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?


1 Answer 1


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

  • $\begingroup$ yes, but one could also argue that $(x\,^\circ\mathrm C)-(y\,^\circ\mathrm C) = (x-y)\,\mathrm K$ $\endgroup$ Commented May 7, 2017 at 17:59
  • 1
    $\begingroup$ Change in temperature is same on kelvin and Celsius scale. How does that matter? @AccidentalFourierTransform $\endgroup$ Commented May 7, 2017 at 18:20
  • $\begingroup$ What are linear units? $\endgroup$
    – technikfe
    Commented May 7, 2017 at 19:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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