Calculate the enthalpy change on freezing of 1.0 mol of water at 10.0°C to ice at -10.0°C.

Δ_fusH = 6.03 kJ/mol at 0°C
C_P [H₂O(l)] = 75.3 J/mol·K
C_P [H₂O(s)] = 36.8 J/mol·K

To calculate the enthalpy change on freezing 1.0 mol of water at 10.0 °C to ice at -10.0 °C, we need to consider two main processes:

1. The cooling of liquid water from 10.0 °C to 0.0 °C.
2. The freezing of water at 0.0 °C to ice.
3. The cooling of ice from 0.0 °C to -10.0 °C.

**Given Data:**
- \(\Delta_{fus}H = 6.03 \, \text{kJ/mol}\) (enthalpy of fusion at 0 °C)
- \(C_P[H_2O(l)] = 75.3 \, \text{J/(mol·K)}\) (heat capacity of liquid water)
- \(C_P[H_2O(s)] = 36.8 \, \text{J/(mol·K)}\) (heat capacity of ice)

### Step 1: Cooling Liquid Water from 10.0 °C to 0.0 °C

To calculate the heat lost by water as it cools from 10.0 °C to 0.0 °C, we use the formula:

\[
q_1 = n \cdot C_P[H_2O(l)] \cdot \Delta T
\]

Where:
- \(n = 1.0 \, \text{mol}\)
- \(C_P[H_2O(l)] = 75.3 \, \text{J/(mol·K)}\)
- \(\Delta T = T_{final} - T_{initial} = 0.0 - 10.0 = -10.0 \, \text{K}\)

Calculating \(q_1\):

\[
q_1 = 1.0 \, \text{mol} \cdot 75.3 \, \text{J/(mol·K)} \cdot (-10.0 \, \text{K})
\]
\[
q_1 = -753 \, \text{J}
\]

### Step 2: Freezing of Water at 0.0 °C

The enthalpy change for the phase change (freezing) is given by:

\[
q_2 = -\Delta_{fus}H
\]

Where:

\[
\Delta_{fus}H = 6.03 \, \text{kJ/mol} = 6030 \, \text{J/mol}
\]

Calculating \(q_2\):

\[
q_2 = -6030 \, \text{J}
\]

### Step 3: Cooling Ice from 0.0 °C to -10.0 °C

To calculate the heat lost by ice as it cools from 0.0 °C to -10.0 °C:

\[
q_3 = n \cdot C_P[H_2O(s)] \cdot \Delta T
\]

Where:
- \(C_P[H_2O(s)] = 36.8 \, \text{J/(mol·K)}\)
- \(\Delta T = -10.0 \, \text{K}\)

Calculating \(q_3\):

\[
q_3 = 1.0 \, \text{mol} \cdot 36.8 \, \text{J/(mol·K)} \cdot (-10.0 \, \text{K})
\]
\[
q_3 = -368 \, \text{J}
\]

### Total Enthalpy Change

The total enthalpy change (\(\Delta H\)) is the sum of all the heat changes:

\[
\Delta H = q_1 + q_2 + q_3
\]

Calculating:

\[
\Delta H = (-753 \, \text{J}) + (-6030 \, \text{J}) + (-368 \, \text{J})
\]
\[
\Delta H = -753 - 6030 - 368
\]
\[
\Delta H = -7151 \, \text{J}
\]

Converting to kJ:

\[
\Delta H = -7.151 \, \text{kJ}
\]

### Final Answer

The enthalpy change on freezing 1.0 mol of water at 10.0 °C to ice at -10.0 °C is approximately -7.15 kJ.