Presentation is loading. Please wait.

Presentation is loading. Please wait.

Energy in Phase Changes. System vs. Surroundings The system is the part of the universe that interests us, i.e. the reactants and products in a chemical.

Similar presentations


Presentation on theme: "Energy in Phase Changes. System vs. Surroundings The system is the part of the universe that interests us, i.e. the reactants and products in a chemical."— Presentation transcript:

1 Energy in Phase Changes

2 System vs. Surroundings The system is the part of the universe that interests us, i.e. the reactants and products in a chemical reaction. The surroundings include everything else in the universe, i.e. the test tube, beaker, air, lab equipment in contact with the system.

3 Exothermic A reaction is exothermic when heat flows “out of” the system and is released to the surroundings. Ex. the combustion of methane. System Energy Surroundings

4 Endothermic A reaction is endothermic when heat flows “into” the system and is absorbed from the surroundings. Ex. the melting of ice. System Energy Surroundings

5 Three Phases Solids – molecules vibrate within their crystal structure; small kinetic energy Liquids – molecules not only vibrate but also rotate and move around; higher kinetic energy Gases – molecules move around more and are faster; highest kinetic energy of the 3 phases.

6 Phase Change A phase change is when a substance changes from one phase to another. IceWater Steam

7 The melting point of a substance is the temperature at which the vibrating motion (kinetic energy) of the molecules becomes large enough to break the crystal structure. The boiling point is the temperature at which the kinetic energy is great enough to overcome the attractive forces between molecules in a liquid.

8 SOLIDLIQUIDGAS meltingboiling condensingfreezing ENDOTHERMIC EXOTHERMIC

9 As heat energy is added to a substance, the molecular motion (kinetic energy) increases. temperature  = kinetic energy  temperature  = kinetic energy  During phase changes however, temperature remains constant.

10 Heat energy added at a constant rate Temperature (ºC) -10 0 10 A BC Heating Curve for Ice

11 A – solid ice is warming; KE is increasing B – phase change of ice melting to water; note temperature remains constant (  no increase in KE) even though energy is still being added. Where is it going? This energy breaks the crystal structure and is stored in the liquid phase as PE. C – liquid water warming; KE is increasing

12 Molar Enthalpy of Fusion is the amount of heat energy required to melt (break the crystal structure) one mole of a pure solid to liquid. This is a constant # and is different for different substances. The molar enthalpy of fusion (  H) for water is +6.03 kJ/mol This same amount would be released if one mole of water freezes (-6.03 kJ/mol)

13 Molar Enthalpy of Vaporization is the amount of heat energy required to boil one mole of a pure liquid to gas. This is a constant # and is different for different substances. The molar enthalpy of vaporization (  H) for water is +40.8 kJ/mol This same amount would be released if one mole of water condensed (-40.8kJ/mol)

14 Calculation Potential Energy We can use the molar enthalpy of fusion or vaporization to calculate the amount of potential energy associated with the phase change. Ep = n  H, where n - # of moles Example: Find the heat energy needed to melt 48.0 g of ice at 0ºC to water at 0ºC.

15 Phase Change Questions

16 Why do you burn your feet walking down the beach, but yet the water is cold??

17 Specific Heat Capacity is the amount of heat required to raise the temperature of one gram of a substance by one degree Celsius. Water has one of the highest specific heat capacities at 4.19 J/g ºC

18 Calculating Kinetic Energy We can use specific heat capacity to calculate the amount of kinetic energy associated with temperature changes. The quantity of heat can be calculated by: q = mc  t where, q = quantity of heat (in J) m = mass of substance (in g) c = specific heat capacity (in J/g ºC)  t = temperature change (in ºC)

19 Example Calculate the amount of heat that flows into 121 g of water to raise its temperature from 10.0ºC to 60.0ºC. q = mc  t = (121g)(4.19J)(+50.0ºC) g ºC = +25,300 J or +25.3 kJ


Download ppt "Energy in Phase Changes. System vs. Surroundings The system is the part of the universe that interests us, i.e. the reactants and products in a chemical."

Similar presentations


Ads by Google