Question

The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law:

$R=R_o[1+\alpha (T-T_o\left)\right]$
The resistance is 101.6 $\Omega$ at the triple-point of water 273.16K, and 165.5 $\Omega$ at the normal melting point of lead (600.5 K). What is the temperature when the resistance is 123.4 $\Omega$ ?

Answer 1

We first find $\alpha$ by putting the given values,
$165.5=101.6(1+\alpha (600.5-273.16\left)\right)$
$\therefore \alpha =1.92\times {10}^{-3}$
Now, Temp when Resistance is 123.4 is
$123.4=101.6(1+\alpha (T-273.16\left)\right)$
$T=384.91K$