Measuring Temperature — Question 5

Question 5

Measuring current can warm a resistance thermometer

A sensor dissipates electrical power $P=I^2R$ and has thermal conductance $G=.020\,\mathrm{W/K}$ to its bath. Assume $R=100\,\Omega$ is constant over the small self-heating rise and steady balance gives $G(T_s-T_b)=P$.

1. Find the temperature rise at $I=1.0$ and $3.0\,\mathrm{mA}$. Determine the largest current for a rise no greater than $.010\,\mathrm K$.

2. Indicated temperatures are $300.005$ K at $1$ mA and $300.045$ K at $3$ mA. Extrapolate linearly in $I^2$ to zero current to infer $T_b$.

Question 5 - Solution

Measuring current can warm a resistance thermometer

(a) Method and calculation

$$T_s-T_b=\frac{I^2R}{G};\quad\Delta T_1=.005\,\mathrm K,\quad\Delta T_3=.045\,\mathrm K.$$$$I_{\max}=\sqrt{\frac{G(.010\,\mathrm K)}R}=1.414\,\mathrm{mA}.$$

(b) Interpretation and further calculation

With $j=I/(1\,\mathrm{mA})$, write $T_s=T_b+A j^2$.$$A=\frac{300.045-300.005}{9-1}=.005\,\mathrm K.$$$$T_b=300.005-.005(1)=300.000\,\mathrm K.$$The extrapolation requires the same bath and the stated quadratic heating model.

Physical check

Triple current gives nine times the heating; reducing current changes both disturbance and electrical signal strength.