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So in equations (for example $Q=mc\Delta T$) one normally has °C divided by °C, giving a result with no temperature units in its dimensionality. So, does °C $-$ °C also give a result with no temperature units? Because in the very same equation I used as an example ($Q=mc\Delta T$) the $\Delta T$ part equals $T_\mathrm{final}-T_\mathrm{initial}$, with both terms in °C.

And shouldn't that equal - whatever number it is going to equal with no °C sign?

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$(x\,^\circ\mathrm C)-(y\,^\circ\mathrm C) = (x-y)\,^\circ\mathrm C$, because it is a linear unit.

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  • $\begingroup$ yes, but one could also argue that $(x\,^\circ\mathrm C)-(y\,^\circ\mathrm C) = (x-y)\,\mathrm K$ $\endgroup$
    – AccidentalFourierTransform
    Commented May 7, 2017 at 17:59
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    $\begingroup$ Change in temperature is same on kelvin and Celsius scale. How does that matter? @AccidentalFourierTransform $\endgroup$
    – Wrichik Basu
    Commented May 7, 2017 at 18:20
  • $\begingroup$ What are linear units? $\endgroup$
    – technikfe
    Commented May 7, 2017 at 19:51

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