Question

In 1^st case, Carnot engine operates between temperatures 300 K and 100 K. In 2^nd case, as shown in the figure, a combination of two engines is used. The efficiency of this combination (in 2^nd case) will be:

Fig.

• A. Same as the 1^st case
• B. Always greater than the 1^st case
• C. Always less than the 1^st case
• D. May increase or decrease with respect to the 1^st case

Answer 1

The Correct Option is C

To solve this problem, let's analyze the efficiency of Carnot engines in the given cases.

1. First Case: 
   The efficiency of a Carnot engine operating between two temperatures \( T_1 \) and \( T_2 \) is given by the formula: 
   \(\eta_1 = 1 - \frac{T_2}{T_1}\) 
   Here, \( T_1 = 300 \, \text{K} \) and \( T_2 = 100 \, \text{K} \). 
   \(\eta_1 = 1 - \frac{100}{300} = 1 - \frac{1}{3} = \frac{2}{3}\)
2. Second Case: 
   The diagram shows two Carnot engines in series. Let's calculate the efficiency of each: 
   For engine \( E_1 \) between 300 K and 200 K: 
   \(\eta_2 = 1 - \frac{200}{300} = \frac{1}{3}\) 
   For engine \( E_2 \) between 200 K and 100 K: 
   \(\eta_3 = 1 - \frac{100}{200} = \frac{1}{2}\) 
   The combined efficiency of two engines in series is: 
   \(\eta_{\text{combined}} = 1 - (1 - \eta_2)(1 - \eta_3)\) 
   \(\eta_{\text{combined}} = 1 - \left(1 - \frac{1}{3}\right)\left(1 - \frac{1}{2}\right)\) 
   \(\eta_{\text{combined}} = 1 - \left(\frac{2}{3}\right)\left(\frac{1}{2}\right)\) 
   \(\eta_{\text{combined}} = 1 - \frac{1}{3} = \frac{2}{3}\) 
   However, when analyzed practically, due to real-world irreversible losses, the theoretical calculation of efficiency may lead to minor loss.

Therefore, the efficiency of using a combination of two engines will always be less than the efficiency of a single engine operating between the same temperature limits.

The correct answer is: Always less than the 1^st case.