Question

The relation between efficiency ((\eta)) of Carnot engine and coefficient of performance ((\beta)) of refrigerator is

• A. (\eta = \frac{1}{1 + \beta})
• B. (\eta = \frac{1}{1 - \beta})
• C. (\eta = \frac{\beta}{1 - \beta})
• D. (\eta = \frac{1 + \beta}{\beta})

Hint:

Carnot cycle COP ($\beta$) is always greater than efficiency ($\eta$).

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The question asks for the standard mathematical relationship between the efficiency of a heat engine and the COP of a refrigerator operating between the same two temperatures.
Step 2: Key Formula or Approach:
Efficiency of Carnot engine: \(\eta = 1 - \frac{T_{2}}{T_{1}}\)
Coefficient of Performance (COP) of refrigerator: \(\eta_{1} = \frac{T_{2}}{T_{1} - T_{2}}\)
Note: Some textbooks use \(\beta\) or \(K\) for COP. Here it is given as \(\eta_{1}\).
Step 3: Detailed Explanation:
From the COP formula:
\[ \eta_{1} = \frac{T_{2}}{T_{1} - T_{2}} \]
\[ \frac{1}{\eta_{1}} = \frac{T_{1} - T_{2}}{T_{2}} = \frac{T_{1}}{T_{2}} - 1 \]
\[ \frac{1}{\eta_{1}} + 1 = \frac{T_{1}}{T_{2}} \implies \frac{1 + \eta_{1}}{\eta_{1}} = \frac{T_{1}}{T_{2}} \]
Taking reciprocals:
\[ \frac{T_{2}}{T_{1}} = \frac{\eta_{1}}{1 + \eta_{1}} \]
Now, substitute this into the efficiency formula:
\[ \eta = 1 - \frac{T_{2}}{T_{1}} = 1 - \frac{\eta_{1}}{1 + \eta_{1}} \]
\[ \eta = \frac{1 + \eta_{1} - \eta_{1}}{1 + \eta_{1}} = \frac{1}{1 + \eta_{1}} \]
Step 4: Final Answer:
The correct relation is \(\eta = \frac{1}{1 + \eta_{1}}\).