The change of energy on freezing `1.00 kg` of liquid water at `0^(@)C` and1 atmis

Options

A. `236 .7 kJ kg^(-1)`
B. ` 333.4kJ kg^(-1)`
C. ` - 333.4kJ kg^(-1)`
D. `- 236 .7 kJ kg^(-1)`

Explanation

To find the change of energy when freezing 1.00 kg of liquid water at 0°C and 1 atm, we need to use the enthalpy of fusion of ice. The enthalpy of fusion is the amount of energy required to convert a unit mass of a solid into a liquid without a change in temperature.

### Step-by-step Solution:

1. **Identify the enthalpy of fusion of ice**:
The enthalpy of fusion of ice (ΔH_fus) is approximately 6.03 kJ/mol at 0°C and 1 atm.

2. **Convert the mass of water to moles**:
The molar mass of water (H₂O) is approximately 18 g/mol.
To find the number of moles in 1 kg (1000 g) of water:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{1000 \text{ g}}{18 \text{ g/mol}} \approx 55.56 \text{ moles} \]

3. **Calculate the total enthalpy change for freezing**:
The energy change (ΔE) when freezing 1 kg of water can be calculated using the formula:
\[ \Delta E = -n \times \Delta H_{fus} \]
where \( n \) is the number of moles and \( \Delta H_{fus} \) is the enthalpy of fusion.
\[ \Delta E = -55.56 \text{ moles} \times 6.03 \text{ kJ/mol} \approx -335.0 \text{ kJ} \]

4. **Interpret the result**:
The negative sign indicates that energy is released when water freezes. Therefore, the change in energy on freezing 1.00 kg of liquid water at 0°C and 1 atm is approximately -335.0 kJ.

### Final Answer:
The change of energy on freezing 1.00 kg of liquid water at 0°C and 1 atm is approximately -335.0 kJ.

---