Phas e Changes. Solid to Liquid Melting Liquid SolidGas.

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Presentation transcript:

Phas e Changes

Solid to Liquid Melting Liquid SolidGas

Liquid to Solid Freezing Liquid SolidGas

Melting & Freezing These changes happen at the: Melting/Freezing Point Liquid SolidGas

Liquid to Gas Vaporization Liquid SolidGas

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

Gas to Liquid Condensation Liquid SolidGas

Condensation & Vaporization Liquid SolidGas These changes happen at the: Boiling Point

Solid to Gas Sublimation Liquid SolidGas

Gas to Solid Deposition Liquid SolidGas

Change of State (H 2 O) Liquid SolidGas M/FBP Gas/ Liquid Liquid /Solid

Heat of Fusion The amount of energy needed to change a material from a Solid to a Liquid (or Liquid to a Solid) Heat of Fusion of H 2 O334 kJ Heat of Fusion of H 2 O = 334 kJkg

Heat of Vaporization The amount of energy needed to change a material from a Liquid to a Gas (or Gas to a Liquid) Heat of Vapor. of H 2 O2260 kJ Heat of Vapor. of H 2 O = 2260 kJkg

Specific Heat The amount of ENERGY needed to raise the temperature of 1 kg of material 1 °C The amount of ENERGY needed to raise the temperature of 1 kg of material 1 °C Every material has its own “SH” Every material has its own “SH” H 2 O has three “SH”: H 2 O has three “SH”: – Solid (ICE) = 2.00 – Liquid (WATER) = 4.18 – Gas (VAPOR) = 2.02 kJ (kg X °C)

Energy needed to change from Ice to Water to Vapor FormulaSAME STATE Formula: Stays in the SAME STATE E = MasskJ (kg X °C) S.HeatX (°C) (kg) Δ Temp X Δ = “Change In”

Energy needed to change from Ice to Water to Vapor E = MasskJkg Heat of (Fusion or Vaporization) X(kg) FormulaCHANGES State Formula: CHANGES State

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

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

Example #2: How much energy is needed to raise the temperature of 25 kg of water from 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 raise the temperature 25 kg 5°C to 50°C Specific Heat Water (Liquid) 4.18 kJ (kg x °C) 4.18 kJ (kg x °C)

Example #2 (Cont.): (1) E = mass X  Temp X S. Heat (3) Energy = 4,702.5 kJ (2) E = 25 kg X 45 °C X 4.18 kJ kg x °C 11

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