Dear Samir, your electrical analogy is formally correct, but conclusion depends on other variables not included in your model, but which can have a relevant importance in real thermal case, e.g. role of evaporation. See the attached paper, where Newton Law is studied with and without considering evaporation.  Gianluca

The heat equation is a diffusion equation, while the time-dependent Schrödinger equation, which underlies the microscopic description of transport of charged particles, is only a diffusion equation in imaginary times (this is because of the complex unit i multiplying the time derivative in the time-dependent Schrödinger equation). Put it differently, heat can only propagate in the presence of matter [1], and this is achieved by the collision of the atoms comprising the matter, thus bringing about local, and ultimately global, thermodynamic equilibrium. While charge transport can be diffusive (in disordered conductors), collision and diffusion are not necessary for charge transport (one can have ballistic transport of charge). In short, the analogies between the heat transfer and the transfer of charge are of limited physical significance.

[1] Electromagnetic waves, or radiation, can propagate through vacuum, but the radiation energy turns only into heat when radiation interacts with matter. This is brought about by the dissipative vibrations of the atoms, or molecules, of the matter with which radiation interacts.

The analogical reasoning is a way to prove the principle stated in my question. Where it is stated that evaporating water by thin layers is more efficient. I think we can use the entropy to prove this principle. By heating thin layers iteratively,the entropy created to reach the vapor state is maximum because the difference of temperature is kept maximum between the hot source of heat and the layer of water (see the attachments Scnario2 and Scenario1 attached to my question). In scenario 1 the entropy created is not maximum because difference of temperature between layers is not maximum, comparing to scenario 2. From this standpoint, i think the analogical reasoning using electricity is correct for this case.

Dear Samir, heat transfer by convection is not the same as heat transfer by applying a temperature difference on two sides of a solid body. Convection makes matters very complicated, since it renders the problem of sound propagation (and therefore heat propagation) non-linear. I did not touch on the subject of convection exactly because a detailed discussion of it would be beyond the scope of this place, in addition to the fact that when an analogy does not apply in the simplest possible case, it does not make much sense to consider the more complex cases. 

Incidentally, I am naturally aware of the analogies that are often used in the context of electrical engineering, where electric potential difference over a conductor is declared as being equivalent to the pressure difference on two sides of a moving body, and that the electric current flow is equivalent to the flow of mass in a tube subjected to a pressure difference, etc. These analogies are based on hydrodynamic considerations and that for instance electrons in a metal can be viewed as forming an incomressible liquid. Hydrodynamic considerations apply only to phenomena at sufficiently long wavelength k, with ||k|| of the order of the inverse of the macroscopic length scale of the sample.

As regards the subject of efficiency of evaporating water by thin layers, it all comes down to the simple fact that water molecules evaporate from the surface (neglecting quantum tunnelling for simplicity), so that the larger the surface, the less time one needs to evaporate a given volume of water. Take a volume V of water in a rectangular container of surface S and depth D (one thus has V = DS). Since rate of evaporation scales like S, from S = V/D it immediately follows that for a given V the smaller the depth D the larger the  value of S and therefore the larger the rate of evaporation.  

Dear Dr Farid, Thank you for your responses. The problem is not evident that it seems. if we consider a volume V and a volume V/100 with the same surface of contact. the time to evaporate V/100 is less than t/100 where t is time to evaporate the Volume V. We can verify that by simple experiment. This is what i mean by the efficiency to evaporate water by thin layers. On the other hand the analogical reasoning exposed in my question is related to convection definitively.

You are welcome Samir.

I did not say that the rate of evaporation were equal to S = V/D, but that it scales like S. In other words, to leading order for the rate of evaporation 1/τ one has 1/τ ~ αS, where α is not a universal constant, but depends on various factors. For instance, when the system is subjected to radiation (say, sun light), absorption of radiation by water leads to its increase of temperature T [1] and thus to the increase of α (therefore, α clearly depends on T). The constant α also depends on the nature of the container (how well it absorbs radiation, what specific heat it has, etc.), etc.  By the bottom of the container getting heated at a faster rate than the water it contains, one will have convection of the water in contact with this surface to the surface in contact with air, so that the smaller the depth D, the larger the constant α and thus the larger the rate of evaporation. For sufficiently small D, to leading order one might have α ~ β/Dδ, where β is a constant (not a universal one) and δ > 0, possibly δ ≈ 1/2 or δ ≈ 1. In such case, to leading order one has 1/τ ~ βS/Dδ ≡ βV/D1+δ.

[1] At constant pressure and around T = 300 K, the heat capacity Cp of water is equal to ~5.65 kJ/(kg K), while that of air is equal to ~1.005 kJ/(kg K).

Dear Samir,

heat for evaporation of a quantity V of water is independent from heat transfer configuration. It depends on the enthalpy difference between water in the system and vapor leaving the system. 

I guess your question was rather on heat transfer rate. When applying electric analogy, equivalent resistance is the reciprocal of heat transfer coefficient times heat tramsfer area. So, as long as total heat transfer area and heat transfer coefficients don't change, the overall heat transfer rate will be the same, considering parallel circuit. Nonetheless, the two proposed scenarios would lead to different heat transfer coefficients.  

Thank you Emanuelle and Dr Farid, for your responses. I think, As Dr Farid said in the last response the more depth is smaller the more evaporation is efficient. I think it respond to my question. but i should consider other parameter as Emanuelle and Dr Gianluca adviced. Thank you for everybody.  It was my pleasure to exchange with you.

You are welcome Samir.