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Cool wax!. Can you stick the sheet in please?. Sketch a predicted graph. Temp (°C). water. ?. wax. Time (mins). Today’s lesson. Describe melting and boiling in terms of energy input without a change of temperature Distinguish between boiling and evaporation

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## Cool wax!

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**Cool wax!**Can you stick the sheet in please?**Sketch a predicted graph**Temp (°C) water ? wax Time (mins)**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**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.**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”)**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)**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**Another formula!**Energy = mass x specific latent heat E = mL**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**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**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**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**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**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**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.**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**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**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**The answer**The sum of these is 16.7 + 20.9 + 113 = 150.6 kJ (151 kJ)**50 kg**Just time for a quick dog accident**Ooops!**50 kg**Evaporation**Consider a beaker of water at room temperature**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**Evaporation**If a molecule is at the surface, and moving fast enough, it may escape the liquid. This is called evaporation. Freedom!**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!**Evaporation**Evaporation can thus take place at any temperature.**Boiling**Boiling occurs when vapour is produced in the body of the liquid. What’s in the bubbles?**Boiling**Boiling occurs when vapour is produced in the body of the liquid. The bubble contains only water vapour, not air!**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!**To summarize:**Evaporation takes place only at the surface of the liquid and can take place at any temperature.**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.**OK, now try to answer some questions!**Page 115 Q 5 Page 119 Qs 1, 2, 3, 4.

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