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Thermal conductivity of copper

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• 1. 15: THERMAL CONDUCTIVITYAim (1) To determine the thermal conductivity k of copper using a modified version of Searles method. (2) Compare values for k obtained using two different techniques; compare theory and experiment.IntroductionIf two opposite faces of a block of material are maintained at a constant difference in temperature, thenthe amount of heat energy flowing through the block in unit time, i.e. the power dissipated, is given by: ΔQ kA ( θ1 − θ 2 ) P= = … (15.1) Δt dwhere A is the area of the faces, d is the length of the block, θ1 and θ2 are the temperatures of the twofaces, and k is a constant, called the thermal conductivity, which depends on the material of the block.Searles apparatus is used to measure k for a material. A modified version of Searles apparatus is shownin Figure 15.1. The material whose thermal conductivity is to be determined (a copper bar in our case) isheated electrically at one end and cooled by water circulating through a cavity at the other. The cavityhas water inlet and outlet tubes which each have a support tube for a thermometer (Th3 and Th4) so thatthe rise in temperature of the water passing through the cavity can be determined.In the steady state, the temperature gradient along the bar is measured by means of two thermometers(Th1 and Th2) placed in support tubes fixed to the surface of the bar. The apparatus is lagged to minimiseheat losses from the bar.In Searles method, it is assumed that all the heat enters one end of the bar, is conducted along the bar,and is extracted at the other end. This gives us two methods for calculating the thermal conductivity kfrom equation (15.1): (A) by determining the rate at which heat is extracted from the bar; and (B) bydetermining the rate at which the heat is supplied to the bar. 15–1
• 2. Figure 15.1. Searles apparatus.Method (A): Heat extracted from bar. Assume that T1, T2, T3 and T4 are the readings of thethermometers Th1, Th2, Th3 and Th4 in a steady state, and that a mass M of water (of specific heat C)emerges from the cavity in a time Δt. Then the rate at which the water extracts energy from the bar is: ΔQ M (C 3T3 − C4T4 ) MC (T3 − T4 ) C + C4 = ≈ , where C = 3 … (15.2). Δt Δt Δt 2Substituting equation (15.2) into equation (15.1), and using A = πD2/4 where D is the diameter of the rod,we obtain: 4MCd (T 3 − T 4 ) kA = … (15.3). π D 2 (T 1 − T 2 ) ΔtMethod (B): Heat supplied to the bar. Since the bar is electrically heated, we can calculate the rate ofsupply of energy to the bar. Thus if a current I passes through the heater under a potential difference V,the rate of supply of energy is: ΔQ = VI … (15.4). ΔtSubstituting equation (15.4) into equation (15.1) and rearranging as before, we get 4VId kB = … (15.5). πD 2 (T 1 − T 2 )Ideally the values for the thermal conductivity of copper obtained by the two methods, kA and kB, shouldbe the same. 15–2