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Phas e Cha nges

Phas e Cha nges. Liquid. Solid. Gas. Solid to Liquid. Melting. Liquid. Solid. Gas. Liquid to Solid. Freezing. Liquid. Solid. Gas. Melting & Freezing. These changes happen at the : Melting/Freezing Point. Liquid. Solid. Gas. Liquid to Gas. Vaporization. Vaporization Types.

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Phas e Cha nges

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  1. Phase Changes

  2. Liquid Solid Gas Solid to Liquid Melting

  3. Liquid Solid Gas Liquid to Solid Freezing

  4. Liquid Solid Gas Melting & Freezing These changes happen at the: Melting/Freezing Point

  5. Liquid Solid Gas Liquid to Gas Vaporization

  6. Vaporization Types • Two TYPES: • Evaporation: - Happens only at the surface of the liquid • Boiling: – Happens throughout the liquid

  7. Liquid Solid Gas Gas to Liquid Condensation

  8. Condensation & Vaporization Liquid These changes happen at the: Boiling Point Solid Gas

  9. Liquid Solid Gas Solid to Gas Sublimation

  10. Liquid Solid Gas Gas to Solid Deposition

  11. Gas/ Liquid 140 120 100 80 Liquid/Solid 60 BP 40 20 M/F 0 -20 2 4 6 8 10 12 14 16 Change of State (H2O) Solid Gas Liquid

  12. Heat of Fusion of H2O = 334 kJ kg Heat of Fusion • The amount of energy needed to change a material from a Solid to a Liquid(or Liquid to a Solid)

  13. Heat of Vapor. of H2O = 2260 kJ kg Heat of Vaporization • The amount of energy needed to change a material from a Liquid to a Gas(or Gas to a Liquid)

  14. kJ (kg X °C) Specific Heat • The amount of ENERGY needed to raise the temperature of 1 kg of material 1 °C • Every material has its own “SH” • H2O has three “SH”: • Solid (ICE) = 2.00 • Liquid (WATER) = 4.18 • Gas (VAPOR) = 2.02

  15. X Δ Temp kJ X S.Heat (kg X °C) Energy needed to change from Ice to Water to Vapor • Formula: Stays in the SAME STATE (kg) E = Mass (°C) Δ = “Change In”

  16. kJ kg X Heatof(FusionorVaporization) Energy needed to change from Ice to Water to Vapor Formula: CHANGES State (kg) E = Mass

  17. Heat of Fusion (H2O) = 334 kJ/kg Example #1: How much energy is needed to melt 150 kg of ice at 0°C? MELT 150 kg (1) Look for key words (2) Choose the correct Formula: Change in States Formula: E = mass X Heat of FUSION (3) Find needed information

  18. 1 Example #1 (cont.): (1) Energy = Mass X Heat of Fusion 150 kg 334 kJ kg (2) E = X (3) Energy = 50,100 kJ

  19. Specific Heat Water (Liquid) 4.18 kJ (kg x °C) Example #2: How much energy is needed to raise the temperature of 25 kg of water from 5°C to 50°C? 25 kg raise the temperature 5°C to 50°C (1) Look for key words (2) Choose the correct Formula: Stays in Same State Formula: E = mass X  Temp X S. Heat

  20. 1 1 Example #2 (Cont.): (1) E = mass X  Temp X S. Heat 25 kg (2) E = 45 °C 4.18 kJ kg x °C X X (3) Energy = 4,702.5 kJ

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