To find the final equilibrium temperature when combining ice at $-20^\circ C$ and liquid water at $100^\circ C$, we need to consider energy transfer. Specifically, we'll apply the principle of conservation of energy, where the heat gained by the ice must equal the heat lost by the water.

Given:

• Mass of ice: 31 moles
• Initial temperature of ice: $-20^\circ C$
• Mass of water: 26.6 moles
• Initial temperature of water: $100^\circ C$

Constants:

• Heat capacity of ice ($c_{ice}$): $2.09 \ \text{J/g} \cdot \text{°C}$
• Heat capacity of water ($c_{water}$): $4.18 \ \text{J/g} \cdot \text{°C}$
• Heat of fusion of ice ($\Delta H_f$): $334 \ \text{J/g}$
• Molar mass of water: $18 \ \text{g/mol}$

To find the final temperature, we'll go through the following steps:

1. Calculate the energy required to warm the ice from $-20^\circ C$ to $0^\circ C$.
2. Calculate the energy required to melt the ice at $0^\circ C$.
3. Consider the cooling of the water from $100^\circ C$ to a lower temperature.
4. Determine if the melted ice and water reach thermal equilibrium.

Let's proceed with these calculations.

1. Heat required to warm the ice to $0^\circ C$:

The formula for heating ice: $q = m \cdot c_{ice} \cdot \Delta T$ where:

• $m$ is the mass of ice in grams.
• $c_{ice}$ is the specific heat capacity of ice.
• $\Delta T$ is the temperature change.

Converting moles of ice to grams: $\text{Mass of ice} = 31 \text{ moles} \times 18 \text{ g/mol} = 558 \text{ g}$ $q_1 = 558 \, \text{g} \times 2.09 \, \text{J/g°C} \times (0 - (-20))^\circ C$ $q_1 = 558 \times 2.09 \times 20 = 23335.2 \text{ J}$

2. Heat required to melt the ice:

The formula for melting ice: $q_2 = m \cdot \Delta H_f$ $q_2 = 558 \text{ g} \times 334 \text{ J/g} = 186372 \text{ J}$

3. Heat released by cooling the water from $100^\circ C$ to $T_f$:

The formula for cooling water: $q_3 = m \cdot c_{water} \cdot \Delta T$ Converting moles of water to grams: $\text{Mass of water} = 26.6 \text{ moles} \times 18 \text{ g/mol} = 478.8 \text{ g}$ The heat released by cooling water to $T_f$: $q_3 = 478.8 \, \text{g} \times 4.18 \, \text{J/g°C} \times (100 - T_f)$ $q_3 = 2002.184 \times (100 - T_f) \, \text{J}$

Energy Balance Equation

For energy conservation: $q_1 + q_2 = q_3$ Substituting the known values: $23335.2 + 186372 = 2002.184 \times (100 - T_f)$ $209707.2 = 2002.184 \times (100 - T_f)$ $100 - T_f = \frac{209707.2}{2002.184}$ $100 - T_f = 104.75$ $T_f = 100 - 104.75 = -4.75^\circ C$

Final Answer:

The final temperature of the system will be $0^\circ C$, as the temperature cannot go below freezing due to phase change considerations, implying that not all the ice melts, and the system stabilizes with ice and water at $0^\circ C$.

Would you like a detailed explanation of these steps or have any questions?

Related Questions:

1. What happens to the energy exchange if more ice is added to the system?
2. How would the final temperature change if the initial temperature of the ice were different?
3. What is the specific role of the heat of fusion in this scenario?
4. Can you explain how specific heat values differ between ice and water?
5. How does the conservation of energy apply to phase changes?

Tip: Always consider the phase change temperature as a key factor when dealing with energy transfer problems involving ice and water.