2. 1 Chapter 2Energy and Matter 2.1 Energy Copyright © 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings

3. 2 Energy makes objects move. makes things stop. is needed to “do work”. Energy

4. 3 Work Work is done when you climb. you lift a bag of groceries. you ride a bicycle. you breathe. your heart pumps blood. water goes over a dam. Copyright © 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings

5. 4 Potential Energy Potential energy is energy stored for use at a later time. Examples are water behind a dam. a compressed spring. chemical bonds in gasoline, coal, or food. Copyright © 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings

6. 5 Kinetic Energy Kinetic energy is the energy of matter in motion. Examples are swimming. water flowing over a dam. working out. burning gasoline. Copyright © 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings

7. 6 Learning Check Identify the energy as potential or kinetic. A. roller blading B. a peanut butter and jelly sandwich C. mowing the lawn D. gasoline in the gas tank

8. 7 Solution Identify the energy as 1) potential or 2) kinetic. A. roller blading (kinetic) B. a peanut butter and jelly sandwich (potential) C. mowing the lawn (kinetic) D. gasoline in the gas tank (potential)

9. 8 Heat is measured in joules or calories. 4.184 Joules (J) = 1 calorie (cal) (exact) 1 kJ = 1000 J 1 kilocalorie (kcal) = 1000 calories (cal) Units for Measuring Energy or Heat

10. 9 Examples of Energy In Joules

11. 10 Learning Check How many calories are obtained from a pat of butter if it provides 150 J of energy when metabolized?

12. 11 Solution How many calories are obtained is a pat of butter if it provides 150 J of energy when metabolized? Given: 150 J Need: calories Plan: J  cal Equality: 1 cal = 4.184 J Set Up:150 J x 1 cal = 36 cal 4.184 J

13. 12 2.2Energy and Matter

14. 13 Calorimeters A calorimeter is used to measure heat transfer. can be made with a coffee cup and a thermometer. indicates the heat lost by a sample indicates the heat gained by water.

15. 14 Energy and Nutrition On food labels, energy is shown as the nutritional Calorie, written with a capital C. In countries other than the U.S., energy is shown in kilojoules (kJ). 1 Cal = 1000 calories 1 Cal = 1 kcal 1 Cal = 1000 cal 1 Cal = 4184 J 1 Cal = 4.184 kJ

16. 15 Caloric Food Values The caloric or energy values for foods indicate the number of kcal(Cal) provided by 1 g of each type of food. Carbohydrate: 4 kcal 1 g Fat (lipid): 9 kcal 1 g Protein: 4 kcal 1 g

17. 16 Energy Values for Some Foods

18. 17 Energy Requirements The amount of energy needed each day depends on age, sex, and physical activity. TABLE 2.3

19. 18 A cup of whole milk contains 12 g of carbohydrate, 9.0 g of fat, and 5.0 g of protein. How many kcal (Cal) does a cup of milk contain? 1) 48 kcal (or Cal) 2) 81 kcal (or Cal) 3) 150 kcal (or Cal) Learning Check

20. 19 A cup of whole milk contains 12 g of carbohydrate, 9.0 g of fat, and 5.0 g of protein. How many kcal (Cal) does a cup of milk contain? 3) 150 kcal (or Cal) 12 g carbohydrates x 4 kcal/g = 50 kcal (1 SF) 9.0 g fat x 9 kcal/g = 80 kcal (1 SF) 5.0 g protein x 4 kcal/g = 20 kcal (1 SF) 150 kcal Solution

21. 20 Chapter 2 Energy and Matter 2.3 Temperature Conversions Copyright © 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings

22. 21 Temperature is a measure of how hot or cold an object is compared to another object. indicates that heat flows from the object with a higher temperature to the object with a lower temperature. is measured using a thermometer. Copyright © 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings

23. 22 Temperature Scales Copyright © 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings are Fahrenheit, Celsius, and Kelvin. have reference points for the boiling and freezing points of water.

24. 23 A. What is the temperature of freezing water? 1) 0°F 2) 0°C 3) 0 K B. What is the temperature of boiling water? 1) 100°F 2) 32°F 3) 373 K C. How many Celsius units are between the boiling and freezing points of water? 1) 1002) 1803) 273 Learning Check

25. 24 A. What is the temperature of freezing water? 2) 0°C B. What is the temperature of boiling water? 3) 373 K C. How many Celsius units are between the boiling and freezing points of water? 1) 100 Solution

26. 25 On the Fahrenheit scale, there are 180°F between the freezing and boiling points and on the Celsius scale, there are 100°C. 180°F = 9°F =1.8°F 100°C 5°C 1°C In the formula for the Fahrenheit temperature, adding 32 adjusts the zero point of water from 0°C to 32°F. T F = 9/5 T C + 32  or T F = 1.8 T C + 32  Fahrenheit Formula

27. 26 T C is obtained by rearranging the equation for T F. T F = 1.8T C + 32 Subtract 32 from both sides. T F - 32 = 1.8T C ( +32 - 32) T F - 32 = 1.8T C Divide by 1.8 =°F - 32 = 1.8 T C 1.8 1.8 T F - 32 = T C 1.8 Celsius Formula

28. 27 Solving A Temperature Problem A person with hypothermia has a body temperature of 34.8°C. What is that temperature in °F? T F = 1.8 T C + 32  T F = 1.8 (34.8°C) + 32° exact tenth's exact = 62.6 + 32° = 94.6°F tenth’s Copyright © 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings

29. 28 The normal temperature of a chickadee is 105.8°F. What is that temperature on the Celsius scale? 1) 73.8°C 2) 58.8°C 3) 41.0°C Learning Check

30. 29 3) 41.0 °C T C = (T F - 32°) 1.8 =(105.8 - 32°) 1.8 =73.8°F = 41.0°C 1.8° Solution

31. 30 A pepperoni pizza is baked at 455°F. What temperature is needed on the Celsius scale? 1) 423°C 2) 235°C 3) 221°C Learning Check

32. 31 A pepperoni pizza is baked at 455°F. What temperature is needed on the Celsius scale? 2) 235°C T F - 32° = T C 1.8 (455 - 32°) = 235°C 1.8 Solution

33. 32 On a cold winter day, the temperature is –15°C. What is that temperature in °F? 1) 19°F 2) 59°F 3) 5°F Learning Check

34. 33 3) 5°F T F = 1.8 T C + 32  T F = 1.8(–15°C) + 32° = – 27 + 32° = 5°F Note: Be sure to use the change sign key on your calculator to enter the minus – sign. 1.8 x 15 +/ – = –27 Solution

35. 34 The Kelvin temperature scale has 100 units between the freezing and boiling points of water. 100 K = 100°Cor 1 K = 1°C is obtained by adding 273 to the Celsius temperature. T K = T C + 273 contains the lowest possible temperature, absolute zero (0 K). 0 K = –273°C Kelvin Temperature Scale

36. 35 Temperatures TABLE 2.5

37. 36 What is normal body temperature of 37°C in Kelvins? 1) 236 K 2) 310. K 3)342 K Learning Check

38. 37 What is normal body temperature of 37°C in kelvins? 2) 310. K T K = T C + 273 = 37°C + 273 = 310. K Solution

39. 38 Chapter 2Energy and Matter 2.4 Specific Heat

40. 39 Specific heat is different for different substances. is the amount of heat that raises the temperature of 1 g of a substance by 1°C. in the SI system has units of J/g  C. in the metric system has units of cal/g  C. Specific Heat

41. Specific heat capacity Specific heat capacity - amount of heat per gram required to raise the temperature by 1º Cheattemperature C water = 1 cal/g°C - higher most common substances (metals)

42. 41 A. When ocean water cools, the surrounding air 1) cools. 2) warms.3) stays the same. B. Sand in the desert is hot in the day, and cool at night. Sand must have a 1) high specific heat. 2) low specific heat. Learning Check

43. 42 A. When ocean water cools, the surrounding air 2) warms. B. Sand in the desert is hot in the day, and cool at night. Sand must have a 2) low specific heat. Solution

44. 43 Learning Check What is the specific heat of a metal if 24.8 g absorbs 275 J of energy and the temperature rises from 20.2  C to 24.5  C?

45. 44 Solution What is the specific heat of a metal if 24.8 g absorbs 275 J of energy and the temperature rises from 20.2  C to 24.5  C? Given: 24.8 g metal, 275 J of energy, 20.2  C to 24.5  C Need: J/g  C Plan: SH = Heat/g  C ΔT = 24.5  C – 20.2  C = 4.3  C Set Up: 275 J = 2.6 J/g  C (24.8 g)(4.3  C)

46. 45 Heat Equation Rearranging the specific heat expression gives the heat equation. Heat = g x °C x J = J g°C The amount of heat lost or gained by a substance is calculated from the mass of substance (g). temperature change (  T). specific heat of the substance (J/g°C).

47. 46 How many kilojoules are needed to raise the temperature of 325 g of water from 15.0°C to 77.0°C? 1) 20.4 kJ 2) 77.7 kJ 3) 84.3 kJ Learning Check

48. 47 How many kilojoules are needed to raise the temperature of 325 g of water from 15.5°C to 77.5°C? 3) 84.3 kJ 77.0°C – 15.0°C = 62.0°C 325 g x 62.0°C x 4.184 J x 1 kJ g °C 1000 J = 84.3 kJ Solution

49. 48 Chapter 2Energy and Matter 2.5 States of Matter Copyright © 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings

50. 49 Solid, liquid, gas Watch the animations and then describe properties of the 3 states of matter AAAA nnnn iiii mmmm aaaa tttt iiii oooo nnnn

51. 50 Solids Solids have a definite shape. a definite volume. particles that are close together in a fixed arrangement. particles that move very slowly. Copyright © 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings

52. 51 Liquids Liquids have an indefinite shape, but a definite volume. the same shape as their container. particles that are close together, but mobile. particles that move slowly.

53. 52 Gases Gases have an indefinite shape. an indefinite volume. the same shape and volume as their container. particles that are far apart. particles that move very fast.

54. 53 Three States of Matter for Water

55. 54 Summary of the States of Matter Solid Liquid Gas Copyright © 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings

56. 55 Identify each as: 1) solid 2) liquid or 3) gas. ___ A. It has a definite volume, but takes the shape of the container. __ B. Its particles are moving rapidly. __ C. It fills the volume of a container. __ D. It has particles in a fixed arrangement. __ E. It has particles close together that are mobile. Learning Check

57. 56 Identify each as: 1) solid 2) liquid or 3) gas. 2 A. It has a definite volume, but takes the shape of the container. 3 B. Its particles are moving rapidly. 3 C. It fills the volume of a container. 1 D. It has particles in a fixed arrangement. 2 E. It has particles close together that are mobile. Solution

58. 57 Chapter 2Energy and Matter 2.6 Changes of State Copyright © 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings

59. 58 Melting and Freezing A substance is melting while it changes from a solid to a liquid. is freezing while it changes from a liquid to a solid. such as water has a freezing (melting) point of 0°C. Copyright © 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings

60. 59 Calculations Using 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) is 80. cal 1 g water

61. 60 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 80. cal 1 g water Example: How much heat in cal is needed to melt 15.0 g of water? 15.0 g water x 80. cal = 1200 cal 1 g water Calculation Using Heat of Fusion

62. 61 A. How many calories are needed to melt 5.00 g of ice at 0°C? 1) 80. cal2) 400 cal 3) 0 cal B. How many calories are released when 25.0 g of water at 0°C freezes? 1) 80. cal2) 0 cal 3) 2000 cal Learning Check

63. 62 A. How many calories are needed to melt 5.00 g of ice at 0°C? 2) 400 cal 5.00 g x 80. cal 1 g B. How many calories are released when 25.0 g of water at 0°C freezes? 3) 2000 cal 25 g x 80. cal 1 g Solution

64. 63 Sublimation occurs when particles change directly from solid to a gas. is typical of dry ice, which sublimes at -78  C. takes place in frost-free refrigerators. is used to prepare freeze- dried foods for long-term storage.

65. 64 Evaporation and Condensation Water evaporates when molecules on the surface gain sufficient energy to form a gas. condenses when gas molecules lose energy and form a liquid. Copyright © 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings

66. 65 Boiling At boiling, all the water molecules acquire enough energy to form a gas. bubbles appear throughout the liquid. Copyright © 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings

67. 66 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) = 540 cal 1 g water

68. 67 Learning Check How many kilocalories (kcal) are released when 50.0 g of steam from a volcano condenses at 100°C? 1) 27 kcal 2) 540 kcal 3) 2700 kcal

69. 68 Solution How many kilocalories (kcal) are released when 50.0 g of steam in a volcano condenses at 100°C? 1) 27 kcal 50.0 g steam x 540 cal x 1 kcal = 27 kcal 1 g steam 1000 cal

70. 69 Summary of Changes of State

71. Liquid Sublimation Melting Vaporization deposition Condensation Solid Freezing Gas Possible Phase Changes

72. 71 Heating Curve A heating curve illustrates the changes of state as a solid is heated. uses sloped lines to show an increase in temperature. uses plateaus (flat lines) to indicate a change of state. Copyright © 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings

73. 72 A. A flat line on a heating curve represents 1) a temperature change. 2) a constant temperature. 3) a change of state. B. A sloped line on a heating curve represents 1) a temperature change. 2) a constant temperature. 3) a change of state. Learning Check

74. 73 A. A flat line on a heating curve represents 2) a constant temperature. 3) a change of state. B. A sloped line on a heating curve represents 1) a temperature change. Solution

75. 74 Cooling Curve A cooling curve illustrates the changes of state as a gas is cooled. uses sloped lines to indicate a decrease in temperature. uses plateaus (flat lines) to indicate a change of state. Copyright © 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings

76. 75 Use the cooling curve for water to answer each. A. Water condenses at a temperature of 1) 0°C.2) 50°C.3) 100°C. B. At a temperature of 0°C, liquid water 1) freezes.2) melts.3) changes to a gas. C. At 40°C, water is a 1) solid. 2) liquid.3) gas. D. When water freezes, heat is 1) removed.2) added. Learning Check

77. 76 Use the cooling curve for water to answer each. A. Water condenses at a temperature of 3) 100°C. B. At a temperature of 0°C, liquid water 1) freezes. C. At 40 °C, water is a 2) liquid. D. When water freezes, heat is 1) removed. Solution

78. Definitions 1. Vaporization – a direct change of state from a liquid to a gas 2. Condensation – a direct change of state from a gas to a liquid 3. Freezing – a direct change of state form a liquid to a solid 4. Melting/Fusion – a direct change of state from solid to liquid 5. Sublimation – a direct change of state from a solid to a gas 6. Deposition – a direct change of state from a gas to a solid

79. Definitions (restated) Evaporation/Vaporization - process by which molecules escape from the liquid. to gas phase. Evaporation - vaporization below boiling point, which occurs at any temp. when the surface is exposed Evaporation Condensation is the process by which molecules go from the gas into the liquid phase. Freezing is the process by which molecules go from the liquid to solid phase. Melting or Fusion is the process by which molecules go from the solid to liquid phase.

80. Definitions Normal melting/freezing point Freezing pt. - temperature at which a liquid becomes a solid at 1 atm. Melting pt. - temperature at which a solid becomes a liquid at 1 atm. Normal boiling point The temp. at which a liquid boils under 1 atm.

81. Enthalpy of Vaporization (  H vap ) - the enthalpy change that accompanies conversion of one mole of a substance from liquid to gas at constant temperature.  H vap is 40.7 kJ/mol (water). Enthalpy of fusion (  H fus ) - enthalpy change that accompanies the change of one mole of solid into liquid at constant temperature or… Definitions

82. Phase Changes… Heating Curves-- Plot of temperature change versus heat added is a heating curve. Plateaus occur wherever a change of state occurs. The first plateau represents melting (fusion) and the second plateau boiling (vaporization).

83. Phase Changes… Heating Curves-- Plot of temperature change versus heat added is a heating curve. Plateaus occur wherever a change of state occurs. The first plateau represents melting (fusion) and the second plateau boiling (vaporization).

84. Phase Change As you add heat to a solid, its temperature increases until you hit the melting point. While the solid is melting, heat is absorbed at constant temperature. The added heat gives the molecules enough energy to break their bonds with their neighbors.

85. 84 To reduce a fever, an infant is packed in 250. g of ice. If the ice (at 0°C) melts and warms to body temperature (37.0°C), how many calories are removed from the body? Step 1: Diagram the changes 37°C  T = 37.0°C - 0°C = 37.0°C temperature increase 0°C solid liquid melting Combined Heat Calculations

86. 85 Combined Heat Calculations (continued.) Step 2: Calculate the heat to melt ice (fusion) 250. g ice x 80. cal= 20 000 cal 1 g ice Step 3: Calculate the heat to warm the water from 0°C to 37.0°C 250. g x 37.0°C x 1.00 cal = 9 250 cal g °C Total: Step 2 + Step 3 = 29 000 cal (rounded to 2 SF)

87. Gases Liquids Solids Condensation Hº vap = - 40.7 kJ/mol Vaporization Hº vap = 40.7 kJ/mol Freezing Hº fus = - 6.02 kJ/mol Melting Hº fus = 6.02 kJ/mol SublimationDeposition Sublimation - Hº sub Hº sub

88. 50 25 0 100 75 Heat Added Temperature ( °C) Ice Ice - ( heat added to weaken intermolecular forces between ice molecules); Q = mC ice (T f – T i ); C ice = 2.06 J/g °C Liquid - Liquid - (heat added to weaken intermolecular forces between liquid molecules); Q = mC water (T f – T i ) ; C water = 4.18 J/g °C Gas Gas - (heat added to weaken intermolecular forces between gas molecules) Q = mC steam (T f – T i ); C steam = 2.03 J/g °C Liquid and Gas Liquid and Gas - (vaporization—heat added is used to convert liquid to a gas, heat of vaporization); ΔH vap = 40.6 kJ/mol Ice and Liquid Ice and Liquid (melting—heat added is used to convert ice to liquid, heat of fusion); ΔH fus = 6.02 kJ/mol200-50

89. 50 25 0 100 75 Heat Added Temperature ( °C) Q = mC ice (T f – T i ); C ice = 2.06 J/g °C Q = mC water (T f – T i ) C water = 4.18 J/g °C Q = mC steam (T f – T i ) C water = 2.03 J/g °C ΔH vap = 40.6 kJ/mol ΔH vap = 6.02 kJ/mol200-50

90. 50 25 0 75 Heat Added Temperature ( °C) Q = m steam (T f – T i ); C steam = 2.03 J/g °C Calculate the total amount of heat given off when 10 g of steam is cooled from 200ºC to 100º C. Q = mC vapor (T f – T i ) = (10g)(2.03J/gºC)(100ºC – 200ºC) = -2030 J = -2.03 kJ200-50

91. 50 25 0 75 Heat Added Temperature ( °C) Calculate the total amount of heat required to condense 10 g of water at 100ºC. Q = 10 g x 1mol H 2 O x -40.6 kJ 18.02 g mol = -22.6 kJ ΔH vap = 40.6 kJ/mol200-50

92. 50 25 0 75 Heat Added Temperature ( °C) Q = m water (T f – T i ); C water = 4.18 J/g °C Calculate the total amount of heat given off when 10 g of water is cooled from 100C to 0º C. Q = mC water (T f – T i ) = (10g)(4.18J/gºC)(0ºC – 100ºC) = -4180 J = -4.18 kJ200-50

93. 12550 25 0 75 Heat Added Temperature ( °C) Q = 10 g x 1mol H 2 O x -6.02 kJ 18.02 g mol = -3.34 kJ ΔH fus = 6.02 kJ/mol200-50 Calculate the total amount of heat required to freeze 10 g of ice at 0ºC.

94. 50 25 0 75 Heat Added Temperature ( °C) Q = mC ice (T f – T i ); C ice = 2.06 J/g °C Q = mC ice (T f – T i ) = (10g)(2.06J/gºC)(-50ºC – 0ºC) = -1030 J = -1.03 kJ200-50 Calculate the total amount of heat given off when 10 g of ice is cooled from 0ºC to -50º C.

95. 50 25 0 75 Heat Added Temperature ( °C) Calculate the total quantity of heat evolved when 10 g of steam at 200ºC is condensed, cooled, and frozen to ice at -50.ºC. -22.6kJ -1.03kJ -4.18kJ -2.03kJ -3.34kJ Q = 2.03 + 22.6 + 4.18 + 3.34 + 1.03 Q = 33.1 kJ of heat evolved or -33.1 kJ200-50

96. 50 25 0 75 Heat Added Temperature ( °C) How much energy would it take to melt the ice at 0ºC? -22.6kJ -1.03kJ -4.18kJ -2.03kJ -3.34kJ200-50

97. 50 25 0 75 Heat Added Temperature ( °C) How much energy would it take cool ice from 0ºC to -50ºC? -22.6kJ -1.03kJ -4.18kJ -2.03kJ -3.34kJ200-50

98. 50 25 0 75 Heat Added Temperature ( °C) How much energy would it take to cool water from 100ºC to 0ºC? -22.6kJ -1.03kJ -4.18kJ -2.03kJ -3.34kJ200-50

99. 12550 25 0 -25 75 Heat Added Temperature ( °C) Calculate the total quantity of heat evolved when 10 g of steam at 200ºC is condensed, cooled, and frozen to ice at - 50.ºC. ΔH vap = 40.6 kJ/mol Q = mC ice (T f – T i ); C ice = 2.06 J/g °C Q = mC water (T f – T i ) C water = 4.18 J/g °C Q = mC steam (T f – T i ) C water = 2.03 J/g °C ΔH vap = 6.02 kJ/mol

100. 99 Homework due next Wednesday,1/25 Chapter 2, pages 43-67 Questions and problems (1-53, odd numbers only) Study key terms (page 66), do not need to turn in definitions just know the terms.