Question

A resistance thermometer reads R = 20.0 $\Omega$, 27.5 $\Omega$, and 50.0 $\Omega$ at the ice point ($0^{\circ }C$), the steam point (${100}^{\circ }C$) and the zinc point (${420}^{\circ }C$) respectively. Assuming that the resistance varies with temperature as $R_{\theta }=R_0(1+\alpha \theta +\beta {\theta }^2)$, find the values of $R_0,\alpha$ and $\beta$. Here $\theta$ represents the temperature on Celsius scale.

Answer 1

R at ice point $\left(R_0\right)=20\Omega$

R at steam point $\left(R_{100}\right)=27.5\Omega$

R at zinc point $\left(R_{420}\right)=50\Omega$

$R_{\theta }=R_0(1+\alpha \theta +\beta {\theta }^2)$

$\Rightarrow R_{100}=R_0+R_0\alpha \theta +R_0\beta {\theta }^2$

$\Rightarrow \frac{(R_{100}-R_0)}{R_0}=\alpha \theta +\beta {\theta }^2$

$\Rightarrow \frac{27.5-20}{20}=\alpha \theta +\beta {\theta }^2$

$\Rightarrow \frac{7.5}{20}=\alpha 100+\beta 10000\cdots \left(i\right)$

Again $R_{420}=R_0(1+\alpha \theta +\beta {\theta }^2)$

$\Rightarrow \frac{50-R_0}{R_0}=\alpha \theta +\beta {\theta }^2$

$\frac{50-20}{20}=420\alpha +176400\beta$

$\Rightarrow \frac{3}{2}=420\alpha +176400\beta \cdots \left(ii\right)$

After solving (i) and (ii), we get

$\alpha =3.5\times {10}^{-3}{/}^{\circ }C$

$\beta =-5.6\times {10}^{-7}{/}^{\circ }C$