Temperature — Question 4

Question 4

A nonlinear indicator can preserve order but distort intervals

An ideal thermal indicator reports $z=(T/100\,\mathrm K)^2$ for $T>0$. Two baths give $z_A=4$ and $z_B=16$. The indicator is reproducible and strictly increasing.

1. Infer both temperatures and their difference. Does $z_B/z_A$ equal $T_B/T_A$?

2. Find the temperature at the arithmetic mean indicator reading. Compare it with the arithmetic mean temperature and explain the distinction.

Question 4 - Solution

A nonlinear indicator can preserve order but distort intervals

(a) Method and calculation

Invert the positive branch:$$T=100\,\mathrm K\sqrt z,\quad T_A=200\,\mathrm K,\quad T_B=400\,\mathrm K.$$$$T_B-T_A=200\,\mathrm K,\qquad z_B/z_A=4\ne T_B/T_A=2.$$An ordered numerical indicator is not automatically an absolute temperature scale.

(b) Interpretation and further calculation

$$z_{\mathrm{mid}}=10,\qquad T(z_{\mathrm{mid}})=100\sqrt{10}\,\mathrm K\approx316.23\,\mathrm K.$$$$\frac{T_A+T_B}{2}=300\,\mathrm K.$$The $16.23\,\mathrm K$ difference comes from nonlinear conversion. Averaging and inversion do not commute.

Physical check

On $T>0$, $dz/dT>0$, so no ordering is reversed even though interval and ratio interpretations fail.