Phase diagram illustrating the relationship between temperature and pressure for a substance, highlighting phase changes and critical points.

Problem Analysis:

The problem involves a phase diagram for water, which shows the states of water (solid, liquid, gas) under different temperatures and pressures. The task is to analyze the behavior of water when it is heated or cooled under specific conditions.

#### Given Information:
1. Initial Conditions:
- 0.5 kg of ice at $-25^\circ \text{C}$ is heated.
- Specific heat capacity of ice: $c_{\text{ice}} = 2.09 \times 10^3 \, \text{J/kg} \cdot \text{K}$.
- Latent heat of fusion of ice: $L_f = 3.34 \times 10^5 \, \text{J/kg}$.
- Latent heat of vaporization of water: $L_v = 2.26 \times 10^6 \, \text{J/kg}$.
- Triple point of water: $T_{\text{triple}} = 273.16 \, \text{K}$, $P_{\text{triple}} = 611 \, \text{Pa}$.
- Critical point of water: $T_{\text{critical}} = 647.1 \, \text{K}$, $P_{\text{critical}} = 22.06 \, \text{MPa}$.

2. Tasks:
- Task 1: Determine the final temperature and state of the system when 100 kJ of heat is added to the ice at $-25^\circ \text{C}$.
- Task 2: Describe the physical state of the system at the triple point.
- Task 3: Explain the "rule of three" in phase diagrams.
- Task 4: Describe the structure formed by supercooled molecules around a growing crystal.
- Task 5: Identify the following substance in reduced units ($\tilde{T}, \tilde{P}$): $\text{CH}_3\text{OH}$ prepared at $T = 283.15 \, \text{K}$, $P = 101.3 \, \text{kPa}$.

---

Solution:

#### Task 1: Final Temperature and State After Adding 100 kJ of Heat

1. Step 1: Calculate the energy required to raise the temperature of ice from $-25^\circ \text{C}$ to $0^\circ \text{C}$:
- Initial temperature of ice: $T_i = -25^\circ \text{C} = 248.15 \, \text{K}$.
- Final temperature before melting: $T_f = 0^\circ \text{C} = 273.15 \, \text{K}$.
- Mass of ice: $m = 0.5 \, \text{kg}$.
- Specific heat capacity of ice: $c_{\text{ice}} = 2.09 \times 10^3 \, \text{J/kg} \cdot \text{K}$.

The energy required to raise the temperature of ice is:
$$
Q_1 = m \cdot c_{\text{ice}} \cdot \Delta T
$$
$$
Q_1 = 0.5 \, \text{kg} \cdot 2.09 \times 10^3 \, \text{J/kg} \cdot \text{K} \cdot (273.15 - 248.15) \, \text{K}
$$
$$
Q_1 = 0.5 \cdot 2.09 \times 10^3 \cdot 25 = 26125 \, \text{J} = 26.125 \, \text{kJ}
$$

2. Step 2: Calculate the energy required to melt the ice at $0^\circ \text{C}$:
- Latent heat of fusion of ice: $L_f = 3.34 \times 10^5 \, \text{J/kg}$.
- Energy required to melt the ice:
$$
Q_2 = m \cdot L_f
$$
$$
Q_2 = 0.5 \, \text{kg} \cdot 3.34 \times 10^5 \, \text{J/kg} = 167000 \, \text{J} = 167 \, \text{kJ}
$$

3. Step 3: Compare the total energy available with the energy requirements:
- Total energy added: $Q_{\text{total}} = 100 \, \text{kJ}$.
- Energy required to raise the temperature of ice: $Q_1 = 26.125 \, \text{kJ}$.
- Remaining energy after raising the temperature:
$$
Q_{\text{remaining}} = Q_{\text{total}} - Q_1 = 100 - 26.125 = 73.875 \, \text{kJ}
$$
- Since $Q_{\text{remaining}} = 73.875 \, \text{kJ}$ is less than $Q_2 = 167 \, \text{kJ}$, the ice will not completely melt. Instead, only a portion of the ice will melt.

4. Step 4: Calculate the mass of ice that melts:
- Let $m_{\text{melted}}$ be the mass of ice that melts.
- Energy required to melt this mass:
$$
Q_{\text{melt}} = m_{\text{melted}} \cdot L_f
$$
- Setting $Q_{\text{melt}} = Q_{\text{remaining}}$:
$$
m_{\text{melted}} \cdot L_f = 73.875 \times 10^3 \, \text{J}
$$
$$
m_{\text{melted}} = \frac{73.875 \times 10^3}{3.34 \times 10^5} = 0.221 \, \text{kg}
$$

5. Step 5: Determine the final state:
- Mass of ice remaining: $m_{\text{remaining}} = m - m_{\text{melted}} = 0.5 - 0.221 = 0.279 \, \text{kg}$.
- The final state is a mixture of ice and water at $0^\circ \text{C}$.

Final Answer for Task 1:
$$
\boxed{0^\circ \text{C}, \text{mixture of ice and water}}
$$

---

#### Task 2: Physical State at the Triple Point

The triple point of water is the unique point where solid, liquid, and gas phases coexist in equilibrium. At the triple point:
- Temperature: $T_{\text{triple}} = 273.16 \, \text{K}$.
- Pressure: $P_{\text{triple}} = 611 \, \text{Pa}$.

At this point, all three phases (solid, liquid, and gas) are in dynamic equilibrium, meaning they can exist simultaneously without any net change in their amounts.

Final Answer for Task 2:
$$
\boxed{\text{Solid, liquid, and gas phases coexist in equilibrium}}
$$

---

#### Task 3: Explanation of the "Rule of Three"

The "rule of three" in phase diagrams states that at any given point in a phase diagram:
- A maximum of three phases can coexist in equilibrium.
- This rule arises because each phase boundary represents a two-phase equilibrium, and at most three such boundaries can meet at a single point.

Final Answer for Task 3:
$$
\boxed{\text{At most three phases can coexist in equilibrium at any point in a phase diagram.}}
$$

---

#### Task 4: Structure Formed by Supercooled Molecules Around a Growing Crystal

When a solution is supercooled (cooled below its freezing point without crystallization), adding a seed crystal initiates nucleation. Supercooled molecules near the growing crystal surface arrange themselves in a structured manner, forming a layer that matches the crystal lattice of the growing solid. This process is called layer-by-layer growth or epitaxial growth. The molecules align with the crystal lattice, leading to the formation of a well-ordered solid structure.

Final Answer for Task 4:
$$
\boxed{\text{Supercooled molecules form a structured layer around the growing crystal, matching its lattice.}}
$$

---

#### Task 5: Identification of Substance in Reduced Units

Reduced units are defined as:
- Reduced temperature: $\tilde{T} = \frac{T}{T_c}$, where $T_c$ is the critical temperature.
- Reduced pressure: $\tilde{P} = \frac{P}{P_c}$, where $P_c$ is the critical pressure.

For methanol ($\text{CH}_3\text{OH}$):
- Critical temperature: $T_c = 512.6 \, \text{K}$.
- Critical pressure: $P_c = 80.98 \, \text{atm} = 8.19 \times 10^6 \, \text{Pa}$.

Given:
- Temperature: $T = 283.15 \, \text{K}$.
- Pressure: $P = 101.3 \, \text{kPa} = 1.013 \times 10^5 \, \text{Pa}$.

1. Calculate reduced temperature:
$$
\tilde{T} = \frac{T}{T_c} = \frac{283.15}{512.6} \approx 0.552
$$

2. Calculate reduced pressure:
$$
\tilde{P} = \frac{P}{P_c} = \frac{1.013 \times 10^5}{8.19 \times 10^6} \approx 0.0124
$$

Final Answer for Task 5:
$$
\boxed{(0.552, 0.0124)}
$$

---

Final Boxed Answers:

1. $\boxed{0^\circ \text{C}, \text{mixture of ice and water}}$
2. $\boxed{\text{Solid, liquid, and gas phases coexist in equilibrium}}$
3. $\boxed{\text{At most three phases can coexist in equilibrium at any point in a phase diagram.}}$
4. $\boxed{\text{Supercooled molecules form a structured layer around the growing crystal, matching its lattice.}}$
5. $\boxed{(0.552, 0.0124)}$