Heat with Phase Change Worksheet

Here are the step-by-step solutions for each problem on the worksheet.

Note on Constants:
Based on the table provided in the image, we will use the following values:
* Specific Heat ($c$):
* Solid (Ice): $2.0 \text{ J/g}^\circ\text{C}$
* Liquid (Water): $4.184 \text{ J/g}^\circ\text{C}$
* Gas (Steam): $2.02 \text{ J/g}^\circ\text{C}$
* Latent Heat (Phase Change):
* Fusion (Melting/Freezing): The table lists $6.02 \text{ kJ/mol}$. To make calculations easier with grams, we convert this to Joules per gram. Since water's molar mass is approx $18.02 \text{ g/mol}$, the value is $\approx 334 \text{ J/g}$.
* Boiling (Vaporization): The table lists $40.65 \text{ kJ/mol}$. Converted to grams, this is $\approx 2260 \text{ J/g}$.

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1) How many joules are required to heat 250 grams of liquid water from $0^\circ\text{C}$ to $100^\circ\text{C}$?

This problem involves heating a substance without changing its phase (it stays liquid). We use the specific heat formula: $q = mc\Delta T$.

* Mass ($m$): $250 \text{ g}$
* Specific Heat ($c$): $4.184 \text{ J/g}^\circ\text{C}$ (for liquid water)
* Change in Temp ($\Delta T$): $100^\circ\text{C} - 0^\circ\text{C} = 100^\circ\text{C}$

$$q = 250 \times 4.184 \times 100$$
$$q = 104,600 \text{ Joules}$$

2) How many joules are required to melt 100 grams of water?

Melting is a phase change from solid to liquid at $0^\circ\text{C}$. Temperature does not change during this step. We use the heat of fusion.

* Mass ($m$): $100 \text{ g}$
* Heat of Fusion ($H_f$): $\approx 334 \text{ J/g}$ (derived from $6.02 \text{ kJ/mol}$)

$$q = m \times H_f$$
$$q = 100 \times 334$$
$$q = 33,400 \text{ Joules}$$

3) How many joules are required to boil 150 grams of water?

Boiling is a phase change from liquid to gas at $100^\circ\text{C}$. Temperature does not change. We use the heat of vaporization.

* Mass ($m$): $150 \text{ g}$
* Heat of Vaporization ($H_v$): $\approx 2260 \text{ J/g}$ (derived from $40.65 \text{ kJ/mol}$)

$$q = m \times H_v$$
$$q = 150 \times 2260$$
$$q = 339,000 \text{ Joules}$$

4) How many joules are required to heat 200 grams of water from $25^\circ\text{C}$ to $125^\circ\text{C}$?

This requires three steps because the water crosses the boiling point ($100^\circ\text{C}$).
1. Heat liquid from $25^\circ\text{C}$ to $100^\circ\text{C}$.
2. Boil the liquid at $100^\circ\text{C}$.
3. Heat the steam (gas) from $100^\circ\text{C}$ to $125^\circ\text{C}$.

Step 1: Heating Liquid
$$q_1 = 200 \text{ g} \times 4.184 \text{ J/g}^\circ\text{C} \times (100 - 25)^\circ\text{C}$$
$$q_1 = 200 \times 4.184 \times 75 = 62,760 \text{ J}$$

Step 2: Boiling (Phase Change)
$$q_2 = 200 \text{ g} \times 2260 \text{ J/g}$$
$$q_2 = 452,000 \text{ J}$$

Step 3: Heating Steam
$$q_3 = 200 \text{ g} \times 2.02 \text{ J/g}^\circ\text{C} \times (125 - 100)^\circ\text{C}$$
$$q_3 = 200 \times 2.02 \times 25 = 10,100 \text{ J}$$

Total Energy:
$$q_{total} = 62,760 + 452,000 + 10,100 = 524,860 \text{ Joules}$$

5) How many joules are given off when 120 grams of water are cooled from $25^\circ\text{C}$ to $-25^\circ\text{C}$?

"Given off" means energy is released (cooling). We cross the freezing point ($0^\circ\text{C}$), so there are three steps.
1. Cool liquid from $25^\circ\text{C}$ to $0^\circ\text{C}$.
2. Freeze the liquid at $0^\circ\text{C}$.
3. Cool the ice (solid) from $0^\circ\text{C}$ to $-25^\circ\text{C}$.

Step 1: Cooling Liquid
$$q_1 = 120 \times 4.184 \times 25 = 12,552 \text{ J}$$

Step 2: Freezing (Phase Change)
$$q_2 = 120 \times 334 = 40,080 \text{ J}$$

Step 3: Cooling Ice
$$q_3 = 120 \times 2.0 \times 25 = 6,000 \text{ J}$$

Total Energy Released:
$$q_{total} = 12,552 + 40,080 + 6,000 = 58,632 \text{ Joules}$$

6) How many joules are required to heat 75 grams of water from $-85^\circ\text{C}$ to $185^\circ\text{C}$?

This is a long journey crossing both freezing and boiling points. There are 5 steps.
1. Heat Ice: $-85^\circ\text{C}$ to $0^\circ\text{C}$ ($\Delta T = 85$)
2. Melt Ice at $0^\circ\text{C}$
3. Heat Liquid: $0^\circ\text{C}$ to $100^\circ\text{C}$ ($\Delta T = 100$)
4. Boil Liquid at $100^\circ\text{C}$