Yes, one is the pressure difference that I mentioned. The salt will add some energy to the system, but this is not the point.
Yes it is part of the point. Added energy must be counted to see if there is any net energy produced.
As the salt removes H2O, more water will drip at the bottom. …
No, water will not leave the salt, drip out of it. You cannot have your cake and eat it too. Either the salt will be attracting water or it will be releasing water, but not both.
Colder bottom water? A temperature difference? Another perpetual motion machine? Indeed, there will only be a slight drop in the water level before it is replenished at equilibrium.
Every gram of water evaporated will require 540 calories, If memory serves me correctly. The container is assumed perfectly insulated. Thus the heat of evaporation is taken from the water. In the SW of the US many homes were once cooled by “evaporative coolers.” Once you could even buy a window mounting unit for your car – cheap early version of car air-conditioning.
There was a net gain in gravitational potential energy of water due to evaporation.
Yes, initially until the stead states is reached. The energy that caused this was the solar energy stored in the dry salt. As it becomes a salt solution, you are “undoing” some of the stored solar energy – returning the salt to a form before the sun had stored the maximum possible energy in it. I.e. you are decreasing the energy stored in the salt below the max which the sun had stored in it.
Yes, there will be equilibrium. However, a part of this equilibrium will be dripping water, IE kinetic energy. Initially, there is only thermal energy; as the system progresses towards equilibrium gravitational potential and kinetic energy increase at the cost of thermal energy.
This so mixed up and wrong I hardly know where to start! Let start with the completely general fact that if any system has “dripping water” it is NOT in equilibrium.
Yes initially there was thermal energy as system was not at absolute zero. – But, unless there is a colder heat sink to dump waste to, none that energy is available to do ANY work or increase high quality energy like lift something in a gravitational field as that increase in gravitational energy could do work at 100% efficiency in principle. We have assumed the system is perfectly insulated so there is no such colder heat sink, but even if there were such a heat sink, we could only convert at most the Carnot fraction of the heat into work or stored gravitational energy in a gravity field.
Initially, there will be an increase in the gravitational energy stored in the H2O molecules absorbed by the salt in the tray at the top. As this system is less efficient than a Carnot cycle, the chemical energy decrease in the salt as it becomes salt solution will be GREATER than the potential energy gained by the water molecules. Also as 540 calories /gram are being supplied to the water part of them need to be added to the water. Water does not without external help just spontaneously cool itself to evaporate. In this case, the “external help” is the salt, which is reducing the density of H2O molecules immediately over the water surface. This reduces the rate at which these vapor molecules are entering and becoming part of the liquid water. It does not change the rate at which H2O molecules are escaping form the liquid water surface, but these “lucky” molecules are the faster than average ones, so the average speed of H2O molecules is dropping – we call that cooling.
Yes, thermal energy is converted to a different kind of energy, namely gravitational potential. The salt acts as a catalyst, and I'm not sure if it's energy is in any way transformed.
If you understood any thermo, you would think otherwise, be concerned. The salt is the source of the energy initially driving the system – Do you think it would work without the salt??? Just on the “thermal energy” alone????? Again I tell you that without any colder heat sink you cannot even one erg out of a billion BTU of thermal energy! (I forget but an erg is very small - about what a flea needs to jump, I think.)
It is hard to believe you could concentrate so much bad thermo into so few words! (Perhaps there is a prize for that? :shrug
The rest of this post of yours is just more of the same nonsense about dripping water and work from heat with no colder heat sink,
so I move to your other post:
Sorry, but I am incredibly zealous about the topic. I have basically spent the last 2 years thinking and researching this stuff, so I know if I'm right.... In a perfect Carnot cycle work is reversible and entropy is not created....
You have as discussed above many false ideas and need a teacher to correct them. Here you have all wrong ideas about a Carnot cycle. It is neither "perfect" and is not failing to generate entropy as it runs.
Although the Carnot cycle is not "prefect" it is more efficient than all other cycles. (Not hard to prove this but will not bother. Basic way to prove that is to divide the other cycle into a large number of Carnot cycles with the two adiabats of each tiny Carnot cycle almost touching. I.e. completely fill the other cycle’s VT (or PT) curve with multitude of tiny, narrow, Carnot cycles.)
The efficiency of the Carnot cycle is E = (T -t)/T where t is the absolute temp of the heat sink where the "waste heat", h, is dumped. T is the temp of the heat source. Q = W + h. (Conservation of energy.)
W = Q(T-t)/T and h = Q(t/T)
My thermo is rusty as 40+ years have passed, but as I recal the entropy is related to H/T where H is the quantity of heat and T the absolute temperature that heat is at. Thus the original entropy was Q/T and the final entropy is W/T +Q(t/T) as assuming no friction, the work W, a fraction of the origianl Q, is neither an increase not a decrease in the entropy of that part of the energy. I.e.
The final entropy is just this not changed entropy fraction W/T + h/t = Q{(T-t)/T}/T + Q(t/T)/t = (Q/T){(T-t)/T + 1} =(Q/T){(T - t + T)/T} = (Q/T){(2T - t)/T}
Now the first factor (Q/T) is just the original entropy. Thus, the second factor, {(2T - t)/T}, tells if the entropy has increased of decreased.
As t < T then (2T -t) must be GREATER than T. Then when something great than T is divided by T, the result is greater than unity. Thus the second factor is greater than unity.
I.e.
contrary to your assertion, the operation of a Carnot engine INCREASES entropy.
I have shown this mathematically but the idea is simple: Some of the high quality heat available at T has been converted into lower quality heat at t.
As I said earlier, you really need a teacher.
You have wasted two years to only become “arrogantly ignorant” about thermodynamics.
I have wasted too much time and effort trying to help you.
I only hope some others have read and learned something from my posts – you surely have not.
You already knew it all – as you have studied for two years!
You can go back to make energy out of room heat radiation falling on a photo cell - I am thru correcting you.
