Latent Heat Capacity The use of an equation

Transcription

1 CHAPTER 30 Latent Heat Capacity The use of an equation

2 Definition When a solid is heated, it may melt and change its state from solid to liquid. If ice is heated it becomes water. The opposite process of freezing occurs when a liquid solidifies. A pure substance melts at a definite temperature, called the melting point; it solidifies at the same temperature - the freezing point.

3 Experiment: cooling curve of ethanamide Half-fill a test-tube with ethanamide (acetamide) and place it in a beaker of water. Heat the water until all the ethanamide has melted.

4 Remove the test-tube and arrange it as in figure with a thermometer in the liquid ethanamide. Record the temperature every minute until it has fallen to 70 o C. Plot a cooling curve of temperature against time. What is the freezing (melting) point of ethanamide?

5 Types of latent heat There exists two types of latent heat. 1 This occurs when solids becomes liquids or vice-versa and is called specific latent heat of fusion. 2 This occurs when liquids becomes gases or vice-versa and is called specific latent heat of vaporization.

6 Latent heat of fusion The previous experiment shows that the temperature of liquid ethanamide falls until it starts to solidify (at 82 o C) and remains constant till it has all solidified. The cooling curve in the following figure is for a pure substance; The flat part AB occurs at the melting point when the substance is solidifying. During solidification a substance loses heat to its surroundings but its temperature does not fall.

7 Conversely when a solid is melting, the heat supplied does not cause a temperature rise. For example, the temperature of a well-stirred ice-water mixture remains at 0 o C until all the ice is melted.

8 Heat which is absorbed by a solid during melting or given out by a liquid during solidification is called latent heat of fusion. Latent means hidden and fusion means melting. Latent heat does not cause a temperature change; it seems to disappear. The specific latent heat of fusion of a substance is the quantity of heat needed to change unit mass from solid to liquid without temperature change. It is measured in J/kg or J/g.

9 The latent heat equation In general, we can write the latent heat equation as i.e. Heat received or given out = mass E = ml Latent Heat Capacity Note that this equation does not depend on temperature since there is no temperature change during the latent heat.

10 Latent heat graph of water

11 Experiment: specific latent heat of Place a 12V electric immersion heater of known power in a filter funnel and pack small pieces of ice round it. Switch on the heater for 3 minutes and find the mass of water which collects in a beaker. fusion of ice

12 Experiment: specific latent heat of fusion of ice Arrange the results as shown. Power of immersion heater Time heat supplied Heat supplied to ice Mass of beaker empty Mass of beaker + melted ice Mass of melted ice = W(J/s) = s = J = g = g = g Calculate the heat needed to melt 1g of ice. What are causes of error in this experiment?

13 Latent heat of vaporization Latent heat is also needed to change a liquid into a vapour. The reading of a thermometer in boiling water remains constant at 100 o C even though heat, called latent heat of vaporization, is still being absorbed by the water from whatever is heating it. When steam condenses to form water, latent heat is given out. This is why a scald from steam may be more serious than one from boiling water.

14 The specific latent heat of vaporization of a substance is the quantity of heat needed to change unit mass from liquid to vapour without change of temperature.

15 An estimate of its value for water can be made using the apparatus shown. The mains-operated immersion heater is clamped so that it is well covered by the water in the can. When the water is boiling briskly the reading on the balance is noted and a stop clock started. The time for 50g of water to be boiled off is found.

16 Suppose a 500W heater is used and that the time required is 4 minutes (240s), then Heat supplied in 240s = 500J/s 240s = J Latent heat of vaporization = J/50g = 2400 J/g The accepted value is 2300 J/g = J/kg = J/kg = 2.3 MJ/kg. Errors arise because the can and water lose heat to the surroundings.

17 Latent heat and the kinetic theory (a) Fusion. The kinetic thoery explains latent heat of fusion as being the energy which enables the molecules of a solid to change their vibratory motion about a fixed position to the greater range of movement they have as liquid molecules. Their p.e. increases but not their average k.e. as happens when the heat causes a temperature rise.

18 Latent heat and the kinetic theory (b) Vaporization. If liquid molecules are to overcome the forces holding them together and gain the freedom to move around independently as gas molecules, they need a large amount of energy. They receive this as latent heat of vaporization which, like latent heat of fusion, increase the p.e. of the molecules but not their k.e. It also gives the molecules the energy required to push back the surrounding atmosphere in the large expansion that occurs when a liquid vaporizes.

19 Table 1 The following table gives the Latent heat capacity of some common substances. Note carefully the units of the latent heat. They are J/g or J g -1 If you want to change them to kilograms, just multiply by Ex. 334 J g -1 = = J kg -1

20 Substance Specific latent heat of fusion J.g -1 o C Specific latent heat of vaporization J.g -1 Water Ethanol Ethanoic acid o C Chloroform Mercury Sulphur Hydrogen Oxygen Nitrogen

21 Table 2 Product latent heat of fusion J.g -1 Product latent heat of fusion J.g -1 Aluminium 321 Lead 22.4 Ammonia 339 Mercury 11.8 Aniline Nickel 19.4 Carbon Dioxide 184 Silver 88.0 Copper 176 Sulpur 39.2 Glycerin 176 Tin 58.5 Iron, gray cast 96 Zinc 118 Iron, white cast 138 Gold 63 Iron, slag 209

22 Example 1 How much heat is needed to change 20g of ice at 0 o C to water at 0 o C? Answer: Since here we have only a change of state then we are going to use the formula of latent heat of fusion (because it is ice). i.e. E = ml Latent heat of fusion of water = 334 Jg -1 So E = = 6680 J

23 Example 2 How much heat is needed to change 50 g of mercury from liquid at 357 o C to gas at 357 o C? Answer: Again we are going to use the Latent heat equation since this involves only a change of state. Latent heat of vaporization of mercury = 294 J/g E = = J

24 Example 3 How much heat is needed to change 50 g of ice at 0 o C to water at 10 o C? Answer: This time we have to find the heat needed to change 50g of ice at 0 o C to water at 0 o C by using the latent heat equation. Then we have to find the heat needed to change 50g of water at 0 o C to water at 10 o C by using the specific heat capacity.

25 1) Latent heat of fusion of water = 334 J/g E = ml = = J 2) Specific heat capacity of water = 4200J/kg o C Change to grams = = 4.2J/g o C To change from 0 o C to 10 o C E = mc T = = 2100 J Total Energy = J J = J

26 Example 4 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 -1.K -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. Answer. First convert the units to g since we have only 50g to deal with.

27 4.18 kj.kg -1.K -1 = 4180 J.kg -1.K -1 = 4.18 J.g - 1.K -1 Using the same pattern we have L F = 334 kj.kg -1 = 334 J.g -1 L V = 2260 kj.kg -1 = 2260 J.g -1 Heat is taken up in three stages: 1. The melting of the ice, 2. The heating of the water, and 3. The vapourization of the water. The heat taken up in the complete process is the sum of the heat taken up in each stage.

28 1. Heat taken up for converting ice at 0ºC to water at 0ºC 2. Heat taken up heating the water from 0 ºC to the boiling point, 100 ºC = M W x L F = 50 (g) x 334 (J.g -1 ) = J = M W x s.h.c W x T = 50 g x 4.18 J.g -1 K -1 x 100 ºK = J 3. Heat taken up vapourizing the water The sum of these is = M W x L V = 50 g x 2260 J.g -1 = J = = J = kj

29 Example 5 An aluminium can of mass 100g contains 200g of water. Both, initially at 15 o C, are placed in a refrigerator at -5 o C. Calculate the quantity of heat that has to be removed from the water and the can for their temperatures to fall to -5 o C. The specific heat capacity of water is 4.2J/g o C. The specific latent heat of fusion of ice is 340 J/g, and the specific heat capacity of aluminium is 0.9 J/g.

30 Answer There are four stages in the change. 1) Heat lost by can in falling from 15 o C to -5 o C 2) Heat lost by water in falling from 15 o C to 0 o C 3) Heat lost by water at 0 o C freezing to ice at 0 o C 4) Heat lost by ice in falling from 0 o C to -5 o C Total heat removed = M can x s.h.c can x T = 100 x 0.9 x 20 = 1 800J = M water x s.h.c water x T = 200 x 4.2 x 15 = J = M W x L F = 200 x 340 = J = M ice x s.h.c ice x T = 200 x 2 x 5 = 2 000J = J

31 Exercise 1 1) How much heat is needed to change 250 g of Ethanol from solid at -114 o C to liquid at -114 o C? Ans: J 2) How much heat is needed to change 335 g of water at 50 o C to steam at 100 o C? Ans: E S = J E L = J E T = J