States of Matter and Changes of State

6.4 Lesson: States of Matter and Changes of State

  • States of Matter

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 (LfL_f) is related to melting and freezing.
  • Specific latent heat of vaporization (LvL_v) is related to boiling and condensing.

Heat Capacity Formulas

  • Q<em>f=mL</em>fQ<em>f = mL</em>f where:
    • QfQ_f = Latent heat of fusion (in J)
    • mm = mass (in kg)
    • LfL_f = specific latent heat of fusion (in J/kg)
    • Note: QfQ_f is used for both melting and freezing.
  • Q<em>v=mL</em>vQ<em>v = mL</em>v where:
    • QvQ_v = Latent heat of vaporization (in J)
    • mm = mass (in kg)
    • LvL_v = specific latent heat of vaporization (in J/kg)
    • Note: QvQ_v 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=10kgm = 10 \text{kg}
    • Water at 100°C converting to steam at 100°C
    • Lv=2.3×106 J/kgL_v = 2.3 \times 10^6 \text{ J/kg} (from a table)
  • Formula:
    • Q<em>v=mL</em>vQ<em>v = mL</em>v
  • Calculation:
    • Qv=(10kg)(2.3×106 J/kg)=2.3×107 JQ_v = (10 \text{kg}) (2.3 \times 10^6 \text{ J/kg}) = 2.3 \times 10^7 \text{ J}
  • The latent heat of 2.3×107 J2.3 \times 10^7 \text{ 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/(°Ckg)c<em>{\text{ice}} = 2100 \text{J}/(°\text{C} \cdot \text{kg}), c</em>water=4180J/(°Ckg)c</em>{\text{water}} = 4180 \text{J}/(°\text{C} \cdot \text{kg}), c<em>water vapour=2080 J/(°Ckg)c<em>{\text{water vapour}} = 2080 \text{ J}/(°\text{C} \cdot \text{kg}), L</em>v=2255000J/kgL</em>v = 2255000 \text{J/kg}, Lf=333000J/kgL_f = 333000 \text{J/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=mc</em>iceΔTice=(5.0 kg)(2.1×103 J/(kg°C))(0°C(30°C))Q<em>1 = m \cdot c</em>{\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: Q<em>2=mL</em>f=(5.0 kg)(3.4×105 J/kg)Q<em>2 = m L</em>f = (5.0 \text{ kg}) (3.4 \times 10^5 \text{ J/kg})
    • Heat to warm water from 0°C to 100°C: Q<em>3=mc</em>wΔTw=(5.0 kg)(4.18×103 J/(kg°C))(100°C0°C)Q<em>3 = m \cdot c</em>{\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: Q<em>4=mL</em>v=(5.0 kg)(2.3×106 J/kg)Q<em>4 = m L</em>v = (5.0 \text{ kg}) (2.3 \times 10^6 \text{ J/kg})
    • Heat to warm steam from 100°C to 140°C: Q<em>5=mc</em>steamΔTsteam=(5.0 kg)(2.08×103 J/(kg°C))(140°C100°C)Q<em>5 = m \cdot c</em>{\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=Q<em>1+Q</em>2+Q<em>3+Q</em>4+Q5Q = Q<em>1 + Q</em>2 + Q<em>3 + Q</em>4 + Q_5
    • Q=315000 J+1700000 J+2090000 J+11500000 J+416000 J=16021000 JQ = 315000 \text{ J} + 1700000 \text{ J} + 2090000 \text{ J} + 11500000 \text{ J} + 416000 \text{ J} = 16021000 \text{ J}
  • The total thermal energy is 1.6×107 J1.6 \times 10^7 \text{ 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