2. Specific Latent Heat From Key stage 3 you’ve been aware that it takes energy to change a solid into a liquid and a liquid into a gas. The temperature of the substance doesn’t change all the time this transformation is happening. The graph for heating a substance is identical to the cooling curve, with two flat sections where the substance is changing state. However heat energy is still being supplied during this time, but to do what exactly?

3. liquid is heated When a liquid is heated the energy supplied turns into kinetic energy of the molecules and forces them apart. Eventually they gain enough energy for some of them to break free of the liquid, if their energy is sufficient, remembering the random distribution. increases their potential energy This increases their potential energy. energy needed to turn 1 kg of a liquid to a gas is called the latent heat of vaporisation The energy needed to turn 1 kg of a liquid to a gas is called the latent heat of vaporisation. This will be a different value for different liquids and is the reason we sweat – it cools us as we supply the energy to change it’s state.

4. solid is heated When a solid is heated the energy supplied turns into vibrational kinetic energy of the molecules and forces them further apart. Eventually they gain enough energy for some of them to break free from their fixed positions turning the solid into a liquid as they are not totally free of the other molecules. increases their potential energy This increases their potential energy. energy needed to turn 1 kg of a solid to a liquid is called the latent heat of fusion The energy needed to turn 1 kg of a solid to a liquid is called the latent heat of fusion. This will be a different value for different substances, explaining why ice can exist in water at 0 o C.

5. Temperature Time Water warming from 0°C to 100°C Ice melting at 0°C Ice warming up to 0 °C Steam heating up from 100°C Water boiling at 100°C Consider heating a block of ice until it melts and finally boils: 100 °C 0 °C Make sure you can explain a graph like this in terms of average energy of molecules and energy supplied.

6. Note that at 100°C both water and steam exist together and the temperature cannot rise until all the water has boiled away if it is well mixed. In practice (in an open pan) the steam cannot exceed 100°C as it gets its heat from the water which cannot exceed 100°C). At 0°C both ice and melt water exist. Enthalpy amount of energy supplied changing state This is the term use for the amount of energy supplied to a substance when it is changing state. Some of this energy will go into raising the temperature but if the substance expands it will also do some work on the surroundings by “squashing” them. enthalpy “Latent Heat” The word enthalpy can therefore be swapped into the equations used later or instead of the term “Latent Heat”

7. This apparatus can be used to investigate the latent heat of vaporisation of water. Boil water and pour in to the insulated container bring back up to boil with the heater Start the timer and record current and voltage Record how long it takes to boil away 10g of water Evaluate the latent heat as shown – latent heat of vaporisation = 2.3 MJkg -1. W = ITV = l  m

8. 12 Volts V A 12V Supply This apparatus can be used to investigate the latent heat of fusion of water. Control

9. Add crushed ice to the two arrangements shown Leave the experiments for a couple of minutes to allow all pieces to cool to the temp of the ice Turn the heater supply on, start the timer and record current and voltage Record how much water is melted in both experiments in a known time Subtract the mass of water melted without the heater from the other mass of water to establish how much water the heater melted Evaluate the latent heat as shown – latent heat of fusion = 330 kJkg -1. W = ITV = l  m

10. Specific Latent Heat l lJkg -1 The units of l are Jkg -1 Remembering that you must be using the correct Latent Heat for the transformation taking place. The amount of heat required to change the state of 1kg of a substance without changing its temperature. Energy Transfer =  ml Enthalpy supplied = mass * specific enthalpy

11. Questions 1) 200g of water is placed in a can. The temperature of both the water and the can is 20°C. 300g of water at 80°C is added and mixed. What is the final steady temperature of the mixture? (c w = 4200Jkg -1 K -1, Heat capacity of the can =200Jkg -1 ) Heat lost by hot water = heat gained by cold water and can mc w  T hot = mc w  T cold + C  T cold Call the final temperature T f 0.3x4200x(80-T f ) = 0.2x4200x(T f -20) + 200x(T f -20) 100800-1260T f = 840T f - 16800 + 200T f - 4000 2300T f = 121600 T f = 52.9°C

12. 2) 5g of ice at 0°C is now added. What will the final temperature of the mixture be? (l = 3.4x105Jkg -1 ) Heat lost by water and can = heat gained by ice melting & the melt water heating up to T f {mc w  T hot } + {C  T hot } = {l  m ice } + {m ice c w  T cold } 0.5x4200x(52.9-T f ) + 200(52.9-T f ) = 3.4x105x0.005 + 0.005x4200x(T f -0) 111090 - 2100T f + 10580 - 200T f = 1700 +21T f 2321T f = 119970 T f = 51.7°C