exchemist

@swansontpost explains that. Electrical discharge in a gas laser, or electronic excitation in the solid state, are the two methods he refers to. But don't ask me to get deep into the physics of that. (I'm only a chemist, so what I know comes from quantum chemistry.)

This is all mere assertion on your part and I'm afraid that 150 years of physics and engineering says you are wrong. The fluid analogy - and of course it is only an analogy - is very useful. A temperature gradient for heat is like a gradient on a riverbed for water. Heat is a flow of internal energy and the steeper the thermal gradient the faster it flows. Yes, you are of course right that heat is due to the motion of the molecules of the substance in question, but the way that motion spreads itself out through a body, along temperature gradients, is analogous to a flow of a physical fluid. People of Carnot's time were not morons. They were groping towards the understanding we now have. Caloric was an intelligent idea, even if it proved to be a defective model. My point is that it provided useful insight for those that came later. As to the idea of heat being entirely converted to motion of a piston, it is quite interesting to think about that for a moment. There is no way that heat can give up all its energy to make a piston move. Why? Because molecules don't move all at the same speed. There is a bell-shaped velocity distribution* among the molecules. So at whatever speed the piston moves when molecules hit it, it will be moving faster than some and slower than others. So some motion will always be left, among most of the molecules, after the interaction. Furthermore, in order to exert a force on the piston to make it move, molecules have to rebound from it. So they leave the piston surface still in motion. This is the whole point about heat. It is disordered, random kinetic energy. There is no way to order all the molecules neatly, so that they all move together, at one speed, towards the piston in order to exactly give up all their momentum to push it. That is the reason why it is impossible to convert all heat to work. (This inherently disordered character of molecules in motion is captured in the concept of entropy.) * In fact not quite a bell curve: a Maxwell-Boltzmann distribution: https://en.wikipedia.org/wiki/Maxwell–Boltzmann_distribution

They are all in motion. All the stars in our galaxy, and any associated solar systems, obviously including our own, are revolving about the galactic centre. The same is true for the stars in other galaxies. And the galaxies are in motion relative to one another as well. However, the distances are so vast that, over the lifetime of recorded history, the stars don't seem to humanity to have moved in the sky appreciably, relative to one another.

On one point, your comment that greater mass of the molecule is responsible does not sound right to me. CO2 and water molecules have dipoles, (O-C-O being - + - and H-O-H being + - + ) which will couple to the electric vector of radiation at characteristic frequencies, determined by the resonance frequencies of the stretching and bending of their chemical bonds. As a result they both absorb in the IR. By contrast, O2 and N2 have no dipoles and are therefore transparent in the IR. So what happens is sun light of all wavelengths warms the ground, which then re-radiates predominantly in the IR and some of this gets absorbed by water and CO2 instead of being radiated directly out into space. (The molar specific heats of gases are not a function of their molecular weight, but depend on the number of degrees of freedom the molecule has to take up energy. For monatomic gases, which can't rotate and have only 3 translational degrees of freedom, it is 3/2RT. For diatomic gases, which can rotate in 2 dimensions as well, it is 5/2RT.)

Yes. "Stimulated emission" is emission caused by the influence on an excited atom of the electric vector of another photon. The effect is to make the atom emit its own photon, in phase with the one causing the stimulation. What you do is create a "population inversion", a term in statistical mechanics that means you have more atoms in excited states than you would have at thermal equilibrium. That is an unstable, non-equilibrium situation. One emitted photon, passing through this inverted population, will quickly collect more and more other photons as it passes among the excited atoms, leading to a beam of photons all in phase. This in principle will continue until an equilibrium population distribution is regained. In practice mirrors are used, so the photons bounce back and forth, collecting more and more companions with each pass.

Oh, by Metacor do you mean Mercator? As in Mercator's projection, the one used by navigators, as it plots compass bearings as straight lines?

Yes, you make a good point! What Carnot did, as I understand it, was to associate a "work value" with a quantity of caloric, which depended on the temperature of the caloric. So in his analysis, he thought the amount of caloric rejected by a heat engine was the same as the amount absorbed by it from the hot source. Clausius later showed this to be wrong and that work and heat were both energy. However, one benefit of Carnot's way of thinking was the idea that caloric (heat) has a temperature, just as a fluid like water does, and it is that which determines how much work it can do. It also has the benefit of assuming that caloric (heat) has to be rejected from the cycle at the end, i.e. what a heat engine does is allow caloric(heat) to fall through a temperature gradient, like water in a water mill, and thereby get it to do work. So that set the scene for the correct idea that you can't convert all the input heat into work, with no waste heat.

It only seems ridiculous to you, because you don't understand the science behind it. You and I have been down this road before. You are a person who seems incapable of distinguishing between a basic principle of operation and the fiddly incidental details involved in constructing particular machines. In fact, you are now arguing against yourself here. Yes of course there are many incidental factors that go into engine efficiency. That is the whole point of having a theory that tells you what the basic principle is, so that you can set these sources of confusion to one side, determine the theoretical result first - and then look at the confounding factors, as things that cause deviations from the ideal. If you can't do that, you will never see the underlying pattern in anything. You will trap yourself in a medieval world in which all you can do is appeal to God to explain how nature works. To do science, you have to be able to simplify things to their essence, so you can see the pattern, in spite of the complicating factors. As far as Carnot goes, he got it right using the wrong model for what heat is. That can happen. As far as heat flows go, it doesn't affect the result if you think heat is a substance like calorific, or just an energy flow.

The Carnot cycle is a theoretical, ideal thermodynamic cycle, with no losses, that enables the maximum efficiency of any heat engine cycle to be determined. http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/carnot.html All real heat engines are less efficient than this, small ones usually much less, as the incidental losses are large compared to the heat throughput. Any designer or operator of a power plant engine or turbine will be well aware of how far short of Carnot efficiency even such huge machines are in reality. The value of it lies in what it tells you about the gain in efficiency to be had from maximising the temperature difference between hot source and cold sink.

However, Japan lost the Battle of Midway and were then progressively crushed by the United States.

No, it was their cucumber sandwiches.

That's far too glib. You do not know the temperatures I said you would need to know. It is the temperature of the two heat exchangers (the nature of which you have not so far specified) at the hot and cold ends that you need to know. You cannot just presume they will have the temperature of the medium conveying heat to and from them. (Strictly speaking it's not even that: it is the hot and cold temperatures reached by the working fluid, during the cycle, that determine the thermodynamics.) Apparently one of these media is boiling water, though you did not say so. (This illustrates why videos are not a substitute for a written description of the setup.) But the hot end is transferring heat into the engine, so the temperature of the heat exchanger will not be quite 100C (212F). If you are trying to insulate the cold end, the presumption must be that its temperature is not the ambient temperature of the room but significantly higher. If the insulation is perfect, then what will happen is that the waste heat rejected by the engine will progressively raise its temperature, eventually reaching that of the input temperature. As this takes place, the engine will run progressively slower and eventually stop. What is for sure is that for the cold end heat exchanger to reject heat into the room, its temperature must be quite a lot higher than the ambient temperature. Recall Newton's law of cooling, according to which the rate of heat transfer is proportional to the temperature difference. So, no temperature difference, no heat transfer. You need to know the real temperatures of hot and cold ends before trying to do a thermodynamic calculation.

Because you evidently still have a temperature difference, in spite of your efforts. What you really need to do is attach thermometer probes to the engine's contact with the hot source and its contact with the cold sink, to see what the input and output temperatures experienced by the engine actually are.. If the temperature of both becomes equal the engine will stop. If it does not become equal then you have a heat leak - or else the engine uses so little heat that the cold side takes ages to heat up and stop. As is usual with videos, the video shows us nothing worthy of comment.

I think you understand perfectly well.

Religion comes into it when it comes to censuring people for their preferences, which is the second of the two issues I was trying to disentangle. The first point is the sex drive point, in which by "we" I mean the great majority that has a heterosexual sex drive.

Because receiving an unwanted advance from a woman feels qualitatively different to me from receiving one from a man. I do not imagine I am unique.

Yes. It is also worth noting that the beam splits into 2J+1 parts. So if the atom has J>1/2, you get more than two parts, something that the classical idea of little compass needles can't account for at all.

No. It is never "nearly perfect". That is bullshit. The maximum degree of alignment is determined by the uncertainty principle limitation I referred to previously and that limitation applies in any situation. The magnitude of the angular momentum vector is √(J(J+1))h/2π, but the maximum value of its projection along the field direction is J.h/2π . This projection can have values of J.h/2π, (J-1).h/2π,.....0,..... -J.h/2π , and only those values. That is what space quantisation of angular momentum is all about. And no, I am not going to get into a side discussion about a piece of classical physics.

This may be my last post to you on this, unless you can avoid dragging other side issues into the discussion. From what you write, I wonder if perhaps you may may suffer from one basic misunderstanding. In space quantisation of angular momentum, the angular momentum vector does not completely align with the applied field. Ever. It continues to precess (in the semi-classical picture). It is only a component of the vector that is parallel or anti-parallel to the z direction of the field. It is impossible for the angular momentum to align totally, as that would violate the uncertainty principle*. There is always a component of the vector in the x,y plane, whose orientation is indeterminate - that is the real QM meaning of the "precession". So it carries on in this state of partial alignment, without any radiation occurring. So your analogy of a magnet oscillating, and then radiating until the oscillation ceases, does not apply. That is not what we observe in the Stern-Gerlach experiment. *The operators for angular momentum along x, y and z axes do not commute. Therefore only angular momentum along one axis can be precisely defined at a time, at the expense of definition of the others.

Exactly. But my problem with the term "homophobia" is that it may be closer to my own personal discomfort at the prospect of a sexual advance from another man than it is to any inability by some to respect the sexual preferences of others. So what we stigmatise in society as homophobia is actually not that, whereas a feeling of personal discomfort or distaste, which could perhaps be described as a mild kind of "phobia" (though overstating it), is a natural thing and not reprehensible at all!

Atoms "radiate abundant energy" is irrelevant, as well as being untrue. Atoms in the ground state do not and cannot radiate energy. This flatly contradicts the predictions of classical EM theory. That is one of the chief reasons why QM was developed. But you seem to adopting a wild, scattergun approach to this topic now. Unless you can stick to the point it will soon become a waste of time discussing this with you.

I suspect distaste for homosexuality is inborn in many of us. Given that we have a drive to be attracted to the opposite sex, we find the idea of sex with someone of the same sex a big turn-off. Consequently we may find the idea of a sexual approach from somebody of our own sex rather disturbing. If that is homophobia, then I am a homophobe. It seems to me that the blanket term "homophobia" is thrown around too easily. One needs to draw a distinction between personal sexual taste and the attempts by some to condemn different (minority) tastes in others. It is the latter that society should refrain from.

No, that is wrong for this situation, because it is a bound system and therefore quantised. I pointed out to you earlier that electrons in atoms do not radiate and fall into the nucleus. That's because it is a bound system, which restricts the states the electron can occupy. Forget brehmsstralung. It's irrelevant. That is for free, i.e. unbound, particles. The Wiki article on brehmsstralung makes the point: https://en.wikipedia.org/wiki/Bremsstrahlung. An unbound particle can occupy a continuum of states: they are not quantised. So the particle can radiate and lose energy in a classical manner. Electrons in a bound system cannot do this.

No need for a video, you just need to be able to read and take in what you have already been told, several times now. You get a double displacement reaction when one of the possible combinations is insoluble, because it is more stable in the solid state than in solution, and therefore precipitates out. I've told you this, so has @chenbeier, and it is repeated in the Wiki link I went to the trouble of providing. In your latest example it is Ca(OH)₂ that is much less soluble than the other possibilities, so it will precipitate from any solution with those ions in it.

Why can't it just divert a proportion of the existing angular momentum - and kinetic energy - to the precession axis? My understanding is that is what happens in a lossless precessing gyroscope. (Though maybe a proper physicist can comment on whether that is the case.)