thermodynamics
What are the equations used to calculate the amount of energy transferred from a constant heat source "like a stove" to a pot which would then transferred energy to water contained in that pot.
I don't know exactly what equation to ask about and just need some direction to go in for formulas.
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 $T_2 \to T_{\infty}$.
Here is an intuitive way to think about it:
A puddle on the ground will evaporate even if the water is not boiling. The same amount of liquid would take much longer to disappear if it were contained in a cup.
There is a nice microscopic reason for this. Imagine each water molecule in the center of the liquid as being "boxed in" by 26 neighbouring molecules (as if it were in the center of a 3x3 rubrik's cube). All these neighbours attract the central one and lock it in place. A molecule on the surface will have 9 empty sites above it. If it gets knocked around by its neighbours, sooner or later it will be kicked into one of the empty sites and evaporate.
You want to maximize the number of water molecules exposed to empty sites. With a sufficiently wide pan (imagine one that is so big that the sap is only 1 water molecule deep) you could evaporate the water almost instantly even without external heating. The external heating will only give the flat pan a further advantage.
Best Answer
To calculate the amount of heat energy needed to heat a pot with its content of water we can use a simple formula:
$$\Delta H=mc_p(T_2-T_1)$$
Where $\Delta H$ is the heat energy needed to heat an object of mass $m$ and specific heat capacity $c_p$ from $T_1$ to $T_2$. For a pot with water these energies would need to be calculated for both separately and then added. Here we assume the $c_p$ values to be temperature invariant, which tends to be true for reasonable temperature intervals.
But:
...is much harder. Specifically when using a stove, burner or such like because the heat transfer from heat source to pot plus water tends to be quite inefficient: only part of the heat supplied by the heat source actually ends up in the pot plus kettle. The rest escapes as air heat, light and IR radiation.
If we call $\epsilon$ the percentage that gets lost, then the heat that must be provided by the heat source would be:
$$\Delta H_{source}=\frac{mc_p(T_2-T_1)}{100-\epsilon}\times 100$$