A refrigerator works between temperature of melting ice and room temperatures `(17^(@)C)` . The amount of energy (in kWh) that must be supplied to freeze I kg of water at `0^(@)C` is

Options

A. 1.4
B. 1.8
C. 0.058
D. 2.5

Explanation

To solve the problem of calculating the amount of energy that must be supplied to freeze 1 kg of water at 0°C using a refrigerator operating between the temperatures of melting ice (0°C) and room temperature (17°C), we can follow these steps:

### Step 1: Understand the Coefficient of Performance (COP)
The Coefficient of Performance (COP) of a refrigerator is defined as the ratio of the heat removed from the cold reservoir (Q2) to the work input (W):
\[ COP = \frac{Q2}{W} \]
Where:
- \( Q2 \) is the heat removed (in Joules).
- \( W \) is the work done (in Joules).

### Step 2: Convert temperatures to Kelvin
The temperatures need to be converted to Kelvin for calculations:
- Room temperature (T1) = 17°C = 17 + 273 = 290 K
- Melting ice temperature (T2) = 0°C = 0 + 273 = 273 K

### Step 3: Calculate the heat removed (Q2)
To freeze 1 kg of water at 0°C, we need to remove the latent heat of fusion. The latent heat of fusion of water is approximately 334,000 J/kg. Therefore, for 1 kg of water:
\[ Q2 = 1 \, \text{kg} \times 334,000 \, \text{J/kg} = 334,000 \, \text{J} \]

### Step 4: Use the COP formula to find W
From the COP formula, we can rearrange it to find W:
\[ W = \frac{Q2}{COP} \]
The COP can also be expressed in terms of temperatures:
\[ COP = \frac{T2}{T1 - T2} = \frac{273}{290 - 273} = \frac{273}{17} \]

### Step 5: Substitute values into the W equation
Substituting the values into the equation for W:
\[ W = \frac{334,000 \, \text{J}}{\frac{273}{17}} = 334,000 \times \frac{17}{273} \]

### Step 6: Calculate W
Calculating W:
\[ W \approx 334,000 \times 0.0623 \approx 20,800 \, \text{J} \]

### Step 7: Convert W to kWh
To convert Joules to kilowatt-hours:
\[ 1 \, \text{kWh} = 3.6 \times 10^6 \, \text{J} \]
Thus,
\[ W \, \text{(in kWh)} = \frac{20,800}{3.6 \times 10^6} \approx 0.00578 \, \text{kWh} \]

### Step 8: Round to appropriate significant figures
Rounding to three significant figures gives:
\[ W \approx 0.058 \, \text{kWh} \]

### Final Answer
The amount of energy that must be supplied to freeze 1 kg of water at 0°C is approximately **0.058 kWh**.