So, I understand the relationship between K(Kelvin), R(Rankine), C(Celsius), and F(Fahrenheit), and the complications of converting between them (i.e. varying meaning of zero).

However, when performing unit conversion between a quantity with a mixed dimensionality (e.g. $[length]^2/[temperature]$), I'm not sure how to properly perform a conversion where the temperature scale changes.

How do temperature scales affect these scenarios and what is the proper method for handling them?

  • $\begingroup$ Can you give a demonstration? $\endgroup$ Dec 16 '19 at 17:48

There are two different situations to deal with:

A formula or quantity which involves a specific temperature.

  1. A formula which involves a specific temperature, such as the ideal gas law,$$PV=nRT$$ or the average energy per molecule in a gas,$$< E>=(3/2)kT$$
  2. A calculation which involves a temperature difference such as heat to cause a temperature change, $$Q=mc\Delta T.$$

In case #1, you must use an absolute scale, either Kelvin or Rankine. Then the $R$ or $k$ must be converted properly, using the size of the temperature unit. In that case, 1 K (unit) = 1.8 R (unit).

In case #2, you can use any scale you want as long as both temperatures ($\Delta T = T_{\mathrm{final}}-T_{\mathrm{initial}}$) are on the same scale, and you use the specific heat, $c$, with the proper temperature unit size, 1 K = 1 C$^o$ = 1.8 F$^o$ = 1.8 R$^o$.

For example, $c$ = 0.7 $\frac{\mathrm{cal}}{\mathrm{g}\cdot\mathrm{C}^o}$ =0.7 $\frac{\mathrm{cal}}{\mathrm{g}\cdot\mathrm{K}}$ = 0.389 $\frac{\mathrm{cal}}{\mathrm{g}\cdot\mathrm{F}^o}$ = 0.389 $\frac{\mathrm{cal}}{\mathrm{g}\cdot\mathrm{R}^o}$

  • $\begingroup$ I'm attempting to generalize. It would seem to me, then, that the only time you'd use the linear conversions ($C * (9/5) + 32 = F$) would be in situations where the temperature stood alone. Otherwise you wouldn't be able to "extract" the portion of the quantity that the temperature contributes in order to properly rescale. Therefore, in those cases, you can only use the size method. Would you agree with that interpretation? $\endgroup$ Dec 16 '19 at 21:45
  • 1
    $\begingroup$ @digitlworld For the most part. There are a few formulas that engineers and lazy physicists use for things like calculating the speed of sound at a temperature using a linear approximation (v = 331 + 0.6 T in m/s with T being the Celsius temp). If you use the "size" method there it would be wrong(er). If you have a F temp, you would convert it to a C temp. $\endgroup$
    – Bill N
    Dec 16 '19 at 22:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.