 $ operates between $ T_H $ and $ T_M $, so its efficiency is:

$$ \eta_1 = 1 - \frac{T_M}{T_H} = 1 - \frac{373}{473} = \frac{473 - 373}{473} = \frac{100}{473} $$

Engine $ E_2 $ operates between $ T_M $ and $ T_C $, so its efficiency is:

$$ \eta_2 = 1 - \frac{T_C}{T_M} = 1 - \frac{273}{373} = \frac{373 - 273}{373} = \frac{100}{373} $$

We need to compare $ \eta_{12} $ with $ \eta_1 + \eta_2 $ and $ \eta_1 \eta_2 $.

$$ \eta_1 + \eta_2 = \frac{100}{473} + \frac{100}{373} = 100 \left( \frac{1}{473} + \frac{1}{373} \right)$$ 
$$= 100 \left( \frac{373 + 473}{473 \cdot 373} \right) = 100 \left( \frac{846}{473 \cdot 373} \right) = \frac{84600}{176329} \approx 0.4798 $$

$$ \eta_1 \eta_2 = \frac{100}{473} \times \frac{100}{373} = \frac{10000}{176329} \approx 0.0567 $$

$$ \eta_{12} = \frac{200}{473} \approx 0.4228 $$

Comparison:

• Since $ 0.4228 < 0.4798 $, we have $ \eta_{12} < \eta_1 + \eta_2 $.

Now let's check if $ (1 - \eta_1)(1 - \eta_2) = 1 - \eta_{12} $:

$$ (1 - \eta_1)(1 - \eta_2) = \frac{373}{473} \times \frac{273}{373} = \frac{273}{473} = 1 - \eta_{12} $$

This implies:

$$ 1 - \eta_1 - \eta_2 + \eta_1 \eta_2 = 1 - \eta_{12} $$

$$ \Rightarrow \eta_{12} = \eta_1 + \eta_2 - \eta_1 \eta_2 $$

Since $ \eta_1 \eta_2 > 0 $, it follows that $ \eta_{12} < \eta_1 + \eta_2 $.

Final Answer:
The final answer is $ \eta_{12} < \eta_1 + \eta_2 $.