Question

Find the temperature at which the resistance of a wire made of silver will be twice its resistance at $20^\circ\text{C$. Take $20^\circ\text{C}$ as the reference temperature and the temperature coefficient of resistance of silver at $20^\circ\text{C}$ as $4.0 \times 10^{-3} \, \text{K}^{-1}$.}

Hint:

To calculate the temperature at which resistance changes, always use the formula $R_T = R_0 (1 + \alpha \Delta T)$ and solve for $\Delta T$.

Answer 1

The resistance of a material varies with temperature according to the formula: \[ R_T = R_0 (1 + \alpha \Delta T), \] where: \begin{itemize} \item $R_T$ is the resistance at temperature $T$, \item $R_0$ is the resistance at the reference temperature $T_0$, \item $\alpha$ is the temperature coefficient of resistance, \item $\Delta T = T - T_0$ is the temperature difference. \end{itemize} Given that $R_T = 2R_0$, $\alpha = 4.0 \times 10^{-3} \, \text{K}^{-1}$, and $T_0 = 20^\circ\text{C}$, substitute these values: \[ 2R_0 = R_0 (1 + \alpha \Delta T). \] Cancel $R_0$ on both sides: \[ 2 = 1 + \alpha \Delta T. \] Rearrange to find $\Delta T$: \[ \Delta T = \frac{2 - 1}{\alpha} = \frac{1}{4.0 \times 10^{-3}} = 250 \, \text{K}. \] Calculate the temperature $T$: \[ T = T_0 + \Delta T = 20 + 250 = 270^\circ\text{C}. \] Thus, the temperature at which the resistance of the wire is twice its resistance at $20^\circ\text{C}$ is: \[ \boxed{270^\circ\text{C}}. \]