# Converting $F^2$ to $C^2$

I'm trying to convert Fahrenheit squared $F^2$ into Celsius squared $C^2$.

I know how to convert a value $x$ in $F$ into $C$ with:

$\frac{5}{9}(x - 32)$

I also know how to convert a value $x$ in $km^2$ into $m^2$ with:

$x \cdot 1000 \cdot 1000$

I don't know though, how to convert a value $x$ in $F^2$ into $C^2$. It seems possible: WolframAlpha says that $5 F^2 = 1.543 C^2$, but I don't understand how they get there.

You might expect that just squaring the conversion leads to the correct result, but that appears to be wrong (following are two reasons why):

• It doesn't return the same result as WolframAlpha:

$(\frac{5}{9}(5 - 32)) ^ 2 = 225$

• It would be different from the $km^2$ conversion:

$(3km)^2 = 3^2 \cdot 1000^2 \cdot m^2 \neq 3 km^2 = 3 \cdot 1000^2 \cdot m^2$

• Actually, convertall (v 0.5.2) says Cannot combine non-linear units when trying to convert $F^2$ to $C^2$. Mar 31, 2015 at 3:58

It greatly depends on what you need to do with your temperature. In almost all physics applications, the thermodynamic temperature is the only one that is meaningful. A notable exception is in linear heat flow simulations, where temperature differences only are important (e.g. between a point in the simulated region and the "ambient").

The squared temperature application is almost certainly a thermodynamic temperature application, like, for example, the Stefan-Boltzmann law for calculating the intensity of blackbody radiation from a black surface. Here the intensity is proportional to the thermodynamic temperature to the fourth power.

So, back to your application. To calculate the square of 5 Fahrenheit, you can convert to Rankine, square, and then convert back to Fahrenheit. I'll work in SI: $5^\circ {\rm F}$ is $-15^\circ {\rm C}$, which, in Kelvin is $273.15{\rm K} - 15{\rm K}=258.85{\rm K}$ (differences in Celsius are the same as differences in Kelvin). Squared, this is $67000{\rm K^2}$.

In Rankine (the thermodynamic temperature scale corresponding to Fahrenheit), 5 Fahrenheit is $(5 + 459.76)^\circ{\rm R}$ Rankine, so squared this is $216000^\circ {\rm R}^2$. You can get this from the Kelvin squared answer by multiplying by $(9/5)^2$.

• Aha, that's the trick. I'll add that convertall (mentioned in my comment to the question) does allow for $R^2\to K^2$ and vice versa. Apr 3, 2015 at 17:37
• @KyleKanos it sounds like Convertall is pretty well built, then. Apr 3, 2015 at 22:22