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

The temperatures are defined as:

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

The efficiency of the Carnot engine (E) operating between $ T_H $ and $ T_C $ is calculated as:

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

The efficiency of engine $ E_1 $ operating between $ T_H $ and $ T_M $ is:

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

The efficiency of engine $ E_2 $ operating between $ T_M $ and $ T_C $ is:

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

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

The sum of efficiencies $ \eta_1 + \eta_2 $ is:

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

The product of efficiencies $ \eta_1 \eta_2 $ is:

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

The efficiency $ \eta_{12} $ is:

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

Comparison Results:

• $ \eta_{12} $ (approximately 0.4228) is less than $ \eta_1 + \eta_2 $ (approximately 0.4798), as $ 0.4228<0.4798 $.

We verify the relationship $ (1 - \eta_1)(1 - \eta_2) = 1 - \eta_{12} $:

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

Since $ \frac{273}{473} = 1 - \frac{200}{473} $, it follows that:

$$ (1 - \eta_1)(1 - \eta_2) = 1 - \eta_{12} $$

Expanding this equation yields:

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

Rearranging the terms gives:

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

Given that $ \eta_1 \eta_2>0 $, it is confirmed that $ \eta_{12}<\eta_1 + \eta_2 $.

Conclusion:
The final result is $ \eta_{12}<\eta_1 + \eta_2 $.