As I said, I stop answering on the subject of graphene, since the only ones who intervene except me have all confirmed that they are totally ignorant and incapable of solving any thermo problem, and those who know it have nothing to say on the subject and know full well that I am right. The only interesting thing to learn from this thread is to show that there are people who are perfectly capable of forming a firm opinion on a subject on which they admit to knowing nothing. As if by chance, they are all elements of the "JOGER" group, it's astonishing ...
I am nevertheless repeating here the simple problem posed by the graphene type machine, which is indeed a thermal machine, contrary to what some say, since it produces work from heat.
You start from a state A which includes: a perfectly insulated enclosure not allowing heat to pass (super thermos type) and a device to produce current which works thanks to thermal fluctuations (assuming that it exists) which is still in its box and not yet plugged in. This is the initial state.
Then
* you open the enclosure
* you install your device by plugging it into wires coming out of the speaker
* you make it work for a time T, it will produce a direct current of voltage U and a power UI, therefore an energy UIT = W
* with this current that you recover from the other side, you feed a motor which will raise a mass m to a height h, increasing its potential energy by mgh. There may be some efficiency so XW will be dissipated in heat q, and mgh = (1-X) W
* when it is finished, you unplug the device, you take it out, being careful that it only comes out or enters a negligible amount of heat, and you put it back in its box.
This is the final state B. The exterior received an energy W, part of this energy was dissipated in heat q, the other was used to mount the mass m, therefore W = q + mgh.
We ask to calculate the entropy variation between A and B, that is ∆S = S (B) -S (A). For those who do not know anything about thermodynamics, this is an incomprehensible question since they have never learned how to calculate a change in entropy. It's as if I asked them to calculate the determinant of a 3x3 matrix, they wouldn't know how to do it since they don't know what a determinant is and maybe not even what a 3x3 matrix is. no shame in that, when you don't know, you don't know. But in fact it is quite simple to calculate, the rule is that when a large and homogeneous medium is at the temperature T and receives a heat Q, its entropy variation is Q / T, it is as simple as that (all those who studied thermo in their young age will confirm it). When the system returns to its initial state, its entropy variation is zero
therefore :
* the machine itself has returned to its initial state so its variation is zero ∆S = 0
* the isolated enclosure provided work W. As there is no other source of energy than its thermal content, it has necessarily decreased by a heat Q = - W (it has cooled), and therefore its entropy variation is - W / T
* the outside received a heat q = XW so its entropy variation is q / T = XW / T
in total the variation of entropy of the Universe is - (1-X) W / T which is strictly negative. However, the 2nd principle stipulates that whatever the device, the entropy of the Universe can only remain constant or increase, in no case decrease. It is a universal principle verified by 100% of known systems, with zero exceptions, not only of course on all thermal power stations, motors, heat pumps, refrigerators etc ... but even by less trivial thermal machines like black holes or ... living beings like you and me (yes, we too are thermal machines!). Zero known exception.
So: the system described would violate the second principle, it is a certainty;
I replied on the graphene thread that I did not know the details of the properties of graphene and the usual idiots on duty laughed "well how can you give your opinion if you know nothing about graphene".
The reason is however very simple, it is that the reasoning absolutely does not involve the properties of graphene and one does not absolutely need to know them. The machine has returned to its initial state so its contribution to entropy is NULL, however it is made, it doesn't matter and we don't care. The entropy balance is just due to the fact that the enclosure had to cool down and that we mounted with a mass of a height h.
There is no need to know the properties of graphene in anything to calculate ∆S and to demonstrate that it would be negative, and therefore to conclude that it would violate the 2nd principle. It's like in relativity, if you are told that a kryptonite motor can exceed the speed of light, you don't have to know kryptonite to know it's wrong.
Those who believe that just haven't understood what a principle of physics is and what it's for.
Good for those who do not know anything about physics, they are excusable to make a mistake, but for a professional physicist like Thibado, it is just outright swindle, and it is unacceptable. The fact that he is making newspaper articles and the front page of his university website is just a sign of the decay into which science has fallen today, which has become largely a communications affair.
To pass for an idiot in the eyes of a fool is a gourmet pleasure. (Georges COURTELINE)
Mééé denies nui went to parties with 200 people and was not even sick moiiiiiii (Guignol des bois)
