The Celsius and Fahrenheit Scales — Question 9

Question 9

A coefficient per degree changes inversely with degree size

Use numerical readings $F=\frac95C+32$ and $T/\mathrm K=C+273.15$; these conversions are exact. A rod obeys $\Delta L/L_0=\alpha_C\Delta C$ with $\alpha_C=18.0\times10^{-6}$ per Celsius degree. The initial length is $2.00\,\mathrm m$ and the temperature rises from $68$ to $158^\circ\mathrm F$.

1. Find the expansion and the coefficient $\alpha_F$ needed when the numerical temperature difference is in Fahrenheit degrees.

2. A program uses the original coefficient with the Fahrenheit difference. Find the predicted expansion, error and multiplicative error factor.

Question 9 - Solution

A coefficient per degree changes inversely with degree size

(a) Method and calculation

$$\Delta F=90,\quad\Delta C=90/1.8=50,\quad\Delta L=2(18\times10^{-6})(50)=1.80\,\mathrm{mm}.$$$$\alpha_F\Delta F=\alpha_C\Delta C\ \Longrightarrow\ \alpha_F=\alpha_C/1.8=10.0\times10^{-6}/{}^\circ\mathrm F.$$

(b) Interpretation and further calculation

$$\Delta L_{\mathrm{wrong}}=2(18\times10^{-6})(90)=3.24\,\mathrm{mm}.$$$$\Delta L_{\mathrm{wrong}}-\Delta L=1.44\,\mathrm{mm},\qquad\frac{\Delta L_{\mathrm{wrong}}}{\Delta L}=1.8.$$A response coefficient is tied to the temperature-interval unit in its denominator.

Physical check

Kelvin and Celsius intervals have equal sizes, so the numerical coefficient per kelvin equals $\alpha_C$.