To calculate the increase in entropy during this process, we can break it down into several steps:
1. Heating the ice from 0°C to its melting point at 0°C.
2. Melting the ice into liquid water at 0°C.
3. Heating the water from 0°C to its boiling point at 100°C.
4. Vaporizing the water into steam at 100°C.
We can use the following equations:
1. **Entropy change due to heating:**
\[
\Delta S = \frac{q}{T}
\]
Where:
- \( \Delta S \) is the change in entropy,
- \( q \) is the heat added, and
- \( T \) is the temperature in Kelvin.
2. **Entropy change due to phase transition (melting or vaporization):**
\[
\Delta S = \frac{\Delta H}{T}
\]
Where:
- \( \Delta H \) is the enthalpy change, and
- \( T \) is the temperature in Kelvin.
We can look up the values for specific heat capacities and enthalpy changes for each step.
1. **Heating ice from 0°C to its melting point at 0°C:**
- Specific heat capacity of ice, \( c_{\text{ice}} = 2.09 \, \text{J/g} \cdot \text{K} \)
- Temperature change, \( \Delta T = 0°C - 0°C = 0°C \)
- Mass of ice, \( m = 18 \, \text{g} \)
- \( q = m \cdot c_{\text{ice}} \cdot \Delta T \)
2. **Melting the ice into liquid water at 0°C:**
- Enthalpy of fusion of water, \( \Delta H_{\text{fusion}} = 334 \, \text{J/g} \)
- \( q = m \cdot \Delta H_{\text{fusion}} \)
3. **Heating water from 0°C to its boiling point at 100°C:**
- Specific heat capacity of water, \( c_{\text{water}} = 4.18 \, \text{J/g} \cdot \text{K} \)
- Temperature change, \( \Delta T = 100°C - 0°C = 100°C \)
- \( q = m \cdot c_{\text{water}} \cdot \Delta T \)
4. **Vaporizing the water into steam at 100°C:**
- Enthalpy of vaporization of water, \( \Delta H_{\text{vap}} = 2260 \, \text{J/g} \)
- \( q = m \cdot \Delta H_{\text{vap}} \)
Once we have calculated \( q \) for each step, we can sum them up to get the total heat added, and then calculate the total change in entropy using the equations provided above.
Let's go through these steps.
