Hi guys this is quite an interesting question I have because there appears to be two correct answers.

There is an equation described by: $$ Pt = mC_\mathrm{mass}T+m_\mathrm{water}C_\mathrm{water}T+K, $$ where $t$ is time, $P$ is power, $T$ is temperature change, $K$ is a constant, and the $C$s are the specific heat capacity of either the water or the mass. The details about the experiment are not worth talking about. My problem comes when I try to find the unit of $K$.

Here is my logic. If I'm gonna plot $t$ against mass $m$ of the "mass" then:

  • my gradient will be $(C_\mathrm{mass}T)/P$;
  • my $y$ intercept will be $(m_\mathrm{water}C_\mathrm{water}T+K)/P$; and
  • my $y$ intercept will pass through the $y$ axis therefore will be in units of seconds only.

This is where I trip up bigtime. As I calculate things, you have $$ {\rm time} = \frac{{\rm (mass) (specific\ heat) (temperature)}+K}{\rm power} $$ so therefore $$ {\rm power\times time}= {\rm (mass) (specific\ heat) (temperature)}+K $$ or in other words $$ \rm{\frac{work}{time}\times\frac{time}{1}} = \rm{\frac{mass}{1}\frac{work}{mass\times temperature}\times\frac{temperature}{1}} +K, $$ i.e. ${\rm work} = {\rm work}+K$.

As you can see I get to the point: $\rm joules=joules+unknown\ unit$.

Therefore the unknown units can either be unitless or be in joules. It is impossible as far as I can see to make a claim one way or the other?

  • 3
    $\begingroup$ Unitless means "units of 1". You can't add something in joules to something in "units of 1". $\endgroup$ – robphy Nov 6 '17 at 10:57
  • $\begingroup$ I have turned your mathematical formulae into LaTeX notation; you're expected to do this yourself in future posts on this site. A good tutorial is here. $\endgroup$ – Emilio Pisanty Nov 6 '17 at 11:53

Since in your original equation $K$ is a summand in a combination that gives the work $Pt$, it must have dimension of energy and units in joules.

Moreover, there is no real support in your calculations for the notion that $K$ would be unitless ─ it's entirely unclear why you think that that would even be an option.

  • $\begingroup$ I was going along the lines of "since it is unitless it won't contribute to the units on the LHS" in other words, as a constant it won't contribute. Joules plus an arbitrary constant is still joules. So it could still be an option? $\endgroup$ – Scott Simmons Nov 6 '17 at 11:45
  • $\begingroup$ Joules plus anything that's not joules (or commensurate to joules) is meaningless and undefined. You cannot add, compare (as in, greater than or less than) or set as equal two quantities with different units. That should be engraved on the inside of your skull ;-). $\endgroup$ – Emilio Pisanty Nov 6 '17 at 11:52
  • $\begingroup$ Yep gonna chisel that into my head right now, by the way thank you so much for helping me $\endgroup$ – Scott Simmons Nov 6 '17 at 11:55

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