512 ml of ice at -20 C mixes with 0.5 L of liquid water at 100 C. Find the final temperature. Use q(ice) = -q(water)

To find the final temperature when 512 mL of ice at \(-20^\circ C\) mixes with 0.5 L of liquid water at \(100^\circ C\), we use the principle of conservation of energy. The heat gained by the ice must equal the heat lost by the water:

\[
q_{\text{ice}} = -q_{\text{water}}
\]

Here's the detailed process:

### Step 1: Calculate the heat required to raise the temperature of the ice from \(-20^\circ C\) to \(0^\circ C\)
The specific heat capacity of ice is \(c_{\text{ice}} = 2.1 \, \text{J/g}^\circ C\). The mass of ice can be found as follows (assuming the density of ice is approximately 1 g/mL):

\[
m_{\text{ice}} = 512 \, \text{g}
\]

Now calculate the heat required to raise the temperature of ice to 0°C using:

\[
q_1 = m_{\text{ice}} \cdot c_{\text{ice}} \cdot \Delta T
\]

Where:
\[
\Delta T = 0^\circ C - (-20^\circ C) = 20^\circ C
\]

So:

\[
q_1 = 512 \, \text{g} \cdot 2.1 \, \text{J/g}^\circ C \cdot 20^\circ C = 21,504 \, \text{J}
\]

### Step 2: Calculate the heat required to melt the ice at \(0^\circ C\)
The latent heat of fusion of ice is \(L_f = 334 \, \text{J/g}\). The heat required to melt the ice is:

\[
q_2 = m_{\text{ice}} \cdot L_f
\]

So:

\[
q_2 = 512 \, \text{g} \cdot 334 \, \text{J/g} = 171,008 \, \text{J}
\]

### Step 3: Calculate the heat lost by water as it cools from \(100^\circ C\) to the final temperature \(T_f\)
The specific heat capacity of liquid water is \(c_{\text{water}} = 4.18 \, \text{J/g}^\circ C\). The mass of the water is:

\[
m_{\text{water}} = 500 \, \text{g}
\]

The heat lost by water as it cools to \(T_f\) is:

\[
q_{\text{water}} = m_{\text{water}} \cdot c_{\text{water}} \cdot (100^\circ C - T_f)
\]

\[
q_{\text{water}} = 500 \, \text{g} \cdot 4.18 \, \text{J/g}^\circ C \cdot (100^\circ C - T_f) = 2090 \, \text{J/}^\circ C \cdot (100^\circ C - T_f)
\]

### Step 4: Solve for the final temperature
The total heat gained by the ice is \(q_{\text{ice}} = q_1 + q_2\), and it is equal to the heat lost by the water:

\[
21,504 \, \text{J} + 171,008 \, \text{J} = 2090 \, \text{J/}^\circ C \cdot (100^\circ C - T_f)
\]

\[
192,512 \, \text{J} = 2090 \, \text{J/}^\circ C \cdot (100^\circ C - T_f)
\]

Now, solve for \(T_f\):

\[
100^\circ C - T_f = \frac{192,512 \, \text{J}}{2090 \, \text{J/}^\circ C}
\]

\[
100^\circ C - T_f = 92.1^\circ C
\]

\[
T_f = 100^\circ C - 92.1^\circ C = 7.9^\circ C
\]

### Final Answer:
The final temperature of the mixture is approximately \(7.9^\circ C\).

Do you want further details or have any questions?

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Here are 5 related questions to expand your understanding:
1. How would the final temperature change if the initial temperature of the ice was lower, such as \(-30^\circ C\)?
2. What would happen if there was more ice added to the water?
3. How does the specific heat capacity of ice differ from that of water, and why?
4. If the water were initially at \(50^\circ C\) instead of \(100^\circ C\), how would the final temperature change?
5. What is the role of latent heat in phase transitions like melting?

**Tip:** Always check units, especially when dealing with volumes and masses. Water's density is 1 g/mL for liquid but slightly less for ice!

Calculate the final temperature when 512 mL of ice at -20°C mixes with 0.5 L of water at 100°C using the principle of conservation of energy. Learn how to apply heat transfer equations and phase change calculations in this detailed step-by-step solution. Suitable for high school students studying thermodynamics.

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