determine at what temperature aluminum will have the same resistivity as tungsten does at 20°c. express your…

Explanation:

Step1: Recall resistivity - temperature formulas

The resistivity of a material as a function of temperature is given by $\rho=\rho_0(1 + \alpha(T - T_0))$, where $\rho$ is the resistivity at temperature $T$, $\rho_0$ is the resistivity at reference - temperature $T_0$, and $\alpha$ is the temperature coefficient of resistivity. The resistivity of tungsten at $T_{0W}=20^{\circ}C$ is $\rho_{W0}=5.6\times10^{-8}\Omega\cdot m$ and its temperature - coefficient of resistivity $\alpha_W = 0.0045/^{\circ}C$. The resistivity of aluminum at a reference temperature $T_{0Al}=20^{\circ}C$ is $\rho_{Al0}=2.65\times10^{-8}\Omega\cdot m$ and its temperature - coefficient of resistivity $\alpha_{Al}=0.00429/^{\circ}C$. We want to find $T$ such that $\rho_{Al}=\rho_{W0}$. So, $\rho_{Al0}(1+\alpha_{Al}(T - T_{0Al}))=\rho_{W0}$.

Step2: Rearrange the equation to solve for $T$

First, expand the left - hand side: $\rho_{Al0}+\rho_{Al0}\alpha_{Al}(T - T_{0Al})=\rho_{W0}$. Then, isolate the term with $T$: $\rho_{Al0}\alpha_{Al}(T - T_{0Al})=\rho_{W0}-\rho_{Al0}$. Next, $T - T_{0Al}=\frac{\rho_{W0}-\rho_{Al0}}{\rho_{Al0}\alpha_{Al}}$. Finally, $T=T_{0Al}+\frac{\rho_{W0}-\rho_{Al0}}{\rho_{Al0}\alpha_{Al}}$.

Step3: Substitute the known values

Substitute $\rho_{Al0}=2.65\times10^{-8}\Omega\cdot m$, $\alpha_{Al}=0.00429/^{\circ}C$, $\rho_{W0}=5.6\times10^{-8}\Omega\cdot m$, and $T_{0Al}=20^{\circ}C$ into the formula. [ \begin{align*} T&=20+\frac{5.6\times 10^{-8}-2.65\times 10^{-8}}{2.65\times 10^{-8}\times0.00429}\ &=20+\frac{2.95\times 10^{-8}}{2.65\times 10^{-8}\times0.00429}\ &=20+\frac{2.95}{2.65\times0.00429}\ &=20+\frac{2.95}{0.0113685}\ &=20 + 260.3\ &\approx280^{\circ}C \end{align*} ]

Answer:

$280^{\circ}C$