Measuring Temperature — Question 6

Question 6

Two transient readings identify a sensor time constant

A thermometer obeys $T_s(t)=T_b+(T_0-T_b)e^{-t/\tau}$ after entering a maintained bath at $T_b=330\,\mathrm K$. It starts at $T_0=290\,\mathrm K$ and reads $310\,\mathrm K$ at $t=4.0\,\mathrm s$.

1. Infer $\tau$ and predict the reading at $8.0\,\mathrm s$.

2. Find the waiting time required for $|T_s-T_b|\le.50\,\mathrm K$. Does waiting one time constant satisfy this condition?

Question 6 - Solution

Two transient readings identify a sensor time constant

(a) Method and calculation

$$\frac{330-310}{330-290}=e^{-4/\tau}=\frac12,\qquad\tau=\frac4{\ln2}=5.771\,\mathrm s.$$$$T_s(8)=330-40(1/2)^2=320\,\mathrm K.$$

(b) Interpretation and further calculation

$$40e^{-t/\tau}\le.50\ \Longrightarrow\ t\ge\tau\ln80=25.288\,\mathrm s.$$$$|T_s(\tau)-T_b|=40/e\approx14.715\,\mathrm K.$$A time constant measures fractional relaxation, not a universally adequate waiting time.

Physical check

At each additional four seconds the residual halves under the supplied single-time-constant model.