1. Matter and Energy Chapter 12.4 Chapter 15.1 - 15.3 1

2. Chapter Objectives Identify observable characteristics of a chemical reaction Define Energy Show how energy applies to chemical reactions and physical processes Interpret phase diagrams Interpret heating (cooling) curves 2

3. Vocabulary – Ch. 3.1 – 3.2 (SIA Review) Physical property Physical property Extensive property Extensive property Intensive property Intensive property Chemical property Chemical property States of matter States of matter Solid Solid Liquid Liquid Gas Gas Vapor Vapor Physical Change Physical Change Chemical Change Chemical Change Law of Conservation of Mass Law of Conservation of Mass Phase Change Phase Change 3

4. Vocabulary – Chapter 15.1-15.2 Energy Energy Heat Heat Joule Joule Specific Heat Specific Heat Specific Heat Equation (q = mC ∆T) Specific Heat Equation (q = mC ∆T) Heat of Vaporization (∆H vap ) Heat of Vaporization (∆H vap ) Heat of Fusion (∆H fus ) Heat of Fusion (∆H fus ) Heating Curve Heating Curve 4

5. Vocabulary – Ch. 12.4 Pressure Barometer Atmosphere Melting Point Vaporization Evaporation Vapor Pressure Boiling Point Sublimation Freezing point Condensation Deposition Phase diagram Triple Point 5

6. Properties of Matter Chemical Properties: the ability to combine or change into other substances. Chemical Properties: the ability to combine or change into other substances. –Examples: flammability, oxidation, rotting 6

7. States of Matter State of Matter: Its physical form. State of Matter: Its physical form. There are three physical states: There are three physical states: Solid: –Definite shape –Definite volume –Closely packed particles 7

8. States of Matter Liquid: –particles move past each other (flow) – definite volume – takes the shape of its container (indefinite) 8

9. States of Matter Gas: – flows – takes the shape of its container its container (indefinite shape) –Fills the container completely. (indefinite volume) completely. (indefinite volume)  Note: A vapor refers to a gaseous state of a substance that is a solid or liquid at room temperature. 9

10. Ch. 12.4 - Changes in Matter Ch. 12.4 - Changes in Matter Physical changes are those which alter the substance without altering its composition. Physical changes are those which alter the substance without altering its composition. –Change of phase  one physical state to another Melting of ice - composition unchanged, i.e. ice is water in solid form (H 2 O) They generally require energy, the ability to absorb or release heat (or work). 10

11. Phase Changes What are the phase changes of water? 1. M elting – changing of a solid to a liquid (heat of fusion = ∆Hf) 1. V aporization – changing from a liquid to a gas (heat of vaporization = ∆Hvap) 2. S ublimation – Changing from a solid to a gas (heat of sublimation = ∆Hsub) What do these processes have in common? Answer: 11

12. Phase Changes Phase Phase changes in the opposite direction have names too. 1. liquid 1. liquid to a solid: 2. gas 2. gas to a liquid: 3. gas 3. gas to a solid: What do these have in common? Answer: 12

13. UNITS OF ENERGY 1 calorie = heat required to raise temp. of 1.00 g of H 2 O by 1.0 o C. 1000 cal = 1 kilocalorie = 1 kcal 1 kcal = 1 Calorie (a food “calorie”) JOULE But we use the unit called the JOULE 1 cal = 4.184 Joules (exactly) James Joule 1818-1889 13

14. Heats of Fusion & Vaporization Heat of Fusion (∆H fus ) – The amount of heat (in Joules) needed to melt 1 g of substance. Heat of Fusion (∆H fus ) – The amount of heat (in Joules) needed to melt 1 g of substance. For ice: 334 J/g For ice: 334 J/g q (heat) = ∆H fus *m (m= mass of ice/water) q (heat) = ∆H fus *m (m= mass of ice/water) Heat of Vaporization (∆H vap ) – The amount of heat (in Joules) needed to vaporize 1 g of substance Heat of Vaporization (∆H vap ) – The amount of heat (in Joules) needed to vaporize 1 g of substance For water: 2260 J/g For water: 2260 J/g q (heat) = ∆H vap *m q (heat) = ∆H vap *m (m= mass of water/steam) (m= mass of water/steam) 14

15. Example Problems How much heat does it take to melt 20.5 g of ice at 0 ⁰ C? How much heat does it take to melt 20.5 g of ice at 0 ⁰ C? q = 334 J/g * 20.5 = 6850 J (6.85 kJ) q = 334 J/g * 20.5 = 6850 J (6.85 kJ) How much heat is released when 50.0 g of steam at 100 ⁰ C condenses to water at 100 ⁰ C? How much heat is released when 50.0 g of steam at 100 ⁰ C condenses to water at 100 ⁰ C? q = - 2260 J/g * 50.0 g = -113,000 J q = - 2260 J/g * 50.0 g = -113,000 J (-113 kJ) (-113 kJ) 15

16. Specific Heat Capacity Specific Heat Capacity – amount of heat (q) Specific Heat Capacity – amount of heat (q) required to raise the required to raise the temperature of one gram temperature of one gram of a substance by 1 degree. of a substance by 1 degree. C = J (energy gained or lost) C = J (energy gained or lost) mass (g) * Temp Change( ⁰C) mass (g) * Temp Change( ⁰C) 16

17. Heat Capacity Values SubstanceSpec. Heat (J/g ⁰C ) Water4.184 Ethylene glycol2.39 Aluminum0.897 glass0.84 17

18. Calculating Heat Gained or lost The heat, q, gained or lost by a substance can be calculated by knowing the mass of the object, the temperature change, and the heat capacity. The heat, q, gained or lost by a substance can be calculated by knowing the mass of the object, the temperature change, and the heat capacity. q = mC∆T q = mC∆T 18

19. Calculations involving Heat Example 1: A 5.00 g piece of aluminum is heated from 25.0 ⁰ C to 99.5 ⁰ C. How many joules of heat did it absorb? Example 1: A 5.00 g piece of aluminum is heated from 25.0 ⁰ C to 99.5 ⁰ C. How many joules of heat did it absorb? q = m * C * ∆T q = m * C * ∆T = 5.00 g * 0.897 J/g* ⁰ C * 74.5 ⁰ C = 5.00 g * 0.897 J/g* ⁰ C * 74.5 ⁰ C = 334 J = 334 J 19

20. Calculations involving Heat Example 2: 10.2 g of cooking oil at 25.0 ⁰C is placed in a pan and 3.34 kJ of heat is required to raise the temperature to 196.4 ⁰C. What is the specific heat of the oil? q = m*C*∆T C = q/(m ∆T) C = 3340 J/(10.2 g * (196.4-25.0) ⁰C) C = 1.91 J/g* ⁰C 20

21. Calculations involving Heat Important Points! Important Points!  q (heat) is a positive quantity. The sign (+ or -) refers to whether the system you’re looking gained it (+) or lost it (-).  From the previous example, the oil would lose 3340 J of heat upon cooling back to 25.0 ⁰ C. (-3340 J heat lost)  Specific heat capacity is like a bucket. It is a measure of how much energy an object absorbs before the temperature changes. 21

22. Heating Curve for Water Note that T is constant as ice melts 22

23. Heating/Cooling Curve for Water 23

24. Heat of fusion of ice = 333 J/g Specific heat of water = 4.2 J/gK Heat of vaporization = 2260 J/g Heat of fusion of ice = 333 J/g Specific heat of water = 4.2 J/gK Heat of vaporization = 2260 J/g What quantity of heat is required to melt 500. g of ice (at 0 o C) and heat the water to steam at 100 o C? Heat & Changes of State +333 J/g + +2260 J/g 24

25. What quantity of energy as heat is required to melt 500. g of ice (at 0 ⁰ C) and heat the water to steam at 100 o C? 1. To melt ice at 0 ⁰ C q = (500. g)(333 J/g) = 1.67 x 10 5 J 2.To raise water from 0 o C to 100 o C q = (500. g)(4.2 J/gK)(100 - 0)K = 2.1 x 10 5 J q = (500. g)(4.2 J/gK)(100 - 0)K = 2.1 x 10 5 J 3.To vaporize water at 100 o C q = (500. g)(2260 J/g) = 1.13 x 10 6 J q = (500. g)(2260 J/g) = 1.13 x 10 6 J 4. Total energy = 1.51 x 10 6 J = 1510 kJ Heat & Changes of State 25

26. Practice problem If we add 6050 J of heat to 54.2g of ice at -10.0 If we add 6050 J of heat to 54.2g of ice at -10.0 ⁰ C, what will it be at the end? What temperature will it be? The specific heat of ice is 2.03 J/g* ⁰ C. 26

27. Pressure is the force acting on an object per unit area: Gravity exerts a force on the earth’s atmosphere A column of air 1 m 2 in cross section exerts a force of about 10 5 N (101,300 N/m 2 ). 1 Pascal (Pa) = 1 N/m 2. So, 101,300 N/m 2 = 101,300 Pa or 101.3 kPa. Since we are at the surface of the earth, we ‘feel’ 1 atmosphere of pressure. 27 Pressure

28. 28

29. Barometer A barometer measures atmospheric pressure A barometer measures atmospheric pressure The pressure of the atmosphere at sea level will hold a column of mercury 760 mm Hg. The pressure of the atmosphere at sea level will hold a column of mercury 760 mm Hg. 1 atm = 760 mm Hg 1 atm = 760 mm Hg 1 atm Pressure 760 mm Hg Vacuum 29

30. Units of pressure 1 atmosphere (atm) = 760 mm Hg = 760 torr 1 atmosphere (atm) = 760 mm Hg = 760 torr 1 atm = 101,300 Pascals = 101.3 kPa 1 atm = 101,300 Pascals = 101.3 kPa Can make conversion factors from these. Can make conversion factors from these. What is 724 mm Hg in atm ? What is 724 mm Hg in atm ? What is 724 mm Hg in kPa What is 724 mm Hg in kPa 30

31. Phase Changes Vapor pressure is the pressure exerted by a vapor over a liquid. The vapor pressure increases with increasing temperature. This is why water evaporates even though it’s not 212˚F. 31

32. Phase Changes However, when the vapor pressure of the water is the same as the atmospheric pressure the water is … boiling. 32

33. Phase Diagram (Ch. 12.4) A phase diagram is a graph of pressure vs temperature that shows in which phase a substance exists under different conditions of T & P. 33

34. Temperature Solid Liquid Gas 1 Atm Pressure 34 Phase Diagram for water Sublimation Deposition Melting Freezing Boiling Condensation