Cool wax!

1. Cool wax! Can you stick the sheet in please?

2. Sketch a predicted graph Temp (°C) water ? wax Time (mins)

3. Today’s lesson • Describe melting and boiling in terms of energy input without a change of temperature • Distinguish between boiling and evaporation • Describe condensation and solidification • State the meaning of melting point and boiling point • Use the terms latent heat of vaporisation and latent heat of fusion and give a molecular interpretation of latent heat

4. Solids, liquids and gases

5. Cooling!

6. Melting point Temp (°C) Time (mins) Latent heat When pure substances cool, the temperature stops changing as the substance changes from a liquid to a solid.

7. 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”)

8. 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)) liquid Melting point Temp (°C) solid Time (mins)

9. The same happens when boiling

10. 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

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

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

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

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

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

16. Example • The specific latent heat of fusion (melting) of ice is 334 000 J/kg. How much energy is needed to melt 5kg of ice at 0°C to 5 kg of water at 0°C? • Energy = mL = 5 x 334 000 = 1670000 J

17. 50g 0°C IB level calculation Calculate the amount of heat required to completely convert 50 g of ice at 0 ºC to steam at 100 ºC. The specific heat capacity of water is 4.18 kJ/kg/°C. The specific latent heat of fusion of ice is 334 kJ/kg, and the specific heat of vaporization of water is 2260 kJ/kg. 50g 100°C

18. 0°C 0°C 100°C 0°C 100°C 100°C 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.

19. 0°C 0°C Stage 1 1. Heat taken up for converting iceat 0ºC to water at 0ºC mass of water x latent heat of fusion= 0.050 (kg) x 334 (kJ.kg-1) = 16.7 kJ

20. 100°C 0°C Stage 2 2. Heat taken up heating the waterfrom 0 ºC to the boiling point, 100 ºC mass of water x specific heat capacity x temperature change= 0.05 (kg) x 4.18 (kJ.kg-1.°C-1)x 100 (ºC)= 20.9 kJ

21. 100°C 100°C Stage 3 3. Heat taken up vaporising thewater mass of water x latent heat of vaporization0.05 (kg) x 2260 kJ.kg-1= 113 kJ

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

23. Got it?

24. 50 kg Just time for a quick dog accident

25. Ooops! 50 kg

26. Let’s try some questions

27. Boiling and evaporation

28. Evaporation Consider a beaker of water at room temperature

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

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

31. 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!

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

33. Boiling?

34. Boiling Boiling occurs when vapour is produced in the body of the liquid. What’s in the bubbles?

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

36. 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!

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

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

39. OK, now try to answer some questions! Page 115 Q 5 Page 119 Qs 1, 2, 3, 4.