To find the final temperature of the system, we'll use the concept of heat transfer. In this case, the heat gained by the ice will equal the heat lost by the water, expressed as $q_{\text{ice}} = -q_{\text{water}}$.

Here’s a step-by-step outline of the process:

Given Information:

• Moles of ice, $n_{\text{ice}} = 31$ moles at $-20^\circ \text{C}$
• Moles of water, $n_{\text{water}} = 26.6$ moles at $100^\circ \text{C}$
• Molar heat capacity of ice, $C_{\text{ice}} = 36.57 \, \text{J/mol}^\circ \text{C}$
• Molar heat capacity of liquid water, $C_{\text{water}} = 75.4 \, \text{J/mol}^\circ \text{C}$
• Latent heat of fusion (ice to water), $\Delta H_{\text{fusion}} = 6.01 \, \text{kJ/mol} = 6010 \, \text{J/mol}$
• Heat released by cooling water from $100^\circ \text{C}$ to the final temperature $T$
• Heat absorbed by heating ice from $-20^\circ \text{C}$, melting it, and then possibly heating it further to the final temperature $T$.

Step-by-Step Heat Transfer Calculations:

1. Heating the ice from $-20^\circ \text{C}$ to $0^\circ \text{C}$:

The heat required to raise the temperature of the ice from $-20^\circ \text{C}$ to $0^\circ \text{C}$ is: $q_1 = n_{\text{ice}} \cdot C_{\text{ice}} \cdot \Delta T$ $q_1 = 31 \cdot 36.57 \, \text{J/mol}^\circ \text{C} \cdot (0 - (-20))^\circ \text{C}$ $q_1 = 31 \cdot 36.57 \cdot 20 = 22614 \, \text{J}$

2. Melting the ice at $0^\circ \text{C}$:

The heat required to melt the ice at $0^\circ \text{C}$ is: $q_2 = n_{\text{ice}} \cdot \Delta H_{\text{fusion}}$ $q_2 = 31 \cdot 6010 = 186310 \, \text{J}$

3. Heating the melted ice (now liquid water) from $0^\circ \text{C}$ to the final temperature $T$:

The heat required to raise the temperature of the melted ice from $0^\circ \text{C}$ to $T^\circ \text{C}$ is: $q_3 = n_{\text{ice}} \cdot C_{\text{water}} \cdot (T - 0)$ $q_3 = 31 \cdot 75.4 \cdot (T - 0) = 2337.4 \cdot T$

4. Cooling the hot water from $100^\circ \text{C}$ to the final temperature $T$:

The heat lost by the hot water as it cools from $100^\circ \text{C}$ to $T^\circ \text{C}$ is: $q_{\text{water}} = n_{\text{water}} \cdot C_{\text{water}} \cdot (100 - T)$ $q_{\text{water}} = 26.6 \cdot 75.4 \cdot (100 - T) = 2006.64 \cdot (100 - T)$ $q_{\text{water}} = 200664 - 2006.64 \cdot T$

Setting up the heat balance equation:

Since $q_{\text{ice}} = -q_{\text{water}}$, we can write: $q_1 + q_2 + q_3 = q_{\text{water}}$ Substitute the values: $22614 + 186310 + 2337.4 \cdot T = 200664 - 2006.64 \cdot T$

Simplifying: $208924 + 2337.4 \cdot T = 200664 - 2006.64 \cdot T$ $2337.4 \cdot T + 2006.64 \cdot T = 200664 - 208924$ $4344.04 \cdot T = -8260$ $T = \frac{-8260}{4344.04} \approx -1.9^\circ \text{C}$

Final Answer:

The final temperature of the system will be approximately $T = -1.9^\circ \text{C}$.

Would you like more details on any step?

Related Questions:

1. How would the final temperature change if the amount of water was increased?
2. What would happen if the ice started at $0^\circ \text{C}$ instead of $-20^\circ \text{C}$?
3. How does the specific heat capacity of ice and water affect the final temperature?
4. What if the ice melts but doesn’t reach the same final temperature as the water?
5. How would the final temperature change if both ice and water were at $0^\circ \text{C}$?

Tip:

Always remember to check if the phase of the substance changes during heating or cooling—it affects the heat calculations significantly.