Temperature — Question 3

Question 3

A mean reading does not create a uniform temperature

A thin rod of length $L=0.40\,\mathrm m$ has a maintained local temperature field $T(x)=300\,\mathrm K+(50\,\mathrm{K/m})x$, for $0\le x\le L$. Each small region is internally equilibrated; the entire rod is not. Define its length average by $\overline T=L^{-1}\int_0^L T(x)\,dx$.

1. Find the endpoint temperatures, average and temperature difference between $x=0.10$ and $0.30\,\mathrm m$.

2. A report replaces the whole field by $\overline T$. Find its largest pointwise error and explain what state information is lost.

Question 3 - Solution

A mean reading does not create a uniform temperature

(a) Method and calculation

$$T(0)=300\,\mathrm K,\qquad T(L)=320\,\mathrm K.$$$$\overline T=300+\frac{50}{L}\frac{L^2}{2}=310\,\mathrm K,\qquad T(.30)-T(.10)=10\,\mathrm K.$$The integral average is a useful summary, not an equilibrium assertion.

(b) Interpretation and further calculation

The replacement error is$$T(x)-\overline T=50x-10\quad\text{in kelvins}.$$$$\max_{0\le x\le .40}|T(x)-\overline T|=10\,\mathrm K.$$The average loses the gradient and local extremes. A single temperature describes the complete rod only when spatial uniformity is justified.

Physical check

The average lies between 300 and 320 K, as required for an average with positive weights.