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TEMPERATURE AND HEAT
The temperature of a body is the measure of coldness or hotness of a body. In kinetic theory, temperature is the measure of average kinetic energy of the molecules of a substance. If the substance is a liquid or gas, the average transitional energy becomes the kinetic energy, but if the substance is solid, the average kinetic energy becomes the average vibrational energy, since the molecules of solids do not move transitionally. Temperature is a scalar quality, it is also a fundamental quantity, hence, its unit is Kelvin (K) is a fundamental system international (S.I) unit. THERMAL EQUILIBRIUM When two bodies are in contact and there is no net flow of heat between, them, they are said to be in thermal equilibrium. In this case, the bodies have the same temperature. Experiment shows that when two bodies A and B are each in thermal equilibrium with a third body C, then A and B are also in thermal equilibrium with each other. This fact above is called “Zeroth law of thermodynamics”. The measurement of temperature is made possible with the Zeroth law of thermodynamics. In practice, we take the third body as thermometer. We first bring bodies A and C thermal equilibrium, and then B and C thermal equilibrium. If the temperature reading is the same in the two cases, then we say that both bodies A and B are in thermal equilibrium from Zeroth’s law. It follows, from Zeroth law, that temperature is that property of a body which determines whether or not it is in thermal equilibrium with another body. HEAT This is a form of energy transferred from one body into another body, due to temperature difference between the bodies. Heat would also be transferred from a hot body to a cold body. This transfer of heat would continue until the bodies attain thermal equilibrium. In kinetic theory, heat is the measure of the total kinetic energy of the molecules of the substance. Temperature measurement Thermometric property – This is measurable property in a substance which sensitive temperature change. The substance is called thermometric substance. Thermometric properties are used in the construction of thermometers. Some of the thermometric substances, their corresponding thermometric properties and thermometers are shown in the table below: Thermometer substance Thermometer properties Thermometer Mecury Volume of mecury in capillary tube Mecury in glass thermometer Alcohol Volume of alcohol in capillary tube Alcohol in glass thermometer
Metal Electrical resistance of wire Electrical resistance thermometer Junctions of the two dissimilar metals Electromotive force (E.M.F) Thermo couple or thermoelectric thermometer Radiation Colour or wave-length of the radiation emitted Pyrometer Gas out of a constant volume Pressure of the gas Constant volume gas thermometer Constant – volume gas thermometer is the most accurate thermometer. It uses the pressure of a gas at a constant volume. However, it is very bulky and very difficult to operate. In addition, it cannot respond quickly to temperature change, hence, it cannot respond in measuring changing temperatures. Thermoelectric thermometer or thermocouple; is less accurate and responds quickly to temperature change, hence it can be used to measure a rapidily changing temperature. Pyrometer; - This measures very high temperature, like the temperature of the sun or furnance. It uses the colour or wave length or the radiation emitted from the interior of the body whose temperature is being measured. Liquid in glass thermometer; - This uses liquid as thermometric substance. This liquid is called thermometric liquid. CHARACTERRISTICS OF A GOOD THERMOMETRIC LIQUID (i) It must be visible or opaque (ii) It must be a good conductor of heat (iii) It does not wet glass (iv) It must have low specific heat capacity (v) It must not have anomalous expansion How to construct a simple liquid – in – glass thermometer (1) The thermometric liquid is mecury. Place the mecury in a small spherical bulb attached to a long, thin capillary tube at the top. (2) Dip the bulb in a bath or container of pure melting ice at the normal atmosphere 760mmty. When the mecury in the glass tube is at the ice, the level of the mecury in the tube would remain constant. Mark this point on the tube and assign zero degree centigrade (0^0 c). This point is called the ice point and it is the lower fixed point of the thermometer.
Heat capacity and specific heat capacity Generally, when a substance absorbs heat energy, its temperature would rise. The amount of heat needed to raise the temperature of a substance through a certain range is different from different substance. Heat capacity; This is also known as thermal capacity. It is defined as the amount of heat required to change the temperature of a substance by one kelvin (k) Heat capacity, c = Quantity of heat change ∈ temperature Let Q = quantity of heat (J) Ɵ = change in temperature. Cs = Q Ɵ = J K = JK ❑ − 1 The unit of heat capacity in Joules is per Kelvin Q = C Ɵ Examples; Find the quantity of heat absorbed by a calorimeter of heat capacity 205Jk
- , when its temperature rises from 20 o c to 50 o c. Solution , C = 205Jk
let Ɵ 1 = initial temperature, Ɵ 2 = find the temperature Ɵ = Ɵ 2 – Ɵ 2 = change in temperature Q =? By formula C =^ Q Ɵ Q = C Ɵ Q = C (Ɵ 2 – Ɵ 1 ) Q = 205 (50 – 20) Q = 205 (30) = 6150J Q = 6.15 x 10 3 J Specific heat capacity This is also known as the specific thermal capacity. The specific heat capacity is defined as the amount of heat required to change the temperature of a unit mass of a body by one kelvin (k). specific heat capacity = Quantity of Heat Mass x Temperature change C = Q MxƟ
J KgxK C = Jkg-1^ k-1^ (unit) Also Q = McƟ
From the formula, it can be seen that the quantity of heat absorbed by a substance depends on; (i) The mass (ii) Specific heat capacity (iii) Change in temperature Note; When the temperature of a body is rising, Ɵ = Ɵ 2 – Ɵ 1 but when the temperature is falling, Ɵ, = Ɵ 1 – Ɵ 2 E.g. How much heat is given out when a specific iron of mass 50g and specific heat capacity 460Jkg -1^ k -1^ cools from 85^0 c to 25^0 c. Solution m = 50g = 50 1000
5 100 kg, c = 460Jkg -1^ k- Ɵ 1 = 85 0 c, Ɵ 2 = 25 0 c Ɵ = Ɵ 1 – Ɵ 2 (cooling) Ɵ = 85 0 c – 25 0 c = 60 0 c By the aforementioned formula, Q = mc Ɵ Q = 5 100 460 x 60 Q = 5x46x Q = 1380J Q = 1.38x10^3 J 4000J of heat raises the temperature of 50kg of a substance from 20^0 c to 30^0 c. Find the specific heat capacity of the substance. Solution Q = 4000J Ɵ 1 = 20^0 c, Ɵ 2 = 30^0 c Ɵ = Ɵ 2 – Ɵ 1 (rising) c =? m = 50g = 50 1000 = 0.05kg c = Q Mx Ɵ
4000 0.05 x ( 30 − 20 ) c = Q Mx Ɵ = 4000 0.05 x ( 30 − 20 ) c = 4000 0.05 x 10 c = 8000J kg
The thermal capacity of the calorimeter and stirrer is 105Jk
- . If the final temperature of the mixture is 35^0 c, determine the temperature of the oven. Specific heat capacity of water = 4200Jkg
k
Solution Let the required temperature of the oven = Ɵ^0 c change in temperature of metal = (Ɵ –
0 c Heat loss by metal = mcƟ = 400 1000 x 500 x ( Ɵ − 35 ) = 200 (Ɵ – 35) J. Heat capacity of the calorimeter and stirrer = 105Jk- Heat gained by calorimeter and stirrer = cƟ = 105(35 – 30) = 105(5) = 525J Heat gained or absorbed by water = mcƟ = 1000 1000 x 4200 x ( 35 − 30 ) = 1 x 4200(5) = 2100J. but the total heat loss = total heat gained mcƟ = mcƟ + cƟ metal water calorimeter 200(Ɵ – 35) = 2100 + 525 200Ɵ – 7000 = 21525 200Ɵ = 21525 + 7000 200Ɵ = 28525 Ɵ = 28525 200 Ɵ = 142. Ɵ≈ 1430 c (to the nearest whole number/degree) A 300g iron shot at 95 0 c is quickly transferred into a lagged copper can with a stirrer of total heat capacity 60J/k containing 500g of water at 28^0 c. Calculate the final steady temperature of the mixture if the specific thermal capacity of iron is 460J/kgk and the specific heat capacity of wate is 4200J/kgk. Solution Using the equation 1 mcƟ = mcƟ + mcƟ iron water copper can Let the final steady temperature be x 300 1000 x 460 (95^0 – x) = 500 1000 x 4200 (x – 28) + 60 (x – 28) 3 x 46 (95 – x) = 5 x 420 (x – 28) + 60 (x – 28) 138 (95 – x) = 2100 (x – 28) + 60 (x – 28) 13110 – 138x = 2100 – 5880 + 60x (x – 28) 13100 + 58800 + 1680 = 2100x + 138x
73590 = 2298x x = 73590 2298
x = 32 0 c A piece of copper of mass 50g at temperature 250 0 c is placed in a copper calorimeter of mass 70g, containing 60g of water at 20^0 c. Ignoring heat loses, what would be the final steady temperature of the mixture? specific heat of copper = 400J/kgk. specific heat capacity of water = 4200J/kgk Solution Note; the mass of the container is given. Hence, the equation becomes; mcƟ = mcƟ + mcƟ copper water calorimeter d 50 1000 x 400 (250 – x) = 60 100 x 4200 (x - 20 0 ) + 70 100 x 400 (x – 20) Where x is the requires find steady temperature of the mixture. Multiplying the above equation by 1000, we have 50 x 400 (250 – x) = 60 x 4200 (x – 20) + 70 x 400 (x – 20) Dividing through by 1000. 5 x 4 (250 – x) = 6x4 (x – 20) + 7x4 (x – 20) 20 (250 – x) = 252 (x – 20) + 28 (x – 20) 5000 – 20x = 252x – 5040 + 28x – 560 5000 + 5040 + 560 = 252x + 28x + 20x 10600 = 300x x = 10600 300
106 3 x = 35.3^0 c A piece of copper mass 120g is heated in an endosure to a temperature of 125^0 c. It is then taken out of the enclosure and held in the air for half minute and dropped carefully in the copper calorimeter of mass 105g containing 200g of water at 20^0 c. The temperature of water rises to 250 0 c. calculate the rate at which the heat is being lost from the copper when it is held in the air. specific heat capacity of copper = 400Jkg-1k-1. Specific heat capacity of water = 400Jkg
- k - Solution mass of water = 200g = 200 1000 = 0.2kg C for water = 4200Jkg
k
change in temperature of water, Ɵ = 25^0 c – 20^0 c = 5^0 c
