1. 1 2.1 Latent heat The power of tropical cyclones Introduction Latent heat Latent heat and particle motion Check-point 1 Specific latent heat Check-point 2 ? E 2.1 Latent heat E

2. 2 The power of tropical cyclones Tropical cyclones just like a travelling energy converter. Internal energy of water vapour Kinetic energy of the air

3. 3 2.1 Latent heat The power of tropical cyclones The energy change in one day is up to 5.2 x 10 19 J = 20 years’ supply of electricity of HK

4. 4 2.1 Latent heat Introduction Matter exists in 3 states:solid, liquid, gas watersteam fusion vaporization solidification condensation e.g. water at freezing pointat boiling point ice Change of state at melting pointat boiling point

5. 5 2.1 Latent heat Cooling curve Graph of temperature vs time? When a hot liquid is cooled down, its temperature drops. ?? time temperature solidliquidgas

6. 6 2.1 Latent heat Record the temperature of the molten octadecan-1-ol as it cools down. Cooling curve of octadecan-1-ol Experiment 2a Video

7. 7 2.1 Latent heat Cooling curve AB drops steadily — liquid cooling (temperature falling ) temperature /  C A B C D Here is a cooling curve of octadecan-1-ol.

8. 8 2.1 Latent heat temperature /  C A B C D BC is flat — liquid solidifying (temperature unchanged) Cooling curve

9. 9 2.1 Latent heat temperature /  C A B C D CD drops steadily — solid cooling to room temperature (temperature falling) Cooling curve

10. 10 2.1 Latent heat temperature /  C A B C D melting point: read from the flat part BC melting point Cooling curve

11. 11 2.1 Latent heat Cooling curve Simulation

12. 12 2.1 Latent heat When a substance is solidifying, it loses energy continuously but... its temperature remains unchanged The cooling curve shows: 1Latent heat

13. 13 2.1 Latent heat The energy given out/absorbed is called latent heat means ‘hidden’ During change of state: 1Latent heat

14. 14 2.1 Latent heat Ice-water mixture stays at 0  C until all the ice is melted. energy is absorbed temperature unchanged from air to change the ice to water This energy is called latent heat of fusion of ice. 1Latent heat

15. 15 2.1 Latent heat energy is absorbed temperature unchanged to change the water to stream This energy is called latent heat of vaporization of water. Energy supplied continuously to keep water boils… 1Latent heat

16. 16 2.1 Latent heat steam water ice condensation condensation releases latent heat of vaporization

17. 17 2.1 Latent heat steam water ice condensation vaporization absorbs latent heat of vaporization

18. 18 2.1 Latent heat condensationvaporization solidification releases latent heat of fusion steam water ice

19. 19 2.1 Latent heat condensation vaporization fusion solidification absorbs latent heat of fusion steam water ice

20. 20 2.1 Latent heat molecule Regular arrangement breaks up E 2Latent heat and particle motion strong attraction weak attraction

21. 21 2.1 Latent heat PE related to the forces of attraction between the particles Energy has to be supplied to oppose the attractive force of the particles. PE  solid  liquid or liquid  gas average potential energy  2Latent heat and particle motion E

22. 22 2.1 Latent heat The transfer of energy does not change the KE. Temperature does not change. latent heat = change in PE during change of state 2Latent heat and particle motion E Simulation

23. 23 2.1 Latent heat Check-point 1 Q1Describe the energy change of the water molecules at each step below In converting 1 kg of ice at 0  C to steam at 100  C. (use the words increase, decrease, and unchanged) E

24. 24 2.1 Latent heat Check-point 1 – Q1 Kinetic energy____________ Potential energy____________ Internal energy____________ unchanged increase E

25. 25 2.1 Latent heat Check-point 1 – Q1 Kinetic energy____________ Potential energy____________ Internal energy____________ unchanged increase E

26. 26 2.1 Latent heat Check-point 1 – Q1 Kinetic energy____________ Potential energy____________ Internal energy____________ unchanged increase E

27. 27 2.1 Latent heat 3Specific latent heat energy E 1 kg solid X1 kg liquid X E = latent heat for 1 kg of X = specific latent heat of X Specific = for 1 kg of a substance e.g. without temp. change

28. 28 2.1 Latent heat or E = ml symbol: l unit: J kg  1 Energy transferred to change the state of 1 kg of the substance without a change in temperature. 3Specific latent heat

29. 29 2.1 Latent heat l f = energy needed to change 1 kg of ice to water (without temperature change) aSpecific latent heat of fusion of ice (l f ) Find mass melted m and energy transferred E  l f = E/m = 3.34 x 10 5 J kg  1 Video

30. 30 2.1 Latent heat Measuring the specific latent heat of fusion of ice Experiment 2b Simulation Video

31. 31 2.1 Latent heat Experiment 2b experimental apparatus control apparatus Ice also melts at room temperature, so a control is needed. Measuring the specific latent heat of fusion of ice

32. 32 2.1 Latent heat For ice, l f = 3.34  10 5 J kg  1 ice (0  C)water (0  C) Measuring the specific latent heat of fusion of ice Experiment 2b

33. 33 2.1 Latent heat aSpecific latent heat of fusion of ice (l f )

34. 34 2.1 Latent heat l v = energy needed to change 1 kg of water to steam (without change of temperature) bSpecific latent heat of vaporization of water (l v ) Find mass boiled away m and energy transferred E  l v = E/m

35. 35 2.1 Latent heat Measuring the specific latent heat of vaporization of water Experiment 2c Video Simulation

36. 36 2.1 Latent heat For water, l v = 2.26  10 6 J kg  1 water (0  C) steam (0  C) Experiment 2c Measuring the specific latent heat of vaporization of water

37. 37 2.1 Latent heat bSpecific latent heat of vaporization of ice (l v )

38. 38 2.1 Latent heat bSpecific latent heat of vaporization of ice (l v )

39. 39 2.1 Latent heat 1Consider a cup of water...Consider a cup of water... 2When vapour condenses...When vapour condenses... 3-5Jimmy melts three materials...Jimmy melts three materials... 6Fill in the value of energy at…Fill in the value of energy at… Check-point 2

40. 40 2.1 Latent heat Check-point 2 – Q1 Consider a cup of water (mass m) being heated from 0  C until it starts to boil at 100  C. Is the student correct? (Yes/No) Since E = mc  T and E = ml v  mc  T = ml v  l v = c  T = 4200 × 100 = 420 kJ kg  1 Since E = mc  T and E = ml v  mc  T = ml v  l v = c  T = 4200 × 100 = 420 kJ kg  1

41. 41 2.1 Latent heat Check-point 2 – Q2 When vapour condenses, is the surrounding air warmed or cooled? (warmed/cooled)

42. 42 2.1 Latent heat Jimmy melts three materials X, Y and Z of equal mass at the same time and place. Temperature /  C X Y Z time / s 0 0.5 1.2 2.3 3.7 Check-point 2 – Q3

43. 43 2.1 Latent heat Which material(s) has/have the greatest melting point? ( X / Y / Z ) Temperature /  C X Y Z time / s 0 0.5 1.2 2.3 3.7 Check-point 2 – Q3

44. 44 2.1 Latent heat Which material(s) has/have the largest value of specific latent heat of fusion? ( X / Y / Z ) Check-point 2 – Q4 Temperature /  C C X Y Z time / s 0 0.5 1.2 2.3 3.7

45. 45 2.1 Latent heat Which material(s) release(s) largest amount of energy (per kg) when they freeze? ( X / Y / Z ) Check-point 2 – Q5 Temperature /  C X Y Z time / s 0 0.5 1.2 2.3 3.7

46. 46 2.1 Latent heat Check-point 2 – Q6 Fill in the value of energy at each step for changing 1 kg of ice at 0  C to steam at 100  C. Energy needed = _____________ + ____________ + _______________ = _____________ 3.34 x 10 5 J4200 x 100 J 2.26 x 10 6 J 3.014 x 10 6 J

47. 47 2.1 Latent heat The End

48. 48 2.1 Latent heat Result of melting experiment Mass of water in experimental cup: m 1 = 0.050 kg in control cup: m 2 = 0.014 kg Joulemeter reading initial: j 1 = 15 000 J final: j 2 = 29 200 J Example 1 Finding specific latent heat of fusion of ice

49. 49 2.1 Latent heat (a)Find the specific latent heat of fusion of ice. l f = E/m = (j 2 – j 1 )/(m 2 – m 1 ) = (29 200 – 15 000)/(0.050 – 0.014) = 3.94  10 5 J kg  1 Results: m 1 = 0.050 kg m 2 = 0.014 kg j 1 = 15 000 J j 2 = 29 200 J Example 1 Finding specific latent heat of fusion of ice

50. 50 2.1 Latent heat There is a rather large error of 18%. The possible sources of error are: Difficulty in keeping the water dripping down the two funnels at the same rate. Example 1 (b)Account for any difference of the value obtained from the standard value, 3.34  10 5 J kg –1. Experimental value = 3.94  10 5 J kg –1 Finding specific latent heat of fusion of ice Energy lost to the surroundings.

51. 51 2.1 Latent heat Return

52. 52 2.1 Latent heat How much energy is required to melt 0.5 kg of ice at 0  C temperature raised to 80  C? Total energy required = latent heat (ice at 0  C → water at 0  C) + energy (water: 0  C → 80  C) = ml f + mc  T = 0.5  3.34  10 5 + 0.5  4200  80 = 3.35  10 5 J Example 2 Heating ice

53. 53 2.1 Latent heat Return

54. 54 2.1 Latent heat Result of boiling experiment Mass of water boiled away = 0.10 kg KW h meter calibration = 600 turns/kW h Number of rotations counted = 41 Example 3 Finding specific latent heat of vaporization of water (a)Find the specific latent heat of vaporization of water.

55. 55 2.1 Latent heat kW h meter calibration = 600 turns/kW h Energy supplied per revolution of the disc = 3.6  10 6 /600 = 6000 J Example 3 Energy supplied to boil 0.10 kg of water = 6000  41 = 246 000 J Specific latent heat of vaporization of water l v = E/m = 246 000/0.10 = 2.46  10 6 J kg –1 Number of rotations = 41 Finding specific latent heat of vaporization of water 1 kW h = 1 kW  1h = 3.6  10 6 J

56. 56 2.1 Latent heat Example 3 1Steam condensing on the heater drips back into the cup. 2Some water ‘bubbles’ out of the cup. 3Energy lost to the surroundings. There is an error of about 9%. (b)Account for any difference of the value obtained from the standard value, 2.26  10 6 J kg –1. Experimental value = 2.26  10 6 J kg –1 Finding specific latent heat of vaporization of water

57. 57 2.1 Latent heat Return

58. 58 2.1 Latent heat How much energy is required to change 0.5 kg of water at 0  C to stream at 100  C. Example 4 Heating water m = 0.5 kg  T = 100  C – 0  C = 100  C

59. 59 2.1 Latent heat m = 0.5 kg,  T = 100  C – 0  C = 100  C The total energy required = energy (water: 0  C → 100  C) + latent heat (water at 100  C → steam at 100  C) = mc  T + ml v = 0.5 × 4200 × 100 + 0.5 × 2.26 × 10 6 = 1.34 × 10 6 J Example 4 = 2.1 × 10 5 + 11.3 × 10 5 Heating water

60. 60 2.1 Latent heat Return

61. 61 2.1 Latent heat An expresso coffee machine injects 0.025 kg of steam at 100  C into a cup of cold coffee of mass 0.15 kg at 20  C. Find the final temp. of the expresso coffee. specific heat capacity of the coffee = 5800 J kg –1  C Making coffee with steam Example 5

62. 62 2.1 Latent heat Let T be the final temperature of the coffee. Assuming no energy loss to the surroundings, 0.025  2.26  10 6 + 0.025  4200  (100 – T ) energy loss by steam energy gained by coffee = 0.025 kg of steam at 100 o C 0.15 kg of coffee at 20 o C Solving the equation, The temperature T of the coffee is 86.6  C. Example 5 Making coffee with steam = 0.15  5800  (T – 20)

63. 63 2.1 Latent heat Return

64. 64 2.1 Latent heat Example 6 Steam and hot water Assume that the temperature of the skin is 33  C. Find the energy released when (a) water of mass 15 g at 100  C, and (b) steam of mass 15 g at 100  C are each spilt on the hand.

65. 65 2.1 Latent heat Example 6 Steam and hot water (a) The energy released = mc  T = 0.015 x 4200 x (100 – 33) = 4.221 x 10 3 J

66. 66 2.1 Latent heat Example 6 Steam and hot water (b)Steam at 100  C will change to water at the same temperature first and latent heat is released. Total energy released = ml + mc  T = 0.015 x 2.26 x 10 6 + 4221 = 3.81 x 10 4 J

67. 67 2.1 Latent heat Return