1. The temperature of boiling water does not increase even though energy is supplied to it continually. Warm-up Agree. Disagree.

2. Do these require energy supply? Separating magnets that stick to each other. Allowing magnets to stick to each other. Some water molecules taken from liquid water are put in air. Water molecules in air come together and form dew. Warm-up

3. less so they have ________ (more / less) KE than water molecules. Ice molecules moves ________ (more / less) freely than water molecules; Warm-up

4. Introduction Matter exists in 3 states:solid, liquid, gas watersteam Fusion (at melting point) vaporization (takes place at boiling point) solidificationcondensation e.g. water at freezing pointat boiling point ice Change of state

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

6. Record the temperature of the melted octadecan-1-ol as it cools down. Cooling curve of octadecan-1-ol Experiment 3a Video

7. Cooling curve When solid is heated, it melts to a liquid On cooling, the variation of temperature with time from A to B The temperature remains constant from B to C, the energy evolved is called the latent heat The latent heat is used to change from liquid to solid at constant temperature, which is called the melting point

8. Why temperature remains constant? From liquid to solid, energy is released to the surroundings due to the decrease in the kinetic energy of the particles From solid to liquid, energy is absorbed from the surroundings due to the increase in the kinetic energy of the particles

9. Specific latent heat of fusion and vaporization Definition –Latent heat is required is proportional to the mass of the substance that undergoes state change i.e. Latent heat (H)  mass (m) H = L m L is a constant called the specific latent heat –Unit: J kg -1

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

11. temperature /  C A B C D BC is flat — liquid solidifying (temperature unchanged) Cooling curve

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

13. temperature /  C A B C D melting point: read from the flat part BC melting point Cooling curve

14. liquid-solid mixture temperature Cooling curve of water Simulation

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

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

17. Ice-water mixture stays at 0 o 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. Latent heat

18. energy is absorbed temperature unchanged to change the water to steam This energy is called latent heat of vaporization of water. Energy supplied continuously to keep water boils… Latent heat

19. steam water ice condensation condensation vaporization fusion solidification releases latent heat of vaporization

20. steam water ice condensation vaporization fusion solidification absorbs latent heat of vaporization

21. condensationvaporization fusion solidification releases latent heat of fusion steam water ice

22. condensation vaporization fusion solidification absorbs latent heat of fusion steam water ice

23. State Change solid liquid gas

24. molecule Regular arrangement breaks up strong attraction weak attraction 2Latent heat and particle motion

25. 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

26. 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 Simulation

27. Specific 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 temperature change

28. 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. Specific latent heat

29. l f = energy needed to change 1 kg of ice to water (without temperature change) Specific latent heat of fusion of ice (l f ) Find (1) mass melted m and (2) energy transferred E  l f = E/m

30. Measuring the specific latent heat of fusion of ice Experiment 3b Video Simulation

31. Experiment 3b experimental apparatus control apparatus Ice also melts at room temperature, so a control is needed.

32. 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 3b

33. 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 (1) mass boiled away m and (2) energy transferred E  l v = E/m

34. Experiment 3c Video Simulation

35. For water, l v = 2.26  10 6 J kg -1 water (0  C) steam (0  C) Experiment 3c

36. vaporization energy in 2260 kJ fusion energy in 334 kJ condensation energy out 2260 kJ solidification energy out 334 kJ steam (1 kg) water (1 kg) ice (1 kg) Summary: change of state

37. Energy involved in heating 1 kg of water ice stream(ice and water) melting (334) (420) water boiling (water and stream) (2260) energy / kJ Summary: from ice to steam 0 100

38. Consider a cup of water... 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

39. When vapour condenses... When vapour condenses, is the surrounding air warmed or cooled? (warmed/cooled)

40. Jimmy melts three materials X, Y and Z of equal mass at the same time and place. Temperature / o C X Y Z time / s 0 0.5 1.2 2.3 3.7 Jimmy melts...

41. 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...

42. Which material(s) has/have the largest value of specific latent heat of fusion? ( X / Y / Z ) Temperature / °C X Y Z time / s 0 0.5 1.2 2.3 3.7

43. Which material(s) release(s) largest amount of energy (per kg) when they freeze? ( X / Y / Z ) Temperature / °C X Y Z time / s 0 0.5 1.2 2.3 3.7

44. Fusion and Boiling Boiling of water and fusion of ice –when water boils, the temperature is found to remained at 100 o C. The energy supplied is only used to change from water to steam without any change in temperature –When the steam condenses, the temperature remains at 100 o C and energy is given out

45. Fusion and Boiling Fusion of ice –when ice melts, the temperature is found to remained at 0 o C. The energy supplied is only used to change from ice to water without any change in temperature –When the water freezes, the temperature remains at 0 o C and energy is given out

46. Latent Heat of Fusion and Vaporization Energies used in fusion and vaporization are called the latent heat of fusion and vaporization The constant temperatures are called the melting point and boiling point

49. 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

50. (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

51. 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.

52. 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

53. 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 (a)Find the specific latent heat of vaporization of water.

54. 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 1 kW h = 1 kW  1h = 3.6  10 6 J

55. 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

56. How much energy is required to change 0.5 kg of water at 0 °C to stream at 100 °C. Example 4 m = 0.5 kg  T = 100 °C – 0 °C = 100 °C

57. 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

58. 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 temeperature of the expresso coffee. specific heat capacity of the coffee = 5800 J kg –1 °C Example 5

59. 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 = 0.15  5800  (T – 20)

60. Effect of pressure on melting point of water melting point of ice at normal pressure is 0 o C it becomes lower when the external pressure is increased (i.e. below 0 o C ) vice versa

61. Effect of pressure on boiling point of water boiling point of ice at normal pressure is 100 o C it becomes higher when the external pressure is increased (i.e. above 100 o C ) vice versa

62. Effect of impurities on melting and boiling point of water Impurities such as salt is added to ice to lower the melting point, causing the ice to melt.