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

Explanation

To solve the problem, we need to calculate the power required for the refrigerator to remove heat while working between two temperature limits. Here’s a step-by-step solution:

### Step 1: Convert temperatures to Kelvin
The temperatures given in Celsius need to be converted to Kelvin:
- \( T_1 = 30^\circ C = 30 + 273 = 303 \, K \)
- \( T_2 = 4^\circ C = 4 + 273 = 277 \, K \)

**Hint:** Remember that to convert Celsius to Kelvin, you add 273.

### Step 2: Convert heat removal to Joules
The heat removal rate is given as \( 600 \, \text{calories/second} \). We need to convert this to Joules using the conversion factor \( 1 \, \text{calorie} = 4.2 \, \text{J} \):
- \( Q_2 = 600 \, \text{calories/second} \times 4.2 \, \text{J/calorie} = 2520 \, \text{J/s} \)

**Hint:** Use the conversion factor to change calories to Joules.

### Step 3: Calculate the Coefficient of Performance (COP)
The Coefficient of Performance (COP) for a refrigerator is given by the formula:
\[ \beta = \frac{Q_2}{W} = \frac{T_2}{T_1 - T_2} \]
We can rearrange this to find the work done \( W \):
\[ W = \frac{Q_2 \cdot (T_1 - T_2)}{T_2} \]

**Hint:** The COP relates the heat removed and the work done; rearranging the formula helps to isolate \( W \).

### Step 4: Substitute values into the equation
Now, substitute the values we have:
- \( Q_2 = 2520 \, \text{J/s} \)
- \( T_1 = 303 \, K \)
- \( T_2 = 277 \, K \)

Calculating \( T_1 - T_2 \):
\[ T_1 - T_2 = 303 - 277 = 26 \, K \]

Now substituting into the equation for \( W \):
\[ W = \frac{2520 \, \text{J/s} \cdot 26 \, K}{277 \, K} \]

### Step 5: Calculate \( W \)
Now perform the calculation:
\[ W = \frac{65520 \, \text{J/s}}{277} \approx 236.5 \, \text{W} \]

**Hint:** When performing the division, ensure you keep track of significant figures.

### Step 6: Conclusion
The power required for the refrigerator to remove 600 calories of heat every second is approximately \( 236.5 \, \text{W} \).

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

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