A refrigerator works between `4^(@)C` and `30^(@)C`. It is required to remove `600 calories` of heat every second in order to keep the temperature of the refrigerator space constant.The power required is (Take `1calorie= 4.2 J`)

Options

A. 23.65 W
B. 236.5 W
C. 2365 W
D. 2.365 W

Answer

To solve the problem, we need to find the power required by the refrigerator that operates between two temperatures and removes a specific amount of heat. Here’s a step-by-step solution:

### Step 1: Convert temperatures to Kelvin
The temperatures given are:
- \( T_H = 30^\circ C \)
- \( T_L = 4^\circ C \)

To convert Celsius to Kelvin, we use the formula:
\[ T(K) = T(°C) + 273 \]

Calculating:
\[ T_H = 30 + 273 = 303 \, K \]
\[ T_L = 4 + 273 = 277 \, K \]

### Step 2: Identify the heat removal rate
The refrigerator is required to remove \( Q_L = 600 \, \text{calories} \) of heat every second. We need to convert this to joules using the conversion factor \( 1 \, \text{calorie} = 4.2 \, \text{J} \):
\[ Q_L = 600 \, \text{calories} \times 4.2 \, \text{J/calorie} = 2520 \, \text{J} \]

### Step 3: Use the Coefficient of Performance (COP) formula
The Coefficient of Performance (COP) for a refrigerator is given by:
\[ COP = \frac{Q_L}{P} \]
where \( P \) is the power required.

Also, the COP can be expressed in terms of the temperatures:
\[ COP = \frac{T_L}{T_H - T_L} \]

### Step 4: Set the two expressions for COP equal
From the two expressions for COP, we have:
\[ \frac{Q_L}{P} = \frac{T_L}{T_H - T_L} \]

### Step 5: Rearranging the equation to find power
Rearranging the equation gives:
\[ P = \frac{Q_L \cdot (T_H - T_L)}{T_L} \]

### Step 6: Substitute the values
Substituting the values we have:
- \( Q_L = 2520 \, \text{J} \)
- \( T_H = 303 \, K \)
- \( T_L = 277 \, K \)

Calculating \( T_H - T_L \):
\[ T_H - T_L = 303 - 277 = 26 \, K \]

Now substituting into the power equation:
\[ P = \frac{2520 \, \text{J} \cdot 26 \, K}{277 \, K} \]

Calculating:
\[ P = \frac{65520 \, \text{J}}{277} \approx 236.5 \, \text{W} \]

### Final Answer
The power required is approximately \( 236.5 \, \text{W} \).