Question

A platinum resistance thermometer has resistance of 52.5 ohms and 9.75 ohms at 0 degrees Celsius and 100 degrees Celsius respectively. When the resistance is 8.25 ohms, find the temperature

Answer 1

Task. A platinum resistance thermometer has resistance of $R_{0}=52.5$ ohms and $R_{1}=9.75$ ohms at $T_{0}=0$ degrees Celsius and $T_{1}=100$ degrees Celsius respectively. Find the temperature when the resistance is 8.25 ohms.

Solution. It is known that the dependence of the resistance on temperature is given by the formula:

$R=R_{0}(1-\alpha(T-T_{0}),$

where $R_{0}=52.5$ ohms is the resistance at temperature $T_{0}=0$ degrees Celsius. Let $R_{1}=9.75$ ohms be the resistance at $T_{1}=100$ degrees Celsius. So

$R_{1}=R_{0}(1-\alpha(T_{1}-T_{0}))$

This allows to find $\alpha$:

$\frac{R_{1}}{R_{0}}=1-\alpha(T_{1}-T_{0})$

$\alpha=\frac{1-R_{1}/R_{0}}{T_{1}-T_{0}}$

Substituting values we get:

$\alpha=\frac{1-9.75/52.5}{100-0}=\frac{0.81429}{100}=0.0081429.$

Therefore the temperature $T_{2}$ connresponding to resistance $R_{2}=8.25$ ohms can be found by the following formula:

$R_{2}=R_{0}(1-\alpha(T_{2}-T_{0}))$

$T_{2}=T_{0}+\frac{1-R_{2}/R_{0}}{\alpha}$

Thus

$T_{2}=T_{0}+\frac{1-R_{2}/R_{0}}{\alpha}=0+\frac{1-8.25/52.5}{0.0081429}=\frac{0.84286}{0.0081429}=103.51\,{}^{\circ}C.$