Question

When a metal rod is heated, its resistance changes both because of a change in resistivity and because of a change in the length of the rod. If a silver rod has a resistance of 1.70 Ω at 21.0°C, what is its resistance when it is heated to 180.0°C? The temperature coefficient for silver is α = 6.1 ✕ 10−3 °C−1, and its coefficient of linear expansion is 18 ✕ 10−6 °C−1. Assume that the rod expands in all three dimensions.

Answer 1

Answer:

$3.34\Omega$

Explanation:

The resistance of a metal rod is given by

$R=\frac{\rho L}{A}$

where

$\rho$ is the resistivity

L is the length of the rod

A is the cross-sectional area

The resistivity changes with the temperature as:

$\rho(T)=\rho_0 (1+\alpha (T-T_0))$

where in this case:

$\rho_0$ is the resistivity of silver at $T_0=21.0^{\circ}C$

$\alpha=6.1\cdot 10^{-3} ^{\circ}C^{-1}$ is the temperature coefficient for silver

$T=180.0^{\circ}C$ is the current temperature

Substituting,

$\rho(180^{\circ}C)=\rho_0 (1+6.1\cdot 10^{-3}(180-21))=1.970\rho_0$

The length of the rod changes as

$L(T)=L_0 (1+\alpha_L(T-T_0))$

where:

$L_0$ is the initial length at $21.0^{\circ}C$

$\alpha_L = 18\cdot 10^{-6} ^{\circ}C^{-1}$ is the coefficient of linear expansion

Substituting,

$L(180^{\circ}C)=L_0(1+18\cdot 10^{-6}(180-21))=1.00286L_0$

The cross-sectional area of the rod changes as

$A(T)=A_0(1+2\alpha_L(T-T_0))$

So, substituting,

$A(180^{\circ}C)=A_0(1+2\cdot 18\cdot 10^{-6}(180-21))=1.00572A_0$

Therefore, if the initial resistance at 21.0°C is

$R_0 = \frac{\rho_0 L_0}{A_0}=1.70\Omega$

Then the resistance at 180.0°C is:

$R(180^{\circ}C)=\frac{\rho(180)L(180)}{A(180)}=\frac{(1.970\rho_0)(1.00285L_0)}{1.00572A_0}=1.9644\frac{\rho_0 L_0}{A_0}=1.9644 R_0=\\=(1.9644)(1.70\Omega)=3.34\Omega$