Question

A student measures the resistivity of a certain material at three different temperatures and gets the results summarized in the following table. 10-8 12. m 10-3 12-m | T2-100° C | P2-3.036 X 10-3 Ω. m Temperature То Resistivity 2.500 2.346 20° C | Po P1 If an equation of the form with To 20° C, is used to fit the data, determine the temperature (in K) in which this material becomes a super conductor, as predicted by the above equation.

Answer 1

rho = rho [1 + alpha(T - T_0) + beta (T - T_0)^2] T_0 = 293.15 K i.e., 20 degree C T_1 = 273.15 K i.e., 0 degree C T_2 = 373.15 K i.e., 100 degree C rho_1 = rho_0 [1 + alpha(T_1 - T_0) + beta(T_1 - T_0)^2] 2.346 times 10^-8 = 2.5 times 10^-8 [1 + alpha (-20) + beta (20)^2] 20 alpha - 400 beta = 0.0616 alpha - 20 beta = 3.08 times 10^-3 -equation(1) rho_2 = rho_0 [1 + alpha (T_2 - T_0) + beta (T_2 - T_0)^2] 3.036 times 10^-8 = 2.5 times 10^-8 [1 + 80 alpha + 80^2 beta] 80 alpha + 6400 beta = 0.2144 alpha + 80 beta = 2.68 times 10^-3 equation (2) equation(2) - equation (1) = 100 beta = -4 times 10^-4 beta = -4 times 10^-6 alpha = 2.68 times 10^-3 + 80 times 4 times 10^-6 = 3 times 10_3 alpha = 3 times 10^-3 beta = -4 times 10^-6 \r\n Let