# Really weird question about unit analysis

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?

• Unitless means "units of 1". You can't add something in joules to something in "units of 1". – robphy Nov 6 '17 at 10:57
• 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. – 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.