Specific Heat Capacity The use of an equation
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1 CHAPTER 29 Specific Heat Capacity The use of an equation
2 Definition If 1 kg of water and 1 kg of paraffin are heated in turn for the same time by the same heater, the temperature rise of the paraffin is about twice that of the water. Since the heater gives equal amounts of heat energy to each liquid, it seems that different substances require different amounts of heat to cause the same temperature rise in the same mass, say 1 o C in 1 kg.
3 Definition The thirst of a substance for heat is measured by its specific heat capacity (symbol c). The specific heat capacity of a substance is the heat required to produce unit temperature rise in unit mass
4 Definition Heat, like other forms of energy, is measured in Joules (J) and the unit of specific heat capacity is the joule per kilogram degree Celsius. i.e. Unit of c = J/kg o C or Jkg -1 o C -1 In physics the word specific means unit mass is being considered.
5 The heat equation If a substance has a specific heat capacity of 1000 J/kg o C then 1000J raise the temp. of 1kg by 1 o C J raise the temp. of 2kg by 1 o C J raise the temp. of 2kg by 3 o C That is, 6000J will raise the temperature of 2kg of this substance by 3 o C.
6 The heat equation We have obtained this answer by multiplying together: (i) The mass in kg (ii) The temperature rise in o C, and (iii) The specific heat capacity in J/kg o C If the temperature of the substance fell by 3 o C, the heat given out would also be 6000J.
7 The heat equation In general, we can write the heat equation as Heat received or given out = mass Specific Temp. Change Heat Capacity i.e. E = mc T
8 Table 1 The following table gives the specific heat capacity of some common substances. Substance Specific heat capacity c ( J kg -1 o C -1 ) Water 4200 Alcohol 2400 Sand 840 Granite 800 Ice 2100 Glass 670 Copper 390 Iron 460
9 Example Find the heat received, if the temperature of a 5kg mass of copper of specific heat capacity 400 J/kg o C rises from 15 o C to 25 o C. Answer E = m c T E = 5 kg 400 J/kg o C (25 15) o C E = = J
10 Exercise 1 1) Find the heat received, if the temperature of a 15kg mass of aluminium of specific heat capacity 900 J/kg o C rises from -12 o C to 22 o C. 2) How much heat is needed to raise the temperature by 10 o C of 5kg of a substance of specific heat capacity 300 J/kg o C?
11 Experiment: finding specific heat capacities You need to know the power of the 12-volt electric immersion heater to be used. To know; A 40-watt heater converts 40 joules of electrical energy into heat energy per second. Do not forget that heat energy is measured in Joules If the power is not marked on the heater ask about it. Another useful equation is Power in watts = volts amperes
12 A) S.H.C. of WATER (LIQUIDS) Weigh out 1 kg of water into a container, e.g. an aluminium saucepan. Note the temperature of the water, insert the heater and switch on the 12V supply. Stir the water and after 5 minutes switch off, but continue stirring and note the highest temperature reached. Assuming heat supplied by the heater equals the heat received by the water, work out the specific heat capacity of water in J/kg o C.
13 Heat received by water (J) = power of heater (J/s) time heater on (s) Rearranging the heat equation we get E E mc T c m T
14 B) Aluminium (SOLIDS) A cylinder weighing 1 kg and having two holes drilled in it is used. Place the immersion heater in the central hole and a thermometer in the other hole as illustrated in the diagram. Note the temperature, connect the heater to a 12V supply and switch it on for 5 minutes. When the temperature stops rising record its highest value. Calculate the specific heat capacity as before.
15 Importance of high specific heat capacity of water The specific heat capacity of pure water is 4200 J/kg o C and of soil it is about 800 J/kg o C. As a result, the temperature of the sea rises and falls more slowly than that of the land. A certain mass of water needs five times more heat for its temperature to rise by 1 o C than does the same mass of soil. Water also has to give out more heat to fall 1 o C.
16 Since islands are surrounded by water they therefore experience much smaller changes of temperature from summer to winter than do large land masses such as Central Asia.
17 The high specific heat capacity of water (as well as its cheapness and availability) accounts for its use to cool car engines and in the radiators of central heating systems.
18 Worked Examples 1. A tank holding 60kg of water is heated by a 3kW electric immersion heater. If the specific heat capacity of water is 4200 J/kg o C, estimate the time for the temperature to rise from 10 o C to 60 o C. Answer; A 3kW (3000W) heater supplies 3000J of heat energy per second. Let t = time taken in seconds to raise the temperature of the water by (60 10) = 50 o C
19 Therefore heat supplied to water in time t = 3000 t J From the heat equation, we can say Heat received by water = J Assuming heat supplied = heat received 3000 t = J t 4200s 3000 (70 mins)
20 Worked Examples 2. A piece of aluminium of mass 0.5kg is heated to 100 o C and then placed in 0.4kg of water at 10 o C. If the resulting temperature of the mixture is 30 o C, what is the specific heat capacity of aluminium if that of water is 4200J/kg o C? Answer: When two substances at different temperatures are mixed, heat flows from the one at the higher temperature to the one at the same temperature the temperature of the mixture. If there is no loss of heat, then in this case
21 Heat given out by aluminium = Heat taken in by water Using the heat equation and letting c be the sp. Heat cap. Of aluminium in J/kg o C, we have Heat given out = 0.5 c (100 30)J Heat taken in = (30 10)J 0.5 c 70 = c J/kg o C
22 Conservation of mechanical energy to heat (a) p.e. The p.e. of water at the top of a waterfall becomes k.e. as the water falls and when it strikes the ground, heat (and sound) are produced. The temperature of the water at the bottom is higher than that at the top. If we assume all the p.e. becomes heat the temperature rise can be calculated.
23 Consider 1kg of water falling a vertical distance of 105m. We have, taking g = 10m/s 2, p.e. of water = mgh = J = 1050 J
24 Using the heat equation and taking the sp. Heat cap. of water as 4200 J/kg o C, Heat produced = J Where is the temperature rise of the water in o C 4200 = 1050 = 1050/4200 = 0.25 o C (b) k.e. When a car is brought to rest by brakes, k.e. becomes heat and the temperature of the brake linings (or disc pads) rises.
25 Worked Examples 3. The apparatus shown in the figure is used to conduct an experiment on finding the specific heat capacity of water. The joulemeter records the heat energy supplied to the water by the immersion heater. The mass of water is measured by an electronic balance. When J of energy is supplied to 100 g of water, the rise in temperature is found to be 30 o C. Find the specific heat capacity of water.
26 Solution Since E = m c T = 0.1 c 30 c = 4230 J kg -1 o C -1 The specific heat capacity of water found in this experiment is 4230 J kg -1 o C -1. The last example shows the standard method used in a school laboratory to measure the specific heat capacity of water. Now let's try another less accurate but more interesting method - use a domestic electric boiler to find the specific heat capacity of water. In fact, all apparatus used in this activity can be found in an ordinary family.
27 Worked Examples 4. When preparing instant cup noodle, Tommy pours 300 g of water at 90 o C onto 75 g of noodle at 20 o C in a polystyrene cup. a) Estimate the temperature of the mixture soon after mixing. b) Would the actual temperature be higher or lower than the estimated value? Why? Given the specific heat capacity of the noodle and water are 1700 J kg -1 o C -1 and 4200 J kg -1 o C -1 respectively.
28 Solution Let T be the temperature of the mixture soon after mixing. Heat loss by hot water = (90 T) Heat gain by noodle = (T 20) Assuming no heat is lost to the surrounding, we have Heat loss by hot water = Heat gain by noodle (90 T) = (T 20) T = 83.6 o C The temperature of the mixture is 83.6 o C. We expect the actual temperature to be much lower than this value due to heat loss to the surrounding and the heat used to raise the temperature of the polystyrene cup.
29 Worked Examples 5. The initial temperature of 0.1 kg of sand is 25 o C. If 840 J of heat is transferred to it, what is the final temperature of the sand? Repeat the calculation with the same mass of water. Given the specific heat capacity of sand is 840 J kg -1 o C -1 ; and the specific heat capacity of water is 4200 J kg -1 o C -1.
30 Solution Consider the sand first. Since E = m c T = mc(t 2 T 1 ) we have For the same mass of water, we have
31 The final temperature of the sand is 35 o C and the final temperature of the water is 27 o C. We can see that the change in temperature of water is much smaller than that of sand because water has a much higher specific heat capacity.
32 Energy of food When food is eaten it reacts with the oxygen we breathe into our lungs and is slowly burnt. As a result chemical energy stored in food becomes heat to warm the body and mechanical energy for muscular movement. The calorific value of a substance is the amount of energy released when unit mass is completely oxidized.
33 It is measured in J/kg. Dieticians sometimes use kilocalories instead of joules. The calorie was the previous unit of heat and equals 4.2J. Some calorific values are given below in megajoules per kilogram (MJ/kg: 1MJ = J) Fat 38 Sugar 16 Ice cream 9 Milk 2.9 Butter 31 Flour 15 Eggs 7 Apples 2.6 Cheese 21 Beef 10 Potatoes 4 Carrots 1.7
34 Foods with high values are fattening and if more are eaten than the body really needs they are stored as fat. The average person requires about 12 MJ per day. Our muscles change chemical energy into mechanical energy when we exert a force, to lift a weight for example. Unfortunately they are, at best, only about 25% efficient at doing this. Every 100J of mechanical energy supplied by our muscles therefore uses 400J of chemical energy. The other 300J becomes heat, most of which the body has to get rid of.
35 The energy crisis Energy changes from one form to another, it cannot be destroyed. However it usually ends up as heat (as a result of overcoming friction) and raises the temperature of its surroundings. It is then not easy to use. Fossil fuels such as coal and oil are the stores of chemical energy from which most other useful forms of energy come at present. Stocks of these are running down and there is now world-wide concern to make the best of what we still have and to develop other sources, especially nuclear and solar energy.
36 Although there are safety and environmental problems connected with nuclear energy, it seems likely to be the main solution to our fuel problems in the foreseeable future. Solar energy is difficult to capture directly in the amount required. The average family in Britain needs about J/s for all purposes, i.e. 15kW. The average solar power over the whole of the earth s surface is roughly 1.5kW/m 2. Each family would therefore require a collecting area of 10m 2 but, since the efficiency of solar panels is very low, the actual area would be much greater.
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