Calculate the amount of energy required to completely vaporize 120.0 g of liquid water at 

100.00 ºC. 

3. Calculate the amount of energy required to completely melt 120.0 g of ice at 0.00 ºC. 

4. Based on your calculations in question 3 and 4, which process requires more energy, 

vaporizing water or melting water?  Explain why. 

### 1. Energy required to completely vaporize 120.0 g of liquid water at 100.00 ºC:

The energy required to vaporize water depends on the heat of vaporization. The heat of vaporization of water is:
\[
\Delta H_{vap} = 2260 \, \text{J/g}
\]
The formula to calculate the energy required to vaporize the water is:
\[
q = m \times \Delta H_{vap}
\]
where:
- \( q \) = energy required (in joules),
- \( m \) = mass of the water (in grams),
- \( \Delta H_{vap} \) = heat of vaporization (in J/g).

Now, using the given mass of water:
\[
q = 120.0 \, \text{g} \times 2260 \, \text{J/g} = 271200 \, \text{J}
\]
Thus, the energy required to vaporize 120.0 g of water is **271,200 J**.

---

### 2. Energy required to completely melt 120.0 g of ice at 0.00 ºC:

The energy required to melt ice depends on the heat of fusion. The heat of fusion of water (ice) is:
\[
\Delta H_{fus} = 334 \, \text{J/g}
\]
The formula to calculate the energy required to melt the ice is:
\[
q = m \times \Delta H_{fus}
\]
where:
- \( q \) = energy required (in joules),
- \( m \) = mass of the ice (in grams),
- \( \Delta H_{fus} \) = heat of fusion (in J/g).

Now, using the given mass of ice:
\[
q = 120.0 \, \text{g} \times 334 \, \text{J/g} = 40080 \, \text{J}
\]
Thus, the energy required to melt 120.0 g of ice is **40,080 J**.

---

### 3. Which process requires more energy: vaporizing water or melting ice?

From the calculations above:
- Energy required to vaporize 120.0 g of water = **271,200 J**,
- Energy required to melt 120.0 g of ice = **40,080 J**.

Clearly, vaporizing the water requires significantly more energy than melting the ice. 

**Explanation:**
This is because the heat of vaporization is much higher than the heat of fusion. When vaporizing water, you're breaking the intermolecular forces (hydrogen bonds) that hold the liquid together, which requires much more energy than just weakening these forces during melting, where the molecules still remain in close contact in the liquid phase.

---

Would you like further details or have any questions? Here are 5 related questions:

1. How does the specific heat capacity of water affect the energy required to change its temperature?
2. Why is the heat of vaporization much higher than the heat of fusion for most substances?
3. How much energy would be required to vaporize water at a temperature other than 100.00 ºC?
4. How does atmospheric pressure affect the boiling point and heat of vaporization of water?
5. What is the significance of latent heat in phase transitions like vaporization and melting?

**Tip:** The energy required for phase changes (melting, vaporization) does not change the temperature, as it goes into breaking molecular bonds, not raising the temperature.

Calculate the amount of energy required to completely vaporize 120.0 g of liquid water at 100.00 ºC. Calculate the amount of energy required to completely melt 120.0 g of ice at 0.00 ºC. Based on your calculations, which process requires more energy, vaporizing water or melting water? Explain why.

Learn how to calculate the energy required to vaporize 120.0 g of water at 100.00 ºC and melt 120.0 g of ice at 0.00 ºC. This solution covers the concepts of heat of vaporization and heat of fusion, making it suitable for high school students in Grades 9-12.

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