Changing Units — Question 6

Question 6

Changing a temperature is not the same as converting a change

Let $C$ and $F$ denote the numerical readings on the Celsius and Fahrenheit scales for the same temperature, and let $K_T=T/\mathrm{K}$ be its thermodynamic temperature’s numerical value in kelvins. Use the exact relations $$F=\tfrac95 C+32,\qquad K_T=C+273.15.$$

A sample warms from $20.0\,{}^\circ\mathrm{C}$ to $35.0\,{}^\circ\mathrm{C}$. Treat these readings as exact for this exercise.

1. Convert both endpoint readings to Fahrenheit and kelvins. Then express the temperature rise in Celsius degrees, Fahrenheit degrees, and kelvins.

2. A student substitutes the Celsius rise into $F=\tfrac95 C+32$ and reports a rise of $59.0\,{}^\circ\mathrm{F}$. Diagnose the error by subtracting the two endpoint equations symbolically.

3. Another student says $20\,{}^\circ\mathrm{C}$ is twice the thermodynamic temperature of $10\,{}^\circ\mathrm{C}$. Assess this using kelvins. Explain why an offset scale cannot support that ratio argument.

Question 6 - Solution

Changing a temperature is not the same as converting a change

(a) Convert readings first, then subtract.

$$\begin{aligned}20\ ^\circ\mathrm C&\longleftrightarrow68\ ^\circ\mathrm F\longleftrightarrow293.15\ \mathrm K,\\35\ ^\circ\mathrm C&\longleftrightarrow95\ ^\circ\mathrm F\longleftrightarrow308.15\ \mathrm K,\\\Delta T&=15\ ^\circ\mathrm C=27\ ^\circ\mathrm F=15\ \mathrm K\quad\text{(intervals)}.\end{aligned}$$ The rise is $\boxed{15.0\,{}^\circ\mathrm{C}=27.0\,{}^\circ\mathrm{F}=15.0\ \mathrm{K}}$ as temperature intervals. These equalities do not identify the corresponding scale readings.

(b) The offsets cancel in a difference.

$$\Delta T_F=(1.8T_{C,2}+32)-(1.8T_{C,1}+32)=1.8\Delta T_C.$$

$$F_2-F_1=(\tfrac95 C_2+32)-(\tfrac95 C_1+32) =\tfrac95(C_2-C_1).$$ So the rise is $(9/5)(15)=27$ Fahrenheit degrees. Adding 32 incorrectly treats a temperature difference as a Celsius reading. Similarly $K_{T,2}-K_{T,1}=C_2-C_1$, since 273.15 cancels.

(c) A numerical ratio depends on the origin.

The two thermodynamic temperatures are $293.15\ \mathrm{K}$ and $283.15\ \mathrm{K}$. Their ratio is $$\frac{293.15}{283.15}=\frac{5863}{5663}\approx\boxed{1.0353},$$ not 2. Doubling the Celsius numerical reading doubles its difference from the Celsius zero, not the thermodynamic temperature. A common multiplicative unit change preserves ratios; an additive shift of origin generally does not.