The Celsius and Fahrenheit Scales — Question 10

Question 10

A negative Celsius reading is not a negative absolute temperature

Use numerical readings $F=\frac95C+32$ and $T/\mathrm K=C+273.15$; these conversions are exact. For the ordinary equilibrium systems in this exercise require $T\ge0\,\mathrm K$. A logger reports $-500^\circ\mathrm F$. A separate sensor reports $-273.10\pm.10^\circ\mathrm C$ as a hard interval before imposing that physical constraint.

1. Convert the logger value to Celsius and kelvin and assess its admissibility in the stipulated domain.

2. Convert the second interval to kelvins, then intersect it with the allowed domain. Give the resulting interval and explain why it is no longer symmetric about the reported center.

Question 10 - Solution

A negative Celsius reading is not a negative absolute temperature

(a) Method and calculation

$$C=\frac{-500-32}{1.8}=-295.5556,\qquad T=-22.4056\,\mathrm K.$$$$T<0\quad\Longrightarrow\quad\text{outside the stipulated domain}.$$A negative Celsius or Fahrenheit number alone would not be sufficient to reject a reading.

(b) Interpretation and further calculation

$$C\in[-273.20,-273.00]\ \Longrightarrow\ T\in[-.05,.15]\,\mathrm K.$$$$[ -.05,.15]\cap[0,\infty)=[0,.15]\,\mathrm K.$$The physical constraint clips the lower side. The reported center remains $.05$ K, but the feasible deviations are $-.05$ and $+.10$ K.

Physical check

The result is a constrained feasible interval, not a new symmetric measurement error model.