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Matter and Energy

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

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Matter and Energy

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  1. Matter and Energy Chapter 12.4 Chapter 15.1 - 15.3

  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

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

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

  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

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

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

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

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

  10. Ch. 12.4 - Changes in Matter • 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 (H2O) They generally require energy, the ability to absorb or release heat (or work).

  11. Phase Changes • What are the phase changes of water? • Melting – changing of a solid to a liquid (heat of fusion = ∆Hf) • Vaporization – changing from a liquid to a gas (heat of vaporization = ∆Hvap) • Sublimation – Changing from a solid to a gas (heat of sublimation = ∆Hsub) What do these processes have in common? Answer:

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

  13. James Joule 1818-1889 UNITS OF ENERGY 1 calorie = heat required to raise temp. of 1.00 g of H2O by 1.0 oC. 1000 cal = 1 kilocalorie = 1 kcal 1 kcal = 1 Calorie (a food “calorie”) But we use the unit called the JOULE 1 cal = 4.184 Joules (exactly)

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

  15. Example Problems • How much heat does it take to melt 20.5 g of ice at 0⁰C? • q = 334 J/g * 20.5 = 6850 J (6.85 kJ) • How much heat is released when 50.0 g of steam at 100 ⁰C condenses to water at 100 ⁰C? • q = - 2260 J/g * 50.0 g = -113,000 J (-113 kJ)

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

  17. Heat Capacity Values Substance Spec. Heat (J/g•⁰C) Water 4.184 Ethylene glycol 2.39 Al uminum 0.897 glass 0.84

  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. • q = mC∆T

  19. Calculations involving Heat • Example 1: A 5.00 g piece of aluminum is heated from 25.0⁰Cto 99.5⁰C. How many joules of heat did it absorb? • q = m * C * ∆T = 5.00 g * 0.897 J/g*⁰C * 74.5⁰C = 334 J

  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

  21. Calculations involving Heat • 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 ⁰C. (-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.

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

  23. Heating/Cooling Curve for Water

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

  25. Heat & Changes of State What quantity of energy as heat is required to melt 500. g of ice (at 0⁰C) and heat the water to steam at 100 oC? 1. To melt ice at 0⁰C q = (500. g)(333 J/g) = 1.67 x 105 J 2. To raise water from 0 oC to 100oC q = (500. g)(4.2 J/g•K)(100 - 0)K = 2.1 x 105 J 3. To vaporize water at 100oC q = (500. g)(2260 J/g) = 1.13 x 106 J 4. Total energy = 1.51 x 106 J = 1510 kJ

  26. Practice problem If we add 6050 J of heat to 54.2g of ice at -10.0⁰C, whatwillit be at theend? Whattemperaturewillit be? Thespecificheat of ice is 2.03 J/g*⁰C.

  27. Pressure • Pressureis the force acting on an object per unit area: • Gravity exerts a force on the earth’s atmosphere • A column of air 1 m2 in cross section exerts a force of about 105 N (101,300 N/m2). • 1 Pascal (Pa) = 1 N/m2 . So, 101,300 N/m2 = 101,300 Pa or 101.3 kPa. • Since we are at the surface of the earth, we ‘feel’ 1 atmosphere of pressure.

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

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

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

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

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