1. Last lesson?

2. Thermal capacity Thermal capacity is the amount of energy needed to raise the temperature of a substance by 1K.

3. Calculations using Thermal capacity Energy absorbed = Thermal capacity x Temp rise E = QΔT J J.°C -1 °C

4. Specific heat capacity Specific heat capacity is the amount of energy needed to raise the temperature of unit mass of a substance by 1K Specific heat capacity of water = 4186 J.kg -1.°C -1 Specific heat capacity of kerosene = 2010 J.kg -1.°C -1 Specific heat capacity of mercury = 140 J.kg -1.°C -1

5. Calculations using S.H.C. Energy absorbed = Mass x Specific Heat capacity x Temp rise E = mcΔT Jkg J.kg -1.°C -1 °C

6. Today’s lesson Changes of state Latent heat

7. Solids, liquids and gases

8. Melting? Changes in kinetic energy and potential energy?

9. Evaporation? Changes in kinetic energy and potential energy?

10. Condensing? Changes in kinetic energy and potential energy?

11. Freezing? Changes in kinetic energy and potential energy?

12. Boiling and evaporation

13. Evaporation Consider a beaker of water at room temperature

14. Evaporation The molecules of water are moving around at different speeds, some fast, some slow. speed of molecule (m/s) # of molecules at a particular speed Average speed

15. Evaporation If a molecule is at the surface, and moving fast enough, it may escape the liquid. This is called evaporation. Freedom!

16. Evaporation Since the average speed of the remaining molecules must now be lower, the temperature of the liquid drops (since temperature is a measure of the kinetic energy of the molecules). Freedom!

17. Evaporation Evaporation can thus take place at any temperature.

18. Increasing the rate of evaporation Increasing the temperature.

19. Increasing the rate of evaporation Increasing the temperature means that more molecules are moving fast enough to escape.

20. Increasing the rate of evaporation Increasing the surface area

21. Increasing the rate of evaporation Increasing the surface area means that more molecules are at the surface.

22. Increasing the rate of evaporation Increasing the air flow over the surface

23. Increasing the rate of evaporation Increasing the air flow over the surface so that molecules are carried away before they can fall back into the liquid

24. Increasing the rate of evaporation Decreasing the humidity of the surrounding atmosphere

25. Increasing the rate of evaporation Decreasing the humidity of the surrounding atmosphere to stop water molecules from the atmosphere entering the liquid.

26. Boiling Boiling occurs when vapour is produced in the body of the liquid.

27. Boiling Boiling occurs when vapour is produced in the body of the liquid. The bubble contains only water vapour, not air!

28. Boiling Boiling occurs when vapour is produced in the body of the liquid. This only happens at the boiling point of the liquid. The bubble contains only water vapour, not air!

29. To summarize: Evaporation takes place only at the surface of the liquid and can take place at any temperature.

30. To summarize: Boiling occurs when vapour is produced in the body of the liquid. This only happens at the boiling point of the liquid. Boiling means bubbles!

31. Latent heat

32. In last year’s experiment, you will have noticed that the temperature of the salol stopped changing as the salol changed from a liquid to a solid. Temp (°C) Time (mins) Melting point

33. Latent heat Why does this happen?

34. Latent heat When the molecules of a substance settle into the regular patter of a solid, energy is released as bonds are formed. This energy released is called latent heat. This stops the temperature from falling. (“latent” = “hidden”)

35. Latent heat The opposite happens when a solid makes. Heat is needed to break the bonds between the solid particles (increasing their potential energy instead of raising the temperature (kinetic energy)) Temp (°C) Time (mins) Melting point solid liquid

36. Specific Latent heat The specific latent heat of a substance tells us how much energy is needed to change the state of 1 kg of substance at constant temperature. Solid to liquid/liquid to solid or liquid to gas/gas to liquid

37. Specific Latent Heat The specific latent heat of fusion (melting) of ice at 0 º C, for example, is 334000 J.kg -1. This means that to convert 1 kg of ice at 0 º C to 1 kg of water at 0 º C, 334000 J of heat must be absorbed by the ice. 1 kg 334000 J absorbed All at 0°C

38. Specific Latent Heat Conversely, when 1 kg of water at 0 º C freezes to give 1 kg of ice at 0 º C, 334000 J of heat will be released to the surroundings. 334000 J released 1 kg All at 0°C

39. Specific Latent Heat of Vaporisation For water at its normal boiling point of 100 º C, the latent specific latent heat of vaporization is 2260000 J.kg -1. This means that to convert 1 kg of water at 100 º C to 1 kg of steam at 100 º C, 2260000 J of heat must be absorbed by the water. All at 100°C 2260000 J input 1 kg

40. Latent heat Conversely, when 1 kg of steam at 100 º C condenses to give 1 kg of water at 100 º C, 2260 kJ of heat will be released to the surroundings. All at 100°C 2260000 J released 1 kg

41. Another formula! Energy = mass x specific latent heat E = mL

42. An example calculation Calculate the amount of heat required to completely convert 50 g of ice at 0 º C to steam at 100 º C. The specific heat capacity of water is 4.18 kJ.kg -1.°C -1. The specific latent heat of fusion of ice is 334 kJ.kg -1, and the specific heat of vaporization of water is 2260 kJ.kg -1. 50g 0°C 100°C

43. An example calculation Heat is taken up in three stages: 1. The melting of the ice. 2. The heating of the water. 3. The vaporization of the water. 0°C 100°C

44. Stage 1 1. Heat taken up for converting ice at 0 º C to water at 0 º C mass of water x latent heat of fusion = 0.050 (kg) x 334 (kJ.kg -1 ) = 16.7 kJ 0°C

45. Stage 2 2. Heat taken up heating the water from 0 º C to the boiling point, 100 º C mass of water x specific heat capacity x temperature change = 0.05 (kg) x 4.18 (kJ.kg -1.°C -1 )x 100 ( º C) = 20.9 kJ 0°C 100°C

46. Stage 3 3. Heat taken up vaporising the water mass of water x latent heat of vaporization 0.05 (kg) x 2260 kJ.kg -1 = 113 kJ 100°C

47. The answer The sum of these is 16.7 + 20.9 + 113 = 150.6 kJ (151 kJ)

48. Got it?

49. OK, now try to answer some questions! Page 172 Qs 5, 6, 7, 8 and 14.

50. Just time for a quick dog accident 50 kg

51. Ooops! 50 kg