Question

A silver wire has a resistance of 40 ohms at 19 degrees Celsius. After several hours of heating, the resistance of the coil has increased to 60 ohms. What is the new temperature of the wire?

Answer 1

The electrical resistance of a conductor at temperature $T$ :

$R_T=R_r(1+\alpha(T-T_r))$

where $R_r$ is resistance of conductor at reference temperature $T_r$ ,

$\alpha$ is temperature coefficient of resistance at reference temperature $T_r$

$\alpha=0.003819$ for silver at $T_r=20\degree C$

Then:

$R_r=\frac{R_{19}}{1+\alpha(19-20)}$

$R_r=\frac{40}{1+0.003819\cdot(19-20)}=40.15$ ohms

The new temperature of the wire:

$T=(R_T/R_r-1)/\alpha+T_r$

$T=(60/40.15-1)/0.003819+20=149.46\degree C$