Question

The resistance of a metal wire at \(20^\circ\text{C}\) is \(1.05\,\Omega\) and at \(100^\circ\text{C}\) is \(1.38\,\Omega\). Determine the temperature coefficient of resistivity of this metal.

Answer 1

Temperature Coefficient of Resistivity

Step 1: Use the relation between resistance and temperature. \[ R_t = R_0(1+\alpha t). \] For two temperatures, \[ R_2 = R_1[1+\alpha(T_2-T_1)]. \] Given, \[ R_1=1.05\Omega \] at \[ T_1=20^\circ C, \] and \[ R_2=1.38\Omega \] at \[ T_2=100^\circ C. \]

Step 2: Substitute values. \[ 1.38 = 1.05[1+\alpha(100-20)]. \] \[ 1.38 = 1.05(1+80\alpha). \] Dividing by \(1.05\), \[ 1.3143 = 1+80\alpha. \] \[ 80\alpha = 0.3143. \] \[ \alpha = \frac{0.3143}{80}. \] \[ \alpha = 3.93\times10^{-3}\,^\circ C^{-1}. \] Final Answer: \[ \boxed{ \sigma = \frac{ne^2\tau}{m} } \] and \[ \boxed{ \alpha = 3.93\times10^{-3}\,^\circ C^{-1} } \]