1. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu To View the presentation as a slideshow with effects select “View” on the menu bar and click on “Slide Show.” To advance through the presentation, click the right-arrow key or the space bar. From the resources slide, click on any resource to see a presentation for that resource. From the Chapter menu screen click on any lesson to go directly to that lesson’s presentation. You may exit the slide show at any time by pressing the Esc key. How to Use This Presentation

2. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter Presentation Transparencies Sample Problems Visual Concepts Standardized Test Prep Resources

3. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Heat Chapter 9 Table of Contents Section 1 Temperature and Thermal Equilibrium Section 2 Defining Heat Section 3 Changes in Temperature and Phase

4. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 1 Temperature and Thermal Equilibrium Chapter 9 Objectives Relate temperature to the kinetic energy of atoms and molecules. Describe the changes in the temperatures of two objects reaching thermal equilibrium. Identify the various temperature scales, and convert from one scale to another.

5. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 9 Defining Temperature Temperature is a measure of the average kinetic energy of the particles in a substance. Adding or removing energy usually changes temperature. Internal energy is the energy of a substance due to both the random motions of its particles and to the potential energy that results from the distances and alignments between the particles. Section 1 Temperature and Thermal Equilibrium

6. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 9 Forms of Internal Energy Section 1 Temperature and Thermal Equilibrium

7. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 9 Thermal Equilibrium Thermal equilibrium is the state in which two bodies in physical contact with each other have identical temperatures. –By placing a thermometer in contact with an object and waiting until the column of liquid in the thermometer stops rising or falling, you can find the temperature of the object. –The reason is that the thermometer is in thermal equilibrium with the object. The temperature of any two objects in thermal equilibrium always lies between their initial temperatures. Section 1 Temperature and Thermal Equilibrium

8. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 9 Thermal Equilibrium Section 1 Temperature and Thermal Equilibrium

9. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 9 Thermal Expansion In general, if the temperature of a substance increases, so does its volume. This phenomenon is known as thermal expansion. Different substances undergo different amounts of expansion for a given temperature change. The thermal expansion characteristics of a material are indicated by a quantity called the coefficient of volume expansion. Gases have the largest values for this coefficient. Solids typically have the smallest values. Section 1 Temperature and Thermal Equilibrium

10. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 9 Thermal Expansion Section 1 Temperature and Thermal Equilibrium

11. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 9 Measuring Temperature The most common thermometers use a glass tube containing a thin column of mercury, colored alcohol, or colored mineral spirits. When the thermometer is heated, the volume of the liquid expands. The change in length of the liquid column is proportional to the temperature. Section 1 Temperature and Thermal Equilibrium

12. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 9 Measuring Temperature, continued When a thermometer is in thermal equilibrium with a mixture of water and ice at one atmosphere of pressure, the temperature is called the ice point or melting point of water. This is defined as zero degrees Celsius, or 0°C. When the thermometer is in thermal equilibrium with a mixture of steam and water at one atmosphere of pressure, the temperature is called the steam point or boiling point of water. This is defined as 100°C. Section 1 Temperature and Thermal Equilibrium

13. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 9 Measuring Temperature, continued The temperature scales most widely used today are the Fahrenheit, Celsius, and Kelvin scales. Celsius and Fahrenheit temperature measurements can be converted to each other using this equation: Section 1 Temperature and Thermal Equilibrium The number 32.0 indicates the difference between the ice point value in each scale: 0.0ºC and 32.0ºF.

14. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 9 Measuring Temperature, continued Section 1 Temperature and Thermal Equilibrium Temperature values in the Celsius and Fahrenheit scales can have positive, negative, or zero values. But because the kinetic energy of the atoms in a substance must be positive, the absolute temperature that is proportional to that energy should be positive also. A temperature scale with only positive values is suggested by the graph on the next slide. This scale is called the Kelvin scale.

15. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 9 Measuring Temperature, continued Section 1 Temperature and Thermal Equilibrium The graph suggests that if the temperature could be lowered to –273.15°C, the pressure would be zero. This temperature is designated in the Kelvin scale as 0.00 K, where K represents the temperature unit called the kelvin. Temperatures in the Kelvin scale are indicated by the symbol T.

16. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 9 Measuring Temperature, continued Section 1 Temperature and Thermal Equilibrium A temperature difference of one degree is the same on the Celsius and Kelvin scales. The two scales differ only in the choice of zero point. Thus, the ice point (0.00°C) equals 273.15 K, and the steam point (100.00°C) equals 373.15 K. The Celsius temperature can therefore be converted to the Kelvin temperature by adding 273.15:

17. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 9 Temperature Scales and Their Uses Section 1 Temperature and Thermal Equilibrium

18. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 2 Defining Heat Chapter 9 Objectives Explain heat as the energy transferred between substances that are at different temperatures. Relate heat and temperature change on the macroscopic level to particle motion on the microscopic level. Apply the principle of energy conservation to calculate changes in potential, kinetic, and internal energy.

19. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 2 Defining Heat Chapter 9 Heat and Energy Heat is the energy transferred between objects because of a difference in their temperatures. From a macroscopic viewpoint, energy transferred as heat tends to move from an object at higher temperature to an object at lower temperature. The direction in which energy travels as heat can be explained at the atomic level, as shown on the next slide.

20. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 9 Transfer of Particles’ Kinetic Energy as Heat Section 2 Defining Heat Energy is transferred as heat from the higher-energy particles to the lower-energy particles, as shown on the left. The net energy transferred is zero when thermal equilibrium is reached, as shown on the right.

21. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 9 Temperature and Heat Section 2 Defining Heat

22. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 2 Defining Heat Chapter 9 Heat and Energy, continued The atoms of all objects are in continuous motion, so all objects have some internal energy. –Because temperature is a measure of that energy, all objects have some temperature. Heat, on the other hand, is the energy transferred from one object to another because of the temperature difference between them. –When there is no temperature difference between a substance and its surroundings, no net energy is transferred as heat.

23. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 2 Defining Heat Chapter 9 Heat and Energy, continued Just as other forms of energy have a symbol that identifies them (PE for potential energy, KE for kinetic energy, U for internal energy, W for work), heat is indicated by the symbol Q. Because heat, like work, is energy in transit, all heat units can be converted to joules, the SI unit for energy.

24. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 9 Thermal Units and Their Values in Joules Section 2 Defining Heat

25. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 2 Defining Heat Chapter 9 Thermal Conduction The type of energy transfer that is due to atoms transferring vibrations to neighboring atoms is called thermal conduction. The rate of thermal conduction depends on the substance. Two other mechanisms for transferring energy as heat are convection and electromagnetic radiation. When this burner is turned on, the skillet’s handle heats up because of conduction.

26. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 9 Convection, Conduction, and Radiation Section 2 Defining Heat

27. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 2 Defining Heat Chapter 9 Conservation of Energy If changes in internal energy are taken into account along with changes in mechanical energy, the total energy is a universally conserved property. In other words, the sum of the changes in potential, kinetic, and internal energy is equal to zero. CONSERVATION OF ENERGY  PE +  KE +  U = 0 the change in potential energy + the change in kinetic energy + the change in internal energy = 0

28. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 9 Conservation of Energy Section 2 Defining Heat

29. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 2 Defining Heat Chapter 9 Sample Problem Conservation of Energy An arrangement similar to the one used to demonstrate energy conservation is shown in the figure. A vessel contains water. Paddles that are propelled by falling masses turn in the water. This agitation warms the water and increases its internal energy. The temperature of the water is then measured, giving an indication of the water’s internal energy increase.

30. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 2 Defining Heat Chapter 9 Sample Problem, continued Conservation of Energy, continued If a total mass of 11.5 kg falls 1.3 m and all of the mechanical energy is converted to internal energy, by how much will the internal energy of the water increase? (Assume no energy is transferred as heat out of the vessel to the surroundings or from the surroundings to the vessel’s interior.)

31. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 2 Defining Heat Chapter 9 Sample Problem, continued 1. Define Given: m = 11.5 kg h = 1.3 m g = 9.81 m/s 2 Unknown:  U = ?

32. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 2 Defining Heat Chapter 9 Sample Problem, continued 2. Plan Choose an equation or situation: Use the conservation of energy, and solve for  U.  PE +  KE +  U = 0 (PE f – PE i ) + (KE f – KE i ) +  U = 0  U = –PE f + PE i – KE f + KE i Tip: Don’t forget that a change in any quantity, indicated by the symbol ∆, equals the final value minus the initial value.

33. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 2 Defining Heat Chapter 9 Sample Problem, continued Because the masses begin at rest, KE i equals zero. If we assume that KE f is small compared to the loss of PE, we can set KE f equal to zero also. KE f = 0 KE i = 0 Because all of the potential energy is assumed to be converted to internal energy, PE i can be set equal to mgh if PE f is set equal to zero. PE i = mgh PE f = 0 Substitute each quantity into the equation for ∆U: ∆U = –PE f + PE i – KE f + KE i ∆U = 0 + mgh + 0 + 0 = mgh

34. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 2 Defining Heat Chapter 9 Sample Problem, continued 4. Evaluate The answer can be estimated using rounded values. If m ≈ 10 kg and g ≈ 10 m/s 2, then ∆U ≈ 130 J, which is close to the actual value calculated. 3. Calculate Substitute the values into the equation and solve:  U = mgh  U = (11.5 kg)(9.81 m/s 2 )(1.3 m)  U = 1.5  10 2 J

35. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 3 Changes in Temperature and Phase Chapter 9 Objectives Perform calculations with specific heat capacity. Interpret the various sections of a heating curve.

36. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 3 Changes in Temperature and Phase Chapter 9 Specific Heat Capacity The specific heat capacity of a substance is defined as the energy required to change the temperature of 1 kg of that substance by 1°C. Every substance has a unique specific heat capacity. This value tells you how much the temperature of a given mass of that substance will increase or decrease, based on how much energy is added or removed as heat.

37. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 3 Changes in Temperature and Phase Chapter 9 Specific Heat Capacity, continued Specific heat capacity is expressed mathematically as follows: The subscript p indicates that the specific heat capacity is measured at constant pressure. In this equation,  T can be in degrees Celsius or in degrees Kelvin.

38. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 9 Specific Heat Capacities Section 3 Changes in Temperature and Phase

39. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 3 Changes in Temperature and Phase Chapter 9 Calorimetry Calorimetry is used to determine specific heat capacity. Calorimetry is an experimental procedure used to measure the energy transferred from one substance to another as heat. A simple calorimeter allows the specific heat capacity of a substance to be determined.

40. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 9 Calorimetry Section 3 Changes in Temperature and Phase

41. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 3 Changes in Temperature and Phase Chapter 9 Calorimetry, continued Because the specific heat capacity of water is well known (c p,w = 4.186 kJ/kg°C), the energy transferred as heat between an object of unknown specific heat capacity and a known quantity of water can be measured. energy absorbed by water = energy released by substance Q w = –Q x c p,w m w ∆T w = –c p,x m x ∆T x

42. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 3 Changes in Temperature and Phase Chapter 9 Sample Problem Calorimetry A 0.050 kg metal bolt is heated to an unknown initial temperature. It is then dropped into a calorimeter containing 0.15 kg of water with an initial temperature of 21.0°C. The bolt and the water then reach a final temperature of 25.0°C. If the metal has a specific heat capacity of 899 J/kg°C, find the initial temperature of the metal.

43. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 3 Changes in Temperature and Phase Chapter 9 Sample Problem, continued 1. Define Given: m m = 0.050 kg c p,m = 899 J/kg°C m w = 0.15 kg c p,w = 4186 J/kg°C T w = 21.0°C T f = 25.0°C Unknown: T m = ? Diagram:

44. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 3 Changes in Temperature and Phase Chapter 9 Sample Problem, continued 2. Plan Choose an equation or situation: The energy absorbed by the water equals the energy removed from the bolt. Rearrange the equation to isolate the unknown:

45. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 3 Changes in Temperature and Phase Chapter 9 Sample Problem, continued 3. Calculate Substitute the values into the equation and solve: 4. Evaluate T m is greater than T f, as expected. Tip: Because T w is less than T f, you know that T m must be greater than T f.

46. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 3 Changes in Temperature and Phase Chapter 9 Latent Heat When substances melt, freeze, boil, condense, or sublime, the energy added or removed changes the internal energy of the substance without changing the substance’s temperature. These changes in matter are called phase changes. The energy per unit mass that is added or removed during a phase change is called latent heat, abbreviated as L. Q = mL energy transferred as heat during phase change = mass  latent heat

47. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 9 Latent Heat Section 3 Changes in Temperature and Phase

48. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 3 Changes in Temperature and Phase Chapter 9 Latent Heat, continued During melting, the energy that is added to a substance equals the difference between the total potential energies for particles in the solid and the liquid phases. This type of latent heat is called the heat of fusion, abbreviated as L f. During vaporization, the energy that is added to a substance equals the difference in the potential energy of attraction between the liquid particles and between the gas particles. In this case, the latent heat is called the heat of vaporization, abbreviated as L v.

49. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Multiple Choice 1.What must be true about two given objects for energy to be transferred as heat between them? A. The objects must be large. B. The objects must be hot. C. The objects must contain a large amount of energy. D. The objects must have different temperatures. Standardized Test Prep Chapter 9

50. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Multiple Choice 1.What must be true about two given objects for energy to be transferred as heat between them? A. The objects must be large. B. The objects must be hot. C. The objects must contain a large amount of energy. D. The objects must have different temperatures. Standardized Test Prep Chapter 9

51. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Multiple Choice, continued 2.A metal spoon is placed in one of two identical cups of hot coffee. Why does the cup with the spoon have a lower temperature after a few minutes? F. Energy is removed from the coffee mostly by conduction through the spoon. G. Energy is removed from the coffee mostly by convection through the spoon. H. Energy is removed from the coffee mostly by radiation through the spoon. J. The metal in the spoon has an extremely large specific heat capacity. Standardized Test Prep Chapter 9

52. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Multiple Choice, continued 2.A metal spoon is placed in one of two identical cups of hot coffee. Why does the cup with the spoon have a lower temperature after a few minutes? F. Energy is removed from the coffee mostly by conduction through the spoon. G. Energy is removed from the coffee mostly by convection through the spoon. H. Energy is removed from the coffee mostly by radiation through the spoon. J. The metal in the spoon has an extremely large specific heat capacity. Standardized Test Prep Chapter 9

53. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Multiple Choice, continued Use the passage below to answer questions 3–4. The boiling point of liquid hydrogen is –252.87°C. 3.What is the value of this temperature on the Fahrenheit scale? A. 20.28°F B. –220.87°F C. –423.2°F D. 0°F Standardized Test Prep Chapter 9

54. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Multiple Choice, continued Use the passage below to answer questions 3–4. The boiling point of liquid hydrogen is –252.87°C. 3.What is the value of this temperature on the Fahrenheit scale? A. 20.28°F B. –220.87°F C. –423.2°F D. 0°F Standardized Test Prep Chapter 9

55. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Multiple Choice, continued Use the passage below to answer questions 3–4. The boiling point of liquid hydrogen is –252.87°C. 4.What is the value of this temperature in kelvins? F. 273 K G. 20.28 K H. –423.2 K J. 0 K Standardized Test Prep Chapter 9

56. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Multiple Choice, continued Use the passage below to answer questions 3–4. The boiling point of liquid hydrogen is –252.87°C. 4.What is the value of this temperature in kelvins? F. 273 K G. 20.28 K H. –423.2 K J. 0 K Standardized Test Prep Chapter 9

57. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Multiple Choice, continued 5. A cup of hot chocolate with a temperature of 40°C is placed inside a refrigerator at 5°C. An identical cup of hot chocolate at 90°C is placed on a table in a room at 25°C. A third identical cup of hot chocolate at 80°C is placed on an outdoor table, where the surrounding air has a temperature of 0°C. For which of the three cups has the most energy been transferred as heat when equilibrium has been reached? A. The first cup has the largest energy transfer. B. The second cup has the largest energy transfer. C. The third cup has the largest energy transfer. D. The same amount of energy is transferred as heat for all three cups. Standardized Test Prep Chapter 9

58. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Multiple Choice, continued 5. A cup of hot chocolate with a temperature of 40°C is placed inside a refrigerator at 5°C. An identical cup of hot chocolate at 90°C is placed on a table in a room at 25°C. A third identical cup of hot chocolate at 80°C is placed on an outdoor table, where the surrounding air has a temperature of 0°C. For which of the three cups has the most energy been transferred as heat when equilibrium has been reached? A. The first cup has the largest energy transfer. B. The second cup has the largest energy transfer. C. The third cup has the largest energy transfer. D. The same amount of energy is transferred as heat for all three cups. Standardized Test Prep Chapter 9

59. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Multiple Choice, continued 6.What data are required in order to determine the specific heat capacity of an unknown substance by means of calorimetry? F. c p,water, T water, T substance, T final, V water, V substance G. c p,substance, T water, T substance, T final, m water, m substance H. c p,water, T substance, m water, m substance J. c p,water, T water, Ts ubstance, T final, m water, m substance Standardized Test Prep Chapter 9

60. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Multiple Choice, continued 6.What data are required in order to determine the specific heat capacity of an unknown substance by means of calorimetry? F. c p,water, T water, T substance, T final, V water, V substance G. c p,substance, T water, T substance, T final, m water, m substance H. c p,water, T substance, m water, m substance J. c p,water, T water, Ts ubstance, T final, m water, m substance Standardized Test Prep Chapter 9

61. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Multiple Choice, continued 7.During a cold spell, Florida orange growers often spray a mist of water over their trees during the night. Why is this done? A. The large latent heat of vaporization for water keeps the trees from freezing. B. The large latent heat of fusion for water prevents it and thus the trees from freezing. C. The small latent heat of fusion for water prevents the water and thus the trees from freezing. D. The small heat capacity of water makes the water a good insulator. Standardized Test Prep Chapter 9

62. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Multiple Choice, continued 7.During a cold spell, Florida orange growers often spray a mist of water over their trees during the night. Why is this done? A. The large latent heat of vaporization for water keeps the trees from freezing. B. The large latent heat of fusion for water prevents it and thus the trees from freezing. C. The small latent heat of fusion for water prevents the water and thus the trees from freezing. D. The small heat capacity of water makes the water a good insulator. Standardized Test Prep Chapter 9

63. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Multiple Choice, continued Use the heating curve to answer questions 8–10. The graph shows the change in temperature of a 23 g sample as energy is added to the sample as heat. Standardized Test Prep Chapter 9 8. What is the specific heat capacity of the liquid? F. 4.4  10 5 J/kg°C G. 4.0  10 2 J/kg°C H. 5.0  10 2 J/kg°C J. 1.1  10 3 J/kg°C

64. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Multiple Choice, continued Use the heating curve to answer questions 8–10. The graph shows the change in temperature of a 23 g sample as energy is added to the sample as heat. Standardized Test Prep Chapter 9 8. What is the specific heat capacity of the liquid? F. 4.4  10 5 J/kg°C G. 4.0  10 2 J/kg°C H. 5.0  10 2 J/kg°C J. 1.1  10 3 J/kg°C

65. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Multiple Choice, continued Use the heating curve to answer questions 8–10. The graph shows the change in temperature of a 23 g sample as energy is added to the sample as heat. Standardized Test Prep Chapter 9 9. What is the latent heat of fusion? A. 4.4  10 5 J/kg B. 4.0  10 2 J/kg°C C. 10.15  10 3 J D. 3.6  10 7 J/kg

66. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Multiple Choice, continued Use the heating curve to answer questions 8–10. The graph shows the change in temperature of a 23 g sample as energy is added to the sample as heat. Standardized Test Prep Chapter 9 9. What is the latent heat of fusion? A. 4.4  10 5 J/kg B. 4.0  10 2 J/kg°C C. 10.15  10 3 J D. 3.6  10 7 J/kg

67. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Multiple Choice, continued Use the heating curve to answer questions 8–10. The graph shows the change in temperature of a 23 g sample as energy is added to the sample as heat. Standardized Test Prep Chapter 9 10. What is the specific heat capacity of the solid? F. 1.85  10 3 J/kg°C G. 4.0  10 2 J/kg°C H. 5.0  10 2 J/kg°C J. 1.1  10 3 J/kg°C

68. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Multiple Choice, continued Use the heating curve to answer questions 8–10. The graph shows the change in temperature of a 23 g sample as energy is added to the sample as heat. Standardized Test Prep Chapter 9 10. What is the specific heat capacity of the solid? F. 1.85  10 3 J/kg°C G. 4.0  10 2 J/kg°C H. 5.0  10 2 J/kg°C J. 1.1  10 3 J/kg°C

69. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Short Response Base your answers to questions 11–12 on the information below. The largest of the Great Lakes, Lake Superior, contains 1.20  10 16 kg of fresh water, which has a specific heat capacity of 4186 J/kg°C and a latent heat of fusion of 3.33  10 5 J/kg. 11. How much energy would be needed to increase the temperature of Lake Superior by 1.0°C? Standardized Test Prep Chapter 9

70. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Short Response Base your answers to questions 11–12 on the information below. The largest of the Great Lakes, Lake Superior, contains 1.20  10 16 kg of fresh water, which has a specific heat capacity of 4186 J/kg°C and a latent heat of fusion of 3.33  10 5 J/kg. 11. How much energy would be needed to increase the temperature of Lake Superior by 1.0°C? Answer: 5.0  10 19 J Standardized Test Prep Chapter 9

71. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Short Response, continued Base your answers to questions 11–12 on the information below. The largest of the Great Lakes, Lake Superior, contains 1.20  10 16 kg of fresh water, which has a specific heat capacity of 4186 J/kg°C and a latent heat of fusion of 3.33  10 5 J/kg. 12. If Lake Superior were still liquid at 0°C, how much energy would need to be removed from the lake for it to become completely frozen? Standardized Test Prep Chapter 9

72. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Short Response, continued Base your answers to questions 11–12 on the information below. The largest of the Great Lakes, Lake Superior, contains 1.20  10 16 kg of fresh water, which has a specific heat capacity of 4186 J/kg°C and a latent heat of fusion of 3.33  10 5 J/kg. 12. If Lake Superior were still liquid at 0°C, how much energy would need to be removed from the lake for it to become completely frozen? Answer: 5.00  10 21 J Standardized Test Prep Chapter 9

73. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Short Response, continued Standardized Test Prep Chapter 9 13. Ethyl alcohol has about one-half the specific heat capacity of water. If equal masses of alcohol and water in separate beakers at the same temperature are supplied with the same amount of energy, which will have the higher final temperature?

74. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Short Response, continued 13. Ethyl alcohol has about one-half the specific heat capacity of water. If equal masses of alcohol and water in separate beakers at the same temperature are supplied with the same amount of energy, which will have the higher final temperature? Standardized Test Prep Chapter 9 Answer: the ethyl alcohol

75. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Short Response, continued 14. A 0.200 kg glass holds 0.300 kg of hot water, as shown in the figure. The glass and water are set on a table to cool. After the temperature has decreased by 2.0°C, how much energy has been removed from the water and glass? Standardized Test Prep Chapter 9 (The specific heat capacity of glass is 837 J/kg°C, and that of water is 4186 J/kg°C.)

76. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Short Response, continued 14. A 0.200 kg glass holds 0.300 kg of hot water, as shown in the figure. The glass and water are set on a table to cool. After the temperature has decreased by 2.0°C, how much energy has been removed from the water and glass? Standardized Test Prep Chapter 9 (The specific heat capacity of glass is 837 J/kg°C, and that of water is 4186 J/kg°C.) Answer: 2900 J

77. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Extended Response 15. How is thermal energy transferred by the process of convection? Standardized Test Prep Chapter 9

78. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Extended Response 15. How is thermal energy transferred by the process of convection? Answer: The increasing temperature of a liquid or gas causes it to become less dense, so it rises above colder liquid or gas, transferring thermal energy with it. Standardized Test Prep Chapter 9

79. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Extended Response, continued 16. Show that the temperature –40.0° is unique in that it has the same numerical value on the Celsius and Fahrenheit scales. Show all of your work. Standardized Test Prep Chapter 9

80. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Extended Response, continued 16. Show that the temperature –40.0° is unique in that it has the same numerical value on the Celsius and Fahrenheit scales. Show all of your work. Answer: Standardized Test Prep Chapter 9

81. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 9 Measuring Temperature Section 1 Temperature and Thermal Equilibrium

82. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 9 Determining Absolute Zero for an Ideal Gas Section 1 Temperature and Thermal Equilibrium

83. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 3 Changes in Temperature and Phase Chapter 9 Calorimetry