2. Thermochemistry: Energy and Heat

3. The Nature of Energy u Energy is the ability to do work or produce heat. u It exists in two basic forms, potential energy and kinetic energy.

4. The Nature of Energy u Potential energy is energy due to the composition or position of an object.

5. The Nature of Energy u Kinetic energy is energy of motion. u The potential energy of the dammed water is converted to kinetic energy as the dam gates are opened and the water flows out.

6. The Nature of Energy u Chemical systems contain both kinetic energy and potential energy. u As temperature increases, the motion of particles increases, so its average kinetic energy increases.

7. The Nature of Energy u The potential energy of a substance depends upon its composition: the type of atoms in the substance, the number and type of chemical bonds joining the atoms, and the particular way the atoms are arranged.

8. Law of Conservation of Energy u The law of conservation of energy states that in any chemical reaction or physical process, energy can be converted from one form to another, but it is neither created nor destroyed.

9. Law of Conservation of Energy u Heat, which is represented by the symbol q, is energy that is in the process of flowing from a warmer object to a cooler object. u The SI unit of heat and energy is the joule (J).

10. Heat involves a transfer of energy between 2 objects due to a temperature difference. Heat

11. Heat flows from “hot to cold.”

12. Law of Conservation of Energy u When the warmer object loses heat, its temperature decreases and q is negative. u When the cooler object absorbs heat, its temperature rises and q is positive.

13. Specific Heat u The specific heat of any substance is the amount of heat required to raise the temperature of one gram of that substance by one degree Celsius. u Because different substances have different compositions, each substance has its own specific heat.

14. q = m C ΔT Q = heat (J) C p = specific heat (J/(g. ° C) m = mass (g) Δ T = change in temperature = T f – T i ( ° C) Specific Heat

15. Exothermic: Heat flows out of the system (to the surroundings). The value of ‘q’ is negative. Endothermic: Heat flows into the system (from the surroundings). The value of ‘q’ is positive. Exothermic and Endothermic

16. Example u The temperature of a sample of iron with a mass of 10.0 g changed from 50.4°C to 25.0°C with the release of 114 J heat. What is the specific heat of iron? q =∆TC iron m 10.0114 = (50.4 – 25.0 ) C iron = 0.449 J/g°C

17. Problem u A piece of metal absorbs 256 J of heat when its temperature increases by 182°C. If the specific heat of the metal is 0.301 J/g°C, determine the mass of the metal. m = 4.67 g

18. Problem u If 335 g water at 65.5°C loses 9750 J of heat, what is the final temperature of the water? The specific heat of water is 4.18 J/g°C T f = 58.5 °C

19. Problem  As 335 g of aluminum at 65.5°C gains heat, its final temperature is 300. ° C. The specific heat of aluminum is 0.897 J/g°C. Determine the energy gained by the aluminum. q = 7.05 x 10 4 J

20. Measuring Heat u Heat changes that occur during chemical and physical processes can be measured accurately and precisely using a calorimeter. u A calorimeter is an insulated device used for measuring the amount of heat absorbed or released during a chemical or physical process.

21. A coffee-cup calorimeter made of two Styrofoam cups.

22. Example u Suppose you put 125 g of water into a foam-cup calorimeter and find that its initial temperature is 25.6°C.

23. Example, cont. u Then, you heat a 50.0 g sample of the unknown metal to a temperature of 115.0°C and put the metal sample into the water. Both water and metal have attained a final temperature of 29.3°C.

24. Example, cont. u Heat flows from the hot metal to the cooler water and the temperature of the water rises. u The flow of heat stops only when the temperature of the metal and the water is equal.

25. Example, cont. u Assuming no heat is lost to the surroundings, the heat gained by the water is equal to the heat lost by the metal. u Determine the specific heat of the metal.

26. Example, cont. u Suppose you put 125 g of water into a foam- cup calorimeter and find that its initial temperature is 25.6°C. Both water and metal have attained a final temperature of 29.3°C. Find the heat gained by the water. q =∆TC water m 125(4.18)(29.3 – 25.6) q = 1933 J

27. Example, cont. u Then, you heat a 50.0 g sample of the unknown metal to a temperature of 115.0°C and put the metal sample into the water. Both water and metal have attained a final temperature of 29.3°C. Since heat lost equals heat gained, determine the specific heat of the metal. q =∆TC metal m 50.0-1933 =(29.3 – 115.0) C metal = 0.451 J/g°C

28. Problem u You put 352 g of water into a foam- cup calorimeter and find that its initial temperature is 22.0°C. What mass of 134°C lead, C lead = 0.129 J/g°C, can be placed in the water so that the equilibrium temperature is 26.5°C? m = 477 g

29. Problem u You put water into a foam-cup calorimeter and find that its initial temperature is 25.0°C. What is the mass of the water if 14.0 grams of 125°C nickel, C Ni = 0.444 J/g°C, can be placed in the water so that the equilibrium temperature is 27.5°C? m = 58.0 g

30. Phase Changes Review Solid Liquid Gas Melting Vaporization CondensationFreezing

31. Liquid Sublimation Melting Vaporization Deposition Condensation Solid Freezing Gas

33. q = m H v H v = latent heat of vaporization (J/g) H f = latent heat of fusion (J/g) q = m H f Latent Heat

34. Water and Ice Ice Water and Steam Steam -20 0 20 40 60 80 100 120 0 40120 220760800 Heating Curve for Water The heating curve has 5 distinct regions. 1 2 3 4 5

35. Water and Ice Ice Water and Steam Steam -20 0 20 40 60 80 100 120 0 40120 220760800 Heating Curve for Water The horizontal lines are where phase changes occur. melting vaporization

36. Water and Ice Ice Water and Steam Steam -20 0 20 40 60 80 100 120 0 40120 220760800 Heating Curve for Water Temperature is constant during a phase change!

37. Energy and Phase Change u Heat of vaporization (H v ) is the energy required to change one gram of a substance from liquid to gas.

38. Energy and Phase Change u Notice that during the heat of vaporization region denoted on the next slide, temperature is constant.

39. Water and Ice Ice Water and Steam Steam -20 0 20 40 60 80 100 120 0 40120 220760800 Heating Curve for Water Heat of Vaporization

40. Water and Ice Ice Water and Steam Steam -20 0 20 40 60 80 100 120 0 40120 220760800 Heating Curve for Water Both Water and Steam Exist

41. Energy and Phase Change u Heat of fusion (H f ) is the energy required to change one gram of a substance from solid to liquid.

42. Energy and Phase Change u Notice that during the heat of fusion region denoted on the next slide, temperature is constant.

43. Water and Ice Ice Water and Steam Steam -20 0 20 40 60 80 100 120 0 40120 220760800 Heating Curve for Water Heat of Fusion

44. Water and Ice Ice Water and Steam Steam -20 0 20 40 60 80 100 120 0 40120 220760800 Heating Curve for Water Both Ice and Water Exist

45. Example  How much heat does it take to melt 12.0 g of ice at 0  C? The heat of fusion for water is 334 J/g. q =HfHf m 12.0(334) q = 4010 J

46. Example  How much heat must be removed to condense 5.00 g of steam at 100  C? The heat of vaporization for water is 2260 J/g. q =HvHv m 5.00(2260) q = 11300 J

47. Heating Curve for Water Section A q = m C p q = m C p ΔT (solid) Section B q = mH f Section C q = m C p q = m C p ΔT (liquid) Section E q = m C p q = m C p ΔT (gas) Section D q = mH v

48. Calculating Energy u Three equations can be used in calculating energy.  q = m C p  T u q = m H f u q = m H v u C p = specific heat   T = T final - T initial

49. Solving Problems u The total heat = the sum of all the heats you have to use u Go in order Heat Ice Below 0  C + Melt Ice At 0  C + Heat Water 0  C - 100  C + Boil Water At 100  C + Heat Steam Above 100  C

50. Solving Problems Heat Ice q = m C ice ∆T Melt Ice q = m H f Heat Water q = m C water ∆T

51. Solving Problems Boil Water q = m H v Heat Steam q = m C steam ∆T

52. Numbers Needed For Energy Problems Involving Water  For ice, specific heat = 2.05 J/g  C  For water, specific heat = 4.18 J/g  C  For steam, specific heat = 2.02 J/g  C u Heat of vaporization = 2260 J/g u Heat of fusion = 334 J/g

53. Example  How much heat does it take to heat 12 g of ice at – 6  C to 25  C water? Round to a whole number. You begin at ice below 0 °C. Note: The final temperature for this process cannot exceed 0 °C. q =∆TC ice m 12(2.05)(0 - - 6) q = 148 J

54. Example, cont.  How much heat does it take to heat 12 g of ice at – 6  C to 25  C water? Round to a whole number. Since the temperature needs to rise to 25 °C, you must melt the ice next. q =HfHf m 12(334) q = 4008 J

55. Example, cont.  How much heat does it take to heat 12 g of ice at – 6  C to 25  C water? Round to a whole number. You now have water at 0 °C. The final temperature of the water should be 25 °C. q =∆TC water m 12(4.18)(25 - 0) q = 1254 J

56. Example, cont.  How much heat does it take to heat 12 g of ice at – 6  C to 25  C water? Round to a whole number. Finally, add all of the Q values together. q = 148 +4008 +1254 q = 5410 J

57. Another Example  How much heat does it take to heat 35 g of ice at 0  C to steam at 150  C? Round to a whole number. You begin with ice at 0 °C, so you should melt it first. q =HfHf m 35(334) q = 11690 J

58. Another Example, cont.  How much heat does it take to heat 35 g of ice at 0  C to steam at 150  C? Round to a whole number. You now have water at 0 °C. The final temperature of the steam is to be 150 °C. You must take the water to 100 °C before you can even convert it to steam. q =∆TC water m 35(4.18)(100 - 0) q = 14630 J

59. Another Example, cont.  How much heat does it take to heat 35 g of ice at 0  C to steam at 150  C? Round to a whole number. Now you have water at 100 °C. Convert this to steam. q =HvHv m 35(2260) q = 79100 J

60. Another Example, cont.  How much heat does it take to heat 35 g of ice at 0  C to steam at 150  C? Round to a whole number. You must now take the steam at 100 °C to 150 °C. q =∆TC ice m 35(2.02)(150 - 100) q = 3535 J

61. Another Example, cont.  How much heat does it take to heat 35 g of ice at 0  C to steam at 150  C? Round to a whole number. Finally, add all of the Q values together. q = 11690 +14630 +79100 + q = 108955 J 3535

62. Problem  How much heat does it take to convert 16.0 g of ice to water at 0  C? (5340 J)

63. Problem  How much heat does it take to heat 21.0 g of water at 12.0  C to water at 75.0  C? (5530 J)

64. Problem  How much heat does it take to heat 14.0 g of water at 12.0  C to steam at 122.0  C? (37400 J)