unit conversion issue I have the following equation
where $ T_0(x) $ measures the temperature in Celsius at point x.  
The parameter values are as follows. 
My question is how to handle the second term $Q_m \over \omega_bp_bc_b$, which should give a unit of Celsius right?  
What conversions in these terms do i have to make in order to measure the temperature in Celsius   
I am trying to include the effect from exercises as well to this equation and modify the equation by adding a extra $Q_e\over \omega_bp_bc_b$ term. The exercise intensity $Q_e$ is measured by the unit MET. 1 MET is also defined as $58.2W/m^2$ . Can someone please tell me how to handle the units of $Q_e\over \omega_bp_bc_b$ to come up with Celsius temperature
 A: Writing $(Q_m)$ (and so on) for the numerical value of the constants in the units stated,
\begin{align}
\frac{Q_m}{\omega_b \rho_b c_b}
&=
\frac{(Q_m)\mathrm W/\mathrm m^3}{(\omega_b)\frac{\mathrm{ml}}{\mathrm{ml\: s}} \times (\rho_b) \frac{\mathrm {kg}}{\mathrm m^3} \times (c_b)\frac{\mathrm J}{\mathrm{kg}\:^{\circ}\mathrm C}}
\\ & =
\frac{(Q_m)}{(\omega_b)(\rho_b)(c_b)}
\frac{\mathrm W}{\mathrm m^3}
\frac{\mathrm{ml\: s}}{\mathrm{ml}}
\frac{\mathrm m^3}{\mathrm {kg}}
\frac{\mathrm{kg}\:^{\circ}\mathrm C}{\mathrm J}
\\ & =
\frac{(Q_m)}{(\omega_b)(\rho_b)(c_b)}
\frac{\mathrm W\:\mathrm{s}}{\mathrm J}
{}^{\circ}\mathrm C
\end{align}
from the obvious cancellations. Here $1\mathrm W\:\mathrm s=1\mathrm J$, so the fraction cancels out, and you're left with ${}^{\circ}\mathrm C$ as desired.
Quick notes:


*

*The notations $\mathrm J/\mathrm {kg}{}^\circ \mathrm C$ and $\mathrm {ml}/\mathrm s/\mathrm {ml}$ are both deprecated and should not be used, because it is not clear whether the third symbol is in the global denominator or not. In these situations one should use brackets such as $\mathrm J/(\mathrm {kg}{}^\circ \mathrm C)$ and $\mathrm {ml}/(\mathrm s\:\mathrm {ml})$ or $\mathrm J/\mathrm {kg}{}^\circ \mathrm C$ and $\mathrm {ml}/\mathrm s/\mathrm {ml}$ explicit powers, such as $\mathrm J\:\mathrm {kg}^{-1}{}^\circ \mathrm C^{-1}$ and $\mathrm {ml}\:\mathrm s^{-1}\mathrm {ml}^{-1}$. What one should definitely never do is use both notations, with the same meaning, in the same document. Wherever you got this stuff from has some pretty reprehensible dealings with units and you would do well not to imitate them.

*While the unit $\mathrm {ml}/(\mathrm s\:\mathrm {ml})$ can sort of make sense on occasion (here, milliliters of blood perfused per milliliter of blood pumped, per second), in general it is simply better to cancel those out and leave an explicit $\mathrm s^{-1}$. Sometimes it's worth it to keep un-cancelled units of the same dimension (such as with Hubble's constant, $H_0\approx 70\mathrm {km}\:\mathrm s^{-1}\mathrm{Mpc}^{-1}$, where a $\mathrm{Mpc}$ is a length, so the whole thing simplifies to an impracticably clunky value $H_0\approx 2\times 10^{-18}\mathrm s^{-1}$ with an astronomical number in it) but otherwise it's best to cancel those out.
Note that these are physicist opinions, and that biologists might disagree (or agree in principle but not implement them in practice). You came here to ask so this is the viewpoint you're getting.

*It's $\rho_b$, not $p_b$ (Greek letter rho), as befits a density. Minor point but I thought I'd point it out.
A: Since mL cancels with mL, J/s and N-m are compatible and therefore you should need no scaling.
