A refrigerator works between 0°C and 27°C. Heat is to be removed from the refrigerated space at the rate of 50 kcal/ minute, the power of the motor of the refrigerator is

Options

A. 0.346 kW
B. 3.46 kW
C. 34.6 kW
D. 346 kW

Explanation

To solve the problem, we need to determine the power of the motor of the refrigerator based on the heat removal rate and the temperatures involved. Here’s a step-by-step solution:

### Step 1: Identify the temperatures
The refrigerator operates between:
- Lower temperature (T2) = 0°C = 273 K
- Higher temperature (T1) = 27°C = 300 K

### Step 2: Convert heat removal rate to appropriate units
The heat removal rate (QL) is given as 50 kcal/min. We need to convert this to joules per second (watts):
1 kcal = 4184 J
Thus,
\[ QL = 50 \text{ kcal/min} = 50 \times 4184 \text{ J/min} \]

To convert minutes to seconds:
\[ QL = \frac{50 \times 4184 \text{ J}}{60 \text{ s}} \]

### Step 3: Calculate QL in joules per second
Calculating QL:
\[ QL = \frac{50 \times 4184}{60} \]
\[ QL = \frac{209200}{60} \]
\[ QL \approx 3486.67 \text{ J/s} \]
Thus,
\[ QL \approx 3486.67 \text{ W} \]

### Step 4: Use the Coefficient of Performance (COP)
The Coefficient of Performance (COP) for a refrigerator is given by:
\[ COP = \frac{QL}{W} \]
Where W is the work done by the refrigerator.

From the Carnot efficiency, we know:
\[ COP = \frac{T2}{T1 - T2} \]
Substituting the values:
\[ COP = \frac{273}{300 - 273} = \frac{273}{27} \]

### Step 5: Calculate COP
Calculating COP:
\[ COP = \frac{273}{27} \approx 10.11 \]

### Step 6: Relate COP to work done
From the definition of COP:
\[ W = \frac{QL}{COP} \]
Substituting the values:
\[ W = \frac{3486.67}{10.11} \]

### Step 7: Calculate W
Calculating W:
\[ W \approx 344.5 \text{ W} \]

### Step 8: Convert W to kilowatts
To convert watts to kilowatts:
\[ W \approx 0.3445 \text{ kW} \]

### Final Answer
The power of the motor of the refrigerator is approximately **0.3445 kW**.

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