Answer:
The temperature of silver at this given resistivity is 2971.1 ⁰C
Explanation:
The resistivity of silver is calculated as follows;
![R_t = R_o[1 + \alpha(T-T_o)]\\\\](https://tex.z-dn.net/?f=R_t%20%3D%20R_o%5B1%20%2B%20%5Calpha%28T-T_o%29%5D%5C%5C%5C%5C)
where;
Rt is the resistivity of silver at the given temperature
Ro is the resistivity of silver at room temperature
α is the temperature coefficient of resistance
To is the room temperature
T is the temperature at which the resistivity of silver will be two times the resistivity of iron at room temperature
![R_t = R_o[1 + \alpha(T-T_o)]\\\\\R_t = 1.59*10^{-8}[1 + 0.0038(T-20)]](https://tex.z-dn.net/?f=R_t%20%3D%20R_o%5B1%20%2B%20%5Calpha%28T-T_o%29%5D%5C%5C%5C%5C%5CR_t%20%3D%201.59%2A10%5E%7B-8%7D%5B1%20%2B%200.0038%28T-20%29%5D)
Resistivity of iron at room temperature = 9.71 x 10⁻⁸ ohm.m
When silver's resistivity becomes 2 times the resistivity of iron, we will have the following equations;
![R_t,_{silver} = 2R_o,_{iron}\\\\1.59*10^{-8}[1 + 0.0038(T-20)] =(2 *9.71*10^{-8})\\\\\ \ (divide \ through \ by \ 1.59*10^{-8})\\\\1 + 0.0038(T-20) = 12.214\\\\1 + 0.0038T - 0.076 = 12.214\\\\0.0038T +0.924 = 12.214\\\\0.0038T = 12.214 - 0.924\\\\0.0038T = 11.29\\\\T = \frac{11.29}{0.0038} \\\\T = 2971.1 \ ^0C](https://tex.z-dn.net/?f=R_t%2C_%7Bsilver%7D%20%3D%202R_o%2C_%7Biron%7D%5C%5C%5C%5C1.59%2A10%5E%7B-8%7D%5B1%20%2B%200.0038%28T-20%29%5D%20%3D%282%20%2A9.71%2A10%5E%7B-8%7D%29%5C%5C%5C%5C%5C%20%5C%20%28divide%20%5C%20through%20%5C%20by%20%5C%201.59%2A10%5E%7B-8%7D%29%5C%5C%5C%5C1%20%2B%200.0038%28T-20%29%20%3D%2012.214%5C%5C%5C%5C1%20%2B%200.0038T%20-%200.076%20%3D%2012.214%5C%5C%5C%5C0.0038T%20%2B0.924%20%3D%2012.214%5C%5C%5C%5C0.0038T%20%20%3D%2012.214%20-%200.924%5C%5C%5C%5C0.0038T%20%3D%2011.29%5C%5C%5C%5CT%20%3D%20%5Cfrac%7B11.29%7D%7B0.0038%7D%20%5C%5C%5C%5CT%20%3D%202971.1%20%5C%20%5E0C)
Therefore, the temperature of silver at this given resistivity is 2971.1 ⁰C