PhasPhase Chae Changesnges

Solid to LiquidSolid to Liquid

MeltingMelting

Liquid

Solid Gas

Liquid to SolidLiquid to Solid

FreezingFreezing

Liquid

Solid Gas

Melting & FreezingMelting & Freezing

These changes happen at theThese changes happen at the::

Melting/Freezing PointMelting/Freezing Point

Liquid

Solid Gas

Liquid to GasLiquid to Gas

VaporizationVaporization

Liquid

Solid Gas

Vaporization TypesVaporization TypesTwo TYPESTwo TYPES:EvaporationEvaporation: - Happens only

at the surface of the liquidBoiling:Boiling: – Happens

throughout the liquid

Gas to LiquidGas to Liquid

CondensationCondensation

Liquid

Solid Gas

Condensation & VaporizationCondensation & VaporizationLiquid

Solid Gas

These changes happen at theThese changes happen at the::

Boiling PointBoiling Point

Solid to GasSolid to Gas

SublimationSublimation

Liquid

Solid Gas

Gas to SolidGas to Solid

DepositionDeposition

Liquid

Solid Gas

Change of State (HChange of State (H22O)O)LiquidSolid Gas

-20

120

10010080

60

40

2000

140

2 10 168 1464 12

M/F

BP

Gas/ Gas/ LiquidLiquid

LiquidLiquid/Solid/Solid

Heat of FusionHeat of FusionThe amount of energy

needed to change a material from a Solid to a Liquid (or Liquid to a Solid)

Heat of Fusion of HHeat of Fusion of H22OO = 334 kJ334 kJkgkg

Heat of VaporizationHeat of VaporizationThe amount of energy

needed to change a material from a Liquid to a Gas (or Gas to a Liquid)

Heat of Vapor. of HHeat of Vapor. of H22OO = 2260 kJ2260 kJkgkg

Specific HeatSpecific HeatThe amount of The amount of ENERGYENERGY needed needed

to raise the temperature of 1 kg to raise the temperature of 1 kg of material 1 of material 1 °C°C

Every material has its own “SH”Every material has its own “SH”HH22O has three “SH”:O has three “SH”:

– SolidSolid (ICE) (ICE) = 2.00= 2.00– LiquidLiquid ( (WATERWATER) ) = 4.18= 4.18– GasGas ( (VAPORVAPOR) ) = 2.02= 2.02

kJkJ

(kg X (kg X °C)°C)

Energy needed to change Energy needed to change from from IceIce to to WaterWater to to VaporVapor

FormulaFormula: Stays in the SAME STATESAME STATE

E = E = MassMass

kJkJ

(kg X (kg X °C)°C)S.HeatS.HeatXX

((°C)°C)

(kg)(kg)

Δ TempΔ TempXX

Δ = “Change In”Δ = “Change In”

Energy needed to change Energy needed to change from from IceIce to to WaterWater to to VaporVapor

E = E = MassMass

kJkJ

kgkgHeatHeat ofof ((FusionFusion oror

VaporizationVaporization))XX

(kg)(kg)

FormulaFormula: CHANGES StateCHANGES State

Example #1Example #1::How much energy is needed to How much energy is needed to

melt 150 kg of ice at 0melt 150 kg of ice at 0°C?°C?(1) Look for key words(1) Look for key words

(2) Choose the correct Formula:(2) Choose the correct Formula:

(3) Find needed information(3) Find needed information

Change in States FormulaChange in States Formula::

E = mass X Heat ofE = mass X Heat of

MELTMELT 150 kg150 kg

Heat of FusionHeat of Fusion

(H(H22O) O)

= 334 kJ/kg= 334 kJ/kg

FUSIONFUSION

Example #1 (cont.)Example #1 (cont.)::(1) (1) Energy = Mass X Heat of FusionEnergy = Mass X Heat of Fusion

(2) E = (2) E = 150 kg150 kg X X 334 kJ334 kJ

kgkg 1 1

(3) (3) Energy Energy = = 50,100 kJ50,100 kJ

Example #2Example #2::How much energy is needed to How much energy is needed to

raise the temperature of 25 kg of raise the temperature of 25 kg of water from 5water from 5°C to 50°C?°C to 50°C?

(1) Look for key words(1) Look for key words

(2) Choose the correct Formula:(2) Choose the correct Formula:Stays in Same State FormulaStays in Same State Formula::

E = mass X E = mass X Temp X S. Heat Temp X S. Heat

raise the temperatureraise the temperature 25 kg25 kg55°C to 50°C°C to 50°C

Specific HeatSpecific Heat

Water (Liquid) Water (Liquid)

4.18 kJ4.18 kJ (kg x (kg x °C)°C)

Example #2 (Cont.)Example #2 (Cont.)::

(1) E = mass X (1) E = mass X Temp X S. Heat Temp X S. Heat

(3) Energy =(3) Energy = 4,702.5 kJ 4,702.5 kJ

(2) E =(2) E = 25 kg25 kg XX

45 45 °C°C XX 4.18 kJ4.18 kJ kg x kg x °C°C 1 1 1 1