States of Matter and Changes of State

6.4 Lesson: States of Matter and Changes of State

The heating graph shows the relationship between temperature and thermal energy absorbed as ice is heated to become water and then steam.
Ice warms up from -10°C to 0°C.
Ice melts (at 0°C) into liquid water.
Liquid water warms up from 0°C to 100°C.
Liquid water boils (at 100°C) into water vapor.
Water vapor warms up above 100°C.

The cooling graph illustrates temperature changes as water vapor cools to become liquid water and then ice.
Water vapor cools down.
Water vapor condenses into liquid water.
Liquid water cools down.
Liquid water freezes into ice.
Ice cools down.

Latent Heat: The total energy absorbed or released during a change of state.
Specific Latent Heat: The amount of thermal energy required for 1 kg of a substance to change from one state to another, measured in J/kg.

Specific latent heat of fusion (L_f) is related to melting and freezing.
Specific latent heat of vaporization (L_v) is related to boiling and condensing.

Qf = mLf where:
- Q_f = Latent heat of fusion (in J)
- m = mass (in kg)
- L_f = specific latent heat of fusion (in J/kg)
- Note: Q_f is used for both melting and freezing.
Qv = mLv where:
- Q_v = Latent heat of vaporization (in J)
- m = mass (in kg)
- L_v = specific latent heat of vaporization (in J/kg)
- Note: Q_v is used for both boiling and condensing.

Given:
- m = 10 \text{kg}
- Water at 100°C converting to steam at 100°C
- L_v = 2.3 \times 10^6 \text{ J/kg} (from a table)
Formula:
- Qv = mLv
Calculation:
- Q_v = (10 \text{kg}) (2.3 \times 10^6 \text{ J/kg}) = 2.3 \times 10^7 \text{ J}
The latent heat of 2.3 \times 10^7 \text{ J} is needed.

How much thermal energy is required to raise 5.0 kg of -30°C ice to a temperature of 140°C?
- [Note: c{\text{ice}} = 2100 \text{J}/(°\text{C} \cdot \text{kg}), c{\text{water}} = 4180 \text{J}/(°\text{C} \cdot \text{kg}), c{\text{water vapour}} = 2080 \text{ J}/(°\text{C} \cdot \text{kg}), Lv = 2255000 \text{J/kg}, L_f = 333000 \text{J/kg}]

Heating stages:
- Ice: -30°C → 0°C
- Ice to water (at 0°C)
- Water: 0°C → 100°C
- Water to steam (at 100°C)
- Steam: 100°C → 140°C
Calculations:
- Heat to warm ice from -30°C to 0°C: Q1 = m \cdot c{\text{ice}} \cdot \Delta T_{\text{ice}} = (5.0 \text{ kg}) (2.1 \times 10^3 \text{ J/(kg} \cdot °\text{C)}) (0°\text{C} - (-30°\text{C}))
- Heat for ice to water: Q2 = m Lf = (5.0 \text{ kg}) (3.4 \times 10^5 \text{ J/kg})
- Heat to warm water from 0°C to 100°C: Q3 = m \cdot c{\text{w}} \cdot \Delta T_w = (5.0 \text{ kg}) (4.18 \times 10^3 \text{ J/(kg} \cdot °\text{C)}) (100°\text{C} - 0°\text{C})
- Heat for water to steam: Q4 = m Lv = (5.0 \text{ kg}) (2.3 \times 10^6 \text{ J/kg})
- Heat to warm steam from 100°C to 140°C: Q5 = m \cdot c{\text{steam}} \cdot \Delta T_{\text{steam}} = (5.0 \text{ kg}) (2.08 \times 10^3 \text{ J/(kg} \cdot °\text{C)}) (140°\text{C} - 100°\text{C})
Total heat required:
- Q = Q1 + Q2 + Q3 + Q4 + Q_5
- Q = 315000 \text{ J} + 1700000 \text{ J} + 2090000 \text{ J} + 11500000 \text{ J} + 416000 \text{ J} = 16021000 \text{ J}
The total thermal energy is 1.6 \times 10^7 \text{ J}.

Ch. 5 Review: Page 261 Q1-5,8-11,13,15; Page 262 Q1-4,8-12,15,16,23,24a,29-32,42,44,47,49,50,53,57
Ch. 6 Review: Page 309 Q3-6,13-16; Page 310 Q1-5,8,18,19,28,29,33-35,40,41,44,45,66,67