Thermal Conductivity of a Good Conductor - Searle's Method
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Heat is transferred from one end of the medium to another end by conduction. The rate of heat
flow is proportional to the area, A and the temperature gradient, \(\frac{dT}{dx}\), i.e,
$$\frac{Q}{t} = kA(\frac{dT}{dx})$$
Here, k is the proportionality constant (Thermal Conductivity).
Q is the heat supplied to the bar in time t,
A is the cross-sectional area of the bar,
dT is the difference in temperature between two points in the bar dx apart,
k is the coefficient of thermal conductivity of the bar.
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In general, the heat flown through the medium is used to raise the temperature of water of mass
M by \(\Delta T\) \(^\circ\) C, i.e,
$$Q = M.\Delta T$$
In the apparatus used by you, the heat Q warms up a mass m (in kilograms) of water from
temperature \(T_4\) to \(T_3\) according to the formula:
\(Q = mc (T_3 - T_4))
where ‘c' is the specific heat capacity of water (c = 4190 \(J kg^{-1} K^{-1}).
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At equilibrium, heat transferred is equal to the heat flown.
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Using : \(-dT = T_1 - T_2, dx=d\) , (d in meters), and \(A = \frac{\pi D^2}{4}\) (A in meters
squared) we obtain:
$$mc(T_3 - T_4) = k (\frac{\pi D^2}{4})(\frac{T_1-T_2}{d})t$$
$$k = \frac{4mcd(T_3 - T_4)}{\pi D^2 (T_1 - T_2)} (in Wm^{-1}K^{-1})$$