0.512 mL of ice at -20 C mixes with 0.5L of liquid water at 100 C. What is the final temperature. use q(ice) = q(water)

To solve this problem, we can use the principle of energy conservation, where the heat lost by the hot water is equal to the heat gained by the ice until thermal equilibrium is reached.

### Steps:
1. **Heat gained by ice**: Ice needs to be warmed from \(-20^\circ \text{C}\) to \(0^\circ \text{C}\), then melted into liquid, and finally warmed to the final temperature (if it doesn't all melt).

2. **Heat lost by water**: Water cools from \(100^\circ \text{C}\) to the final temperature.

The formula for heat transfer is \(q = mc\Delta T\), where:
- \(m\) is the mass,
- \(c\) is the specific heat capacity,
- \(\Delta T\) is the change in temperature.

### Data we need:
- **Specific heat of ice**: \(c_{\text{ice}} = 2.09 \ \text{J/g°C}\)
- **Specific heat of water**: \(c_{\text{water}} = 4.18 \ \text{J/g°C}\)
- **Heat of fusion of ice (melting)**: \(L_f = 334 \ \text{J/g}\)
- **Density of water**: \(1 \ \text{g/mL}\) (so 0.512 mL of ice is approximately 0.512 g of ice)

### Heat gained by the ice (in steps):
1. **Heating the ice from \(-20^\circ \text{C}\) to \(0^\circ \text{C}\)**:
   \[
   q_1 = m_{\text{ice}} c_{\text{ice}} \Delta T_{\text{ice}} = (0.512 \, \text{g}) (2.09 \, \text{J/g°C}) (0 - (-20) \, \text{°C})
   \]
   \[
   q_1 = 0.512 \times 2.09 \times 20 = 21.4 \, \text{J}
   \]

2. **Melting the ice at \(0^\circ \text{C}\)**:
   \[
   q_2 = m_{\text{ice}} L_f = (0.512 \, \text{g}) (334 \, \text{J/g}) = 171.0 \, \text{J}
   \]

3. **Heating the melted water from \(0^\circ \text{C}\) to the final temperature \(T_f\)**:
   \[
   q_3 = m_{\text{water}} c_{\text{water}} \Delta T_{\text{water}} = (0.512 \, \text{g}) (4.18 \, \text{J/g°C}) (T_f - 0)
   \]
   \[
   q_3 = 2.14 \times T_f \, \text{J}
   \]

### Heat lost by the water (cooling from \(100^\circ \text{C}\) to \(T_f\)):
\[
q_{\text{water}} = m_{\text{water}} c_{\text{water}} \Delta T_{\text{water}} = (500 \, \text{g}) (4.18 \, \text{J/g°C}) (100 - T_f)
\]
\[
q_{\text{water}} = 2090 \times (100 - T_f) = 209,000 - 2090 \times T_f \, \text{J}
\]

### Energy balance:
Heat gained by the ice = Heat lost by the water:
\[
q_1 + q_2 + q_3 = q_{\text{water}}
\]
\[
21.4 + 171.0 + 2.14 \times T_f = 209,000 - 2090 \times T_f
\]
\[
192.4 + 2.14 \times T_f = 209,000 - 2090 \times T_f
\]
Simplifying:
\[
192.4 + 2.14 T_f + 2090 T_f = 209,000
\]
\[
2092.14 T_f = 208,807.6
\]
\[
T_f = \frac{208,807.6}{2092.14} \approx 99.8^\circ \text{C}
\]

### Final Temperature:
The final temperature is approximately **99.8°C**, meaning most of the water barely cools, and the ice almost fully melts.

---

Do you want further details or have any questions?

Here are 5 related questions:
1. How does the mass of the ice affect the final temperature in this scenario?
2. What if the initial temperature of the ice was lower than \(-20^\circ \text{C}\)?
3. What happens if the amount of ice was larger than the water?
4. Why do we assume no heat is lost to the surroundings?
5. How would the final temperature change if we started with more water at a lower temperature?

**Tip:** When solving energy transfer problems, always break it into parts: temperature change, phase change, and equilibrium, and account for energy conservation.

0.512 mL of ice at -20°C mixes with 0.5L of liquid water at 100°C. What is the final temperature? Use q(ice) = q(water).

Calculate the final temperature when 0.512 mL of ice at -20°C is mixed with 0.5L of water at 100°C using energy conservation. Learn step-by-step how to apply heat transfer and phase change formulas. Ideal for Grades 11-12.

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