States of Matter and Changes of State
6.4 Lesson: States of Matter and Changes of State
6. 4 Changes of States
- Heat transfer occurs during a change in state.
- Solid to liquid: Absorbs thermal energy.
- Liquid to gas: Absorbs thermal energy.
- Gas to liquid: Releases thermal energy.
- Liquid to solid: Releases thermal energy.
- Fusion: The process of liquid changing into a solid.
- Vaporization: The process of a liquid changing into a vapor.
Heating Graph for Water
- 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.
Cooling Graph for Water
- 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
- 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
- Specific latent heat of fusion (Lf) is related to melting and freezing.
- Specific latent heat of vaporization (Lv) is related to boiling and condensing.
- Q<em>f=mL</em>f where:
- Qf = Latent heat of fusion (in J)
- m = mass (in kg)
- Lf = specific latent heat of fusion (in J/kg)
- Note: Qf is used for both melting and freezing.
- Q<em>v=mL</em>v where:
- Qv = Latent heat of vaporization (in J)
- m = mass (in kg)
- Lv = specific latent heat of vaporization (in J/kg)
- Note: Qv is used for both boiling and condensing.
Example #1
- How much heat is required to bring 10 kg of 100°C liquid water to 100°C steam?
Example #1 Solution
- Given:
- m=10kg
- Water at 100°C converting to steam at 100°C
- Lv=2.3×106 J/kg (from a table)
- Formula:
- Q<em>v=mL</em>v
- Calculation:
- Qv=(10kg)(2.3×106 J/kg)=2.3×107 J
- The latent heat of 2.3×107 J is needed.
Example #2
- How much thermal energy is required to raise 5.0 kg of -30°C ice to a temperature of 140°C?
- [Note: c<em>ice=2100J/(°C⋅kg), c</em>water=4180J/(°C⋅kg), c<em>water vapour=2080 J/(°C⋅kg), L</em>v=2255000J/kg, Lf=333000J/kg]
Example #2 solution
- 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: Q<em>1=m⋅c</em>ice⋅ΔTice=(5.0 kg)(2.1×103 J/(kg⋅°C))(0°C−(−30°C))
- Heat for ice to water: Q<em>2=mL</em>f=(5.0 kg)(3.4×105 J/kg)
- Heat to warm water from 0°C to 100°C: Q<em>3=m⋅c</em>w⋅ΔTw=(5.0 kg)(4.18×103 J/(kg⋅°C))(100°C−0°C)
- Heat for water to steam: Q<em>4=mL</em>v=(5.0 kg)(2.3×106 J/kg)
- Heat to warm steam from 100°C to 140°C: Q<em>5=m⋅c</em>steam⋅ΔTsteam=(5.0 kg)(2.08×103 J/(kg⋅°C))(140°C−100°C)
- Total heat required:
- Q=Q<em>1+Q</em>2+Q<em>3+Q</em>4+Q5
- Q=315000 J+1700000 J+2090000 J+11500000 J+416000 J=16021000 J
- The total thermal energy is 1.6×107 J.
Practice Questions
- Page 293 Q1-3
- Page 295 Q6-9
- Page 283 Q1-3
- Page 286 Q1,2
- Page 287 Q1-9
Unit Review Questions
- 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