experimental-physicsthermodynamics
I'm trying to figure out which direction the heat flows. After many google searches I've determined it flows from hot (higher energy) to cold (lower energy).
My question tho is, how do we experimentally prove that it's indeed the case? Can it be "seen" flowing somehow?
It seems we have reached the point where simple models are no longer satisfying. Rather than posing ad hoc DEs maybe it's time to try an actual physical model. Short of doing a full hydrodynamic simulation (definitely overkill here) we can try what is called a lumped capacitance model where we divide the system up into a number of "lumps" and energy flows between the lumps:
The fundamental law governing this system is the conservation of energy. Every lump has an equation of the form
$$ \frac{\mathrm{d}}{\mathrm{d}t}(\text{energy in lump}) = \text{rate of energy entering} - \text{rate of energy leaving}. $$
We treat the heat input as a fixed flow and the air (environment) as a heat bath at a fixed temperature. If we let the heat capacity of the $i$-th lump be $C_i(T)$, which can be a function of temperature then
$$ \begin{array}{rcl} \frac{\mathrm{d}}{\mathrm{d}t} \left( C_p(T_p) T_p \right) &=& P - r_1 (T_p - T_w) - r_2 (T_p - T_a)\\ \frac{\mathrm{d}}{\mathrm{d}t} \left( C_w(T_w) T_w \right) &=& r_1 (T_p - T_w) - r_3 (T_w - T_a) \end{array} $$
You can look up how the heat capacity $C_w(T)$ varies with temperature for water (though probably not for the pot material?), but we will simplify dramatically and assume, incorrectly, that the heat capacities are constant.
$$ \begin{array}{rccl} \frac{\mathrm{d}}{\mathrm{d}t} T_p &=& - \frac{r_1 + r_2}{C_p} T_p + \frac{r_1}{C_p} T_w + \frac{P + r_2 T_a}{C_p} &\equiv& a T_p + b T_w + s_p\\ \frac{\mathrm{d}}{\mathrm{d}t} T_w &=& \frac{r_1}{C_w} T_p - \frac{r_1 + r_3}{C_w} T_w + \frac{r_3 T_a}{C_w} &\equiv& c T_p + d T_w + s_w \end{array} $$
where the $a,b,c,d,s_p,s_w$ are shorthands. Note that only six combinations of the seven parameters ($r_1,r_2,r_3,P,T_a,C_p,C_w$) actually enter the problem, so there is some degeneracy of the parameters. You can see, Taro, that this is almost the model you came up with in your answer. The difference is that I'm including the heat input explicitly, so that conservation of energy is guaranteed.
With the obvious matrix shorthand these equations can be written
$$ \dot{T}-MT = s, $$
which, for a constant source, has the solution
$$ T\left(t\right) = \mathrm{e}^{Mt}T_{0}+\mathrm{e}^{Mt}\left(\int_{0}^{t}\mathrm{e}^{-M\tau}\mathrm{d}\tau\right) s. $$
When $M$ is invertible (which it definitely should be for this problem - if it's not then there is a mistake somewhere) the integral can be simplified:
$$ T\left(t\right) = \mathrm{e}^{Mt}\left(T_{0}+M^{-1}s\right)-M^{-1}s. $$
You can check this satisfies the original equation with the proper boundary conditions. There are eight parameters to fit: the four matrix elements, two sources, and two initial temperatures. It is a non-linear regression problem since the matrix exponential depends non-linearly on the fit parameters. So I'm afraid I don't know of a robust and efficient way to fit this model to your data, unless you use some assumptions to simplify the parameter dependence, but this is the physically motivated model for your situation.
The main problem is that all current technologies seem to waste a certain amount of energy on actually displacing some of the heat energy from A to B. If there were no such inefficiency, then we would in a sense obtain "free" energy as we could displace and focus an arbitrarily large amount of energy from a material into a finite space, and then use the laws of thermodynamics to power (e.g.) a stirling engine. If only there exists a structure that can naturally, without energy input, make one side cold and another warm. However, this gives us a contradiction.
Due to the second law of thermodynamics and the conservation of energy, this becomes impossible. As having a perfect "structure" that can transfer heat from A to B would violate these laws. It is required that energy be put into the system to actually perform the displacement. Therefore, even if said system is 100% efficient, energy is still put into the system such that no "free" energy is ever gained.
For reference (2nd Law of Th.) (wikipedia)
When two initially isolated systems in separate but nearby regions of space, each in thermodynamic equilibrium with itself but not necessarily with each other, are then allowed to interact, they will eventually reach a mutual thermodynamic equilibrium. The sum of the entropies of the initially isolated systems is less than or equal to the total entropy of the final combination. Equality occurs just when the two original systems have all their respective intensive variables (temperature, pressure) equal; then the final system also has the same values.
Unfortunately, the question is hard to answer:
My question is that why do we even need a temperature differential?
The reason is that the contrary is made impossible by a law. A law in physics and mathematics doesn't necessarily equate a theorem or theory. It only states "such is so." whilst a theory would actually explain "why such is so".
Best Answer
The definition of "Heat" is energy transferred due to thermal contact.
We know hot things cool down when they touch cold things, and cold things heat up when they touch hot things. So, that means the hot thing has to be losing energy, and the cold thing has to be gaining energy. That means that heat is flowing from hot to cold.