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.

Hint:

Use: \[ \alpha = \frac{R_2 - R_1}{R_1 \Delta T} \] Works directly when reference temperature is known.

Answer 1

Concept: Resistance variation with temperature: \[ R_t = R_0 (1 + \alpha \Delta T) \] Where:

\( \alpha \) = temperature coefficient of resistance
\( \Delta T = T_2 - T_1 \)

Step 1: Given data. \[ R_1 = 1.05 \, \Omega \text{ at } 20^\circ \text{C} \] \[ R_2 = 1.38 \, \Omega \text{ at } 100^\circ \text{C} \] \[ \Delta T = 100 - 20 = 80^\circ \text{C} \]
Step 2: Use linear relation. \[ R_2 = R_1 (1 + \alpha \Delta T) \] \[ 1.38 = 1.05 (1 + 80\alpha) \]
Step 3: Solve for \( \alpha \). \[ \frac{1.38}{1.05} = 1 + 80\alpha \] \[ 1.314 = 1 + 80\alpha \] \[ 80\alpha = 0.314 \] \[ \alpha = \frac{0.314}{80} = 3.93 \times 10^{-3} \, ^\circ\text{C}^{-1} \] Final Answer: \[ \boxed{\alpha \approx 3.9 \times 10^{-3} \, ^\circ\text{C}^{-1}} \]