Energy and Phase Changes. Energy Requirements for State Changes To change the state of matter, energy must be added or removed.

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Presentation on theme: "Energy and Phase Changes. Energy Requirements for State Changes To change the state of matter, energy must be added or removed."— Presentation transcript:

1 Energy and Phase Changes

2 Energy Requirements for State Changes To change the state of matter, energy must be added or removed.

3 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

4 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)

5 Heating curve A 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

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7 Heat of fusion The 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

8 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

9 Heat of vaporization The 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

10 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

11 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.

12 Heating curve calculations For heating (same phase): q = C.m ΔT For phase changes: heat of fusion or vaporization

13 Example How much heat do you need to melt 10 g of ice and then heat that to 10 o C? Melting: 10 g x 334 J/g = 3340 J Heating: 10 g x 4.184 J/(g. o C) x 10 o C = 418.4 J Total heat needed: 3340 J + 418 J = 3758 J


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