Time in the simple Heat Equation I was teaching the simple Heat equation to a pharmacy student, namely the equation 
$$Q = m\mathcal{C} \Delta T$$
And I was thinking about: this equation gives us the amount of energy necessary to raise the temperature of a mass $m$ of a particular substance from $T_1$ to $T_2$.
If I know the density and the volume of the substance I can always find the mass, but let's suppose I know it's a liter of water inside a pot. Ok being water I already know the volume and the mass and the specific heat. 
What about a formula (maybe a modification of this?) that tell me how much time it will take me to make a liter of water to boil? I know it depends on how powerful is the flame of the stove but.. is there some equation as a function of the time?  
 A: The 'go to' time dependent equation for solving transient heat problems is Fourier's heat equation (here in 3D and Cartesian coordinates):
$$\frac{\partial T}{\partial t}=\kappa\Big(\frac{\partial^2 T}{\partial x^2}+\frac{\partial^2 T}{\partial y^2}+\frac{\partial^2 T}{\partial z^2}\Big)$$
Where $\kappa$ is the thermal diffisivity:
$$\kappa=\frac{k}{\rho c_p}$$
Fourier's equation is derived from Fourier's law:
$$q=-k\Big(\frac{\partial T}{\partial x}+\frac{\partial T}{\partial y}+\frac{\partial T}{\partial z}\Big)$$
Where $q$ is the heat flux: heat flow per unit of area through a surface.
In the absence of work done, a change in internal energy per unit volume in the material, $\Delta Q$ becomes (here in 1D):
$$\Delta Q=\rho c_p\Delta T$$

What about a formula (maybe a modification of this?) that tell me how much time it will take me to make a liter of water to boil? I know it depends on how powerful is the flame of the stove but.. is there some equation as a function of the time?

Application of Fourier's law or the heat equation depends highly on the actual real world problem you're trying to solve, so a 'one-fits-all' formula does not exist.
In the case of the simple heating problem of a kettle on a stove we can use the following simple derivation.

We'll assume there is a contact area $A$ between the kettle and stove and that the stove temperature is at a constant $T_{\infty}$. We'll neglect heat losses (convection, radiation) from the kettle.
The rate of heat transfer between stove and kettle is then given by Newton's cooling/heating law:
$$\frac{dQ}{dt}=uA(T_{\infty}-T)\tag{1}$$
Where $T$ is the kettle temperature (assumed homogeneous) and $u$ the overall heat transfer coefficient.
As the kettle heats up by an infinitesimal amount $dT$, by differentiating:
$$Q = mc_p \Delta T$$
$$\implies dQ=mc_pdT$$
Divide both sides by $dt$:
$$\frac{dQ}{dt}=mc_p\frac{dT}{dt}\tag{2}$$
With the identity $(1)=(2)$ we get the differential equation:
$$mc_p\frac{dT}{dt}=uA(T_{\infty}-T)\tag{3}$$
If we integrate $(3)$ between $(0,T_1)$ and $(t,T_2)$ we get:
$$-\frac{uA}{mc_p}t=\ln\Big[\frac{T_{\infty}-T_2}{T_{\infty}-T_1}\Big]\tag{4}$$
$(4)$ then allows to calculate the time $t$ to heat from $T_1$ to $T_2$.
Note that for $t \to \infty$ then $T_2 \to T_{\infty}$.
