The Celsius and Fahrenheit Scales — Question 6

Question 6

Affine conversion commutes with a correctly weighted mean

Use numerical readings $F=\frac95C+32$ and $T/\mathrm K=C+273.15$; these conversions are exact. A process stays at $20^\circ\mathrm C$ for $10$ minutes and at $80^\circ\mathrm C$ for $30$ minutes. Define the time average by $\overline C=(\sum_i C_i\Delta t_i)/(\sum_i\Delta t_i)$.

1. Compute the time-averaged Celsius and Fahrenheit readings in two ways: average first, and convert each stage first.

2. Compare with the unweighted average of the two Fahrenheit readings. Explain algebraically why conversion commutes with normalized weighting but cannot fix incorrect weights.

Question 6 - Solution

Affine conversion commutes with a correctly weighted mean

(a) Method and calculation

$$\overline C=\frac{20(10)+80(30)}{40}=65,\qquad\overline F=1.8(65)+32=149.$$$$F_1=68,\quad F_2=176,\quad\frac{68(10)+176(30)}{40}=149.$$

(b) Interpretation and further calculation

$$\frac{68+176}{2}=122\ne149.$$$$\sum_i w_i(1.8C_i+32)=1.8\sum_iw_iC_i+32\quad\text{if }\sum_iw_i=1.$$The unweighted mean describes equal residence times, not this process. A unit conversion preserves the chosen weighting rather than correcting it.

Physical check

The mean $65^\circ$C lies nearer the longer-lived $80^\circ$C stage.