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Endothermic process Solid to a Liquid Melting (Fusion) particles overcome attractive forces and move around & past other particles Solid to a Gas Sublimation occurs only at conditions far from normal MP Liquid to a Gas Vaporization particles are very spread out – requires a lot of energy evaporation – vaporization at the surface of a liquid
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Energy and Phase Changes
Energy Requirements for State Changes
• To change the state of matter, energy must be added or removed.
Endothermic process• Solid to a Liquid Melting (Fusion)
• particles overcome attractive forces and move around & past other particles
• Solid to a Gas Sublimation• occurs only at conditions far from normal MP
• Liquid to a Gas Vaporization• particles are very spread out – requires a lot of
energy• evaporation – vaporization at the surface of a
liquid
Exothermic processes
• Gas to a Liquid Condensation(equal and opposite of vaporization)
• Liquid to a solid Solidification(equal and opposite of melting)
Gas to a solid Deposition(equal and opposite of sublimation)
Heating curveA heating curve • illustrates the changes of state as a
solid is heated to a gas.• uses sloped lines to show an increase
in temperature.• uses plateaus (flat lines) to indicate a
change of state.• A cooling curve shows the opposite
process
Heat of fusionThe heat of fusion • is the amount of heat released when 1 gram of
liquid freezes (at its freezing point). • is the amount of heat needed to melt 1 gram of
a solid (at its melting point).• For water (at 0°C) =
334 J 1 g water
The heat needed to freeze (or melt) a specific mass of water (or ice) is calculated using the heat of fusion.
Heat = g water x 334 J 1 g water
Example: How much heat is needed to melt 15.0 g of water?
15.0 g water x 334 J = 5.01 kJ
1 g water
Heat of vaporizationThe heat of vaporization is the amount of heat• absorbed to vaporize 1 g of a liquid to gas at the boiling
point.• released when 1 g of a gas condenses to liquid at the
boiling point.
Boiling Point of Water = 100°C
Heat of Vaporization (water) = 2260 J 1 g water
• The heat needed to vaporize (or boil) a specific mass of water (or water vapor/steam) is calculated using the heat of vaporization.
Heat = g water x 2260 J 1 g water
Example: How much heat is needed to boil 12.0 g of water?
12.0 g water x 2260 J = 27.1 kJ
1 g water
Practice• Calculate the heat of vaporization of
25 g of water (in kJ).
• From this calculate the heat of vaporization of water in kJ/mol.
Heating curve calculations
• For heating (same phase): q = C.m ΔT• For phase changes: heat of fusion or
vaporization
Example• How much heat do you need to melt
10 g of ice and then heat that to 10 oC?
• Melting: 10 g x 334 J/g = 3340 J• Heating: 10 g x 4.184 J/(g.oC) x 10 oC =
418.4 J• Total heat needed: 3340 J + 418 J =
3758 J
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