Chapter 28, Problem 75PQ

Review 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 2.00 Ω at 22.0°C, what is its resistance when it is heated to 200.0°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.

To determine

Resistance of the silver rod when it is heated to $200.0\text{°C}$.

Answer to Problem 75PQ

Resistance of the silver rod when it is heated to $200.0\text{°C}$ is $\underset{\_}{4.16\Omega }$.

Explanation of Solution

Given that the initial temperature of the rod is $22.0\text{°C}$, its initial resistance is $2.00\Omega$, the final temperature is $200.0\text{°C}$, the temperature coefficient of resistivity of silver is $6.1\times {10}^{-3}{\text{°C}}^{-\text{1}}$, and the its coefficient of linear expansion is $18\times {10}^{-6}°{\text{C}}^{-1}$.

Write the general expression for the resistance of a conductor.

  $R=\rho \frac{l}{A}$                                                                                                    (I)

Here, $R$ is the resistance, $l$ is the length of the conductor, and $A$ is the cross sectional area of the conductor.

Write the expression for the temperature dependence of resistivity.

  $\rho ={\rho }_0\left[1+\alpha \left(T-T_0\right)\right]$                                                                                (II)

Here, $\rho$ is the resistivity corresponding to a final temperature $T$, ${\rho }_0$ is the resistivity corresponding to a initial temperature $T_0$, and $\alpha$ is the temperature coefficient of resistivity.

Write the expression for the linear thermal expansion of material.

  $l=l_0\left[1+{\alpha }_l\left(T-T_0\right)\right]$                                                                              (III)

Here, $l$ is the length corresponding to a final temperature $T$, $l_0$ is the length corresponding to a initial temperature $T_0$, and ${\alpha }_l$ is the linear expansion coefficient.

Write the expression for the cross sectional area of the conductor.

  $A=\pi r^2$                                                                                                  (IV)

Here, $r$ is the radius of the conductor.

The linear thermal expansion affects the radius of the conductor in accordance with equation (III). Thus, the cross sectional area at a final temperature $T$ can be written as,

  $A=A_0{\left[1+{\alpha }_l\left(T-T_0\right)\right]}^2$                                                                              (V)

Here, $A_0$ is the initial cross sectional area.

Use equation (II), (III), and (V) in (I).

  $\begin{array}{c}R=\frac{{\rho }_0\left[1+\alpha \left(T-T_0\right)\right]l_0\left[1+{\alpha }_l\left(T-T_0\right)\right]}{A_0{\left[1+{\alpha }_l\left(T-T_0\right)\right]}^2}\\ =\left(\frac{{\rho }_0{l}_0}{A_0}\right)\frac{\left[1+\alpha \left(T-T_0\right)\right]}{\left[1+{\alpha }_l\left(T-T_0\right)\right]}\end{array}$                                               (VI)

Write the expression for the resistance of the conductor, corresponding to an initial temperature $T_0$ (analogous to equation (I)).

  $R_0=\frac{{\rho }_0{l}_0}{A_0}$                                                                                                   (VII)

Use equation (VII) in (VI).

  $R=R_0\frac{\left[1+\alpha \left(T-T_0\right)\right]}{\left[1+{\alpha }_l\left(T-T_0\right)\right]}$                                                                            (VIII)

Conclusion:

Substitute $22.0\text{°C}$ for $T_0$, $200.0\text{°C}$ for $T$, $2.00\Omega$ for $R_0$, $6.1\times {10}^{-3}{\text{°C}}^{-\text{1}}$ for $\alpha$, and $18\times {10}^{-6}°{\text{C}}^{-1}$ for ${\alpha }_l$ in equation (VIII) to find $R$.

  $\begin{array}{c}R=\left(2.00\Omega \right)\frac{\left[1+\left(6.1\times {10}^{-3}{\text{°C}}^{-\text{1}}\right)\left(200.0\text{°C}-22.0\text{°C}\right)\right]}{\left[1+\left(18\times {10}^{-6}°{\text{C}}^{-1}\right)\left(200.0\text{°C}-22.0\text{°C}\right)\right]}\\ =4.16\Omega \end{array}$

Therefore, resistance of the silver rod when it is heated to $200.0\text{°C}$ is $\underset{\_}{4.16\Omega }$.