Example 2: Calculate the enthalpy change of fusion when 5 moles of ice is heated from 250 K (i.e., -23 °C) to 273 K (i.e., 0 °C). The specific heat of ice is 38.1 J/K·mol.

To find the enthalpy change of fusion when 5 moles of ice is heated from 250 K to 273.15 K (0°C), we need to apply the concept of specific heat capacity. Here's a step-by-step solution:

1. Identify the Known Values:
- Number of moles of ice: 5 moles
- Initial temperature of ice: 250 K
- Final temperature of ice: 273.15 K (0°C)
- Specific heat capacity of ice: 38.1 J/K/mol

2. Calculate the Temperature Change:
- The change in temperature ([tex]$\Delta T$[/tex]) is equal to the final temperature minus the initial temperature.
[tex]\[
\Delta T = 273.15 \, \text{K} - 250 \, \text{K} = 23.15 \, \text{K}
\][/tex]

3. Calculate the Enthalpy Change:
- The enthalpy change ([tex]$\Delta H$[/tex]) for heating the ice can be calculated using the formula:
[tex]\[
\Delta H = \text{number of moles} \times \text{specific heat capacity} \times \Delta T
\][/tex]

Substituting the known values:
[tex]\[
\Delta H = 5 \, \text{moles} \times 38.1 \, \text{J/K/mol} \times 23.15 \, \text{K}
\][/tex]
[tex]\[
\Delta H \approx 4410.07 \, \text{J}
\][/tex]

This means the enthalpy change when 5 moles of ice is heated from 250 K to 0°C is approximately 4410.07 J.