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Let's talk about $Q = m c \Delta T$. Let's say I didn't know this equation by heart, but I naturally deduce it. How do I know which units to use (if I hadn't made myself clear, I mean units like Celsius or Fahrenheit)?

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    $\begingroup$ You would look at the units of the specific heat $c$, which has dimensions of energy per unit mass per unit temperature. So if it includes K$^{-1}$ then you know $\Delta T$ should be in Kelvin, etc. $\endgroup$
    – lemon
    Commented Dec 30, 2016 at 22:12

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When you deduce it, you start from somewhere. The mathematical deduction begins with some more basic correlations and definitions. The units from there are still the units in use.

If you find the expression empirically from experiments, you have units of measurement devices and data sets. The units of the expression are the same as those.

It all comes form the starting point

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The easiest way to do this is usually dimensional analysis - that is, you look at the dimensions of the different quantities and make sure they match.

If your mass is in $kg$, you need the heat capacity to have units of "per $kg$"; if your temperature is in Kelvin, the heat capacity needs to include units of "per K"; and if you measure $Q$ in Joules, the heat capacity needs to include units of $J$.

If you just stick with the SI system, all is well; the moment you venture into the realm of "other" units, it is up to you to make sure that the units match. The above approach usually works.

Note - for things like the ideal gas law, you cannot really use $F$ as the temperature scale, unless you apply an offset; so if you see $$PV = nRT$$ you know you can't have temperature in F, even if you have an appropriately-scaled universal gas constant; instead you would need $$PV=nR'(T-T_0)$$ with some value of $T_0$ to get the zero in the right place.

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  • $\begingroup$ If you insist on using the Fahrenheit temperature scale as many engineers do then the equivalent to the kelvin temperature scale is the Rankine scale en.m.wikipedia.org/wiki/Rankine_scale where the interval is equal to a degree Fahrenheit. $\endgroup$
    – Farcher
    Commented Dec 31, 2016 at 1:27
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Are you saying in the situation in which you are inventing units? I'll assume that. In short, it makes no difference whether or not you use units of Fahrenheit or Celsius to define temperature.

In practice, though, people didn't think of it like that. They noticed that some objects were more difficult than others to heat up, in a qualitative sense. They called "how hard something is to change temperature" the specific heat.

If I'm understanding your question correctly though, I think it's vital for you to know that units are truly arbitrary. Whether or not we want to use Fahrenheit or Celsius completely depends on what is convention and the most used.

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Physical equations generally assume a coherent unitary system which doesn't require arbitrary conversion constants and allows consistent dimensional analysis.

Equally units should be carried throughout your calculations as this gives you a robust way of checking for errors. So if the units you end up in the answer don't make sense eg you end up with Nm for something which should be energy then there is something wrong somewhere.

So with this is in mind SI/ISO is the safest bet at it is guaranteed to work consistently across all physical properties.

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