exchemist

So this is just an uninformed speculation on your part, then? That's fair enough, but you did put this thread in Biology, in the form of an actual science question. In science, you can't just make stuff up. You can have a wild idea of course, but them you need to support it with evidence, from actual observation of nature. There does not seem to be any evidence that quantum phenomena play a role in neural processes. The last people that went down this road were Hameroff and Penrose, with Orch OR, which in the end went nowhere (i.e. it made predictions that were not borne out by observation).

It’s not “a fundamental issue”. Other people don’t share your preoccupation with this obscure topic. You’ve been given specific references to look up but for some reason you won’t follow them up, as you would if you were genuinely interested. I also notice you have not answered my question about what research you are doing on terrain theory. I’m beginning to wonder if you are posting in good faith, or if you have some undisclosed crank agenda.

OK, I'm now intrigued. How will these studies help you research "Terrain Theory", a.k.a. germ theory denialism? What is there to research, apart from the psychology of the people who affect to believe in it?

No this is incorrect. The ke of electrons in atoms is consistent with a speed <c. In fact, the explanation for the colour of gold is that, with very high nuclear charge in heavy atoms, the notional speed becomes a significant fraction of c, requiring a relativistic correction to the orbital energies. But still <c, obviously. More details here: https://en.wikipedia.org/wiki/Relativistic_quantum_chemistry So I'm not very interested in your idea, I'm afraid.

This is interesting (especially to some like me with some background in oil chemistry). However as far as a I can see it is not an example of a reaction that is accelerated by a drop in temperature. As far as I can recall, it isn't possible for a chemical reaction rate to increase with a fall in temperature. Reaction rate is governed by the Arrhenius equation: k = A exp(-E/kT) in which E is the activation energy for the reaction. The only way you can get a -ve temperature dependence is if E is -ve. But if that were so, instead of an activation energy you would have an energy minimum at the transition state, so that would be a new stable compound and the reaction would stop there, with the rate being diffusion limited (effectively a zero activation energy). But since the rate of diffusion also increases with temperature, even then it would go faster as the temperature increases. However there are plenty of examples where the thermodynamics are favoured by low temperature. A famous example is the Haber process for ammonia synthesis. The reaction is N2 + 3H2 <->2NH3. The free energy change for this is -ve, making the right hand (ammonia) side favoured at room temperature. However the rate is extremely slow as almost none of the molecules have enough energy to break the N-N triple bond. Raising the temperature overcomes that and speeds it up, BUT that shifts the equilibrium towards the left, so there is less tendency to form ammonia. So to get over that problem, 2 things are done. One is to use a catalyst that makes it easier to break the N-N bond (lowering the activation energy) so not such a high temperature is needed. The other is to increase the pressure, which favours the right hand side because 4 molecules are replaced by only 2.

One meaning of the verb "compromise" is to risk impairing.

I was just trying to offer you a bit of support, actually.

Didn’t Sartre speak of Man being “condemned to be free”? He was using oxymoron, but perhaps our poster is, too.

I've never heard of that. Do you mean terrain theory, the ideas of that crank (and jailbird) Robert O Young, and all that?

What is the nature of your research?

OK, thanks for the clarification. Regarding the common cold, it looks as if the British studies were discontinued due to the disappointing lack of efficacy of every remedy they tried.

It is trivially true that macro scale properties are emergent from ensembles of quantum interactions. But that is distinct from specifically quantum effects manifesting themselves at the macro scale. In general they don't. And, as far as I am aware, quantum effects don't ned to be invoked to account for the operation of neural processes, though I'm open to correction from someone with more knowledge of the subject. (The only people I know of who think differently are the proponents of Orch OR, but that theory seemed to have failed and is not taken seriously any more.)

I don't understand your post. You have found the British Common Cold Study, evidently, and another in the USA. Why do you highlight deliberate infection in capital letters? Who is Alan? What is it you are after here? Or have you now answered your own question?

Why do you think quantum effects have an impact on neural processing? I'd have thought that synaptic connections and neural impulses would be macroscopic, rather than quantum scale, processes.

Cardiovascular fitness is not developed this way, though. That is done by long spells well within maximum capacity.

But you can't explain the nature of science without philosophy. I am constantly having to refer to philosophical ideas when trying to explain to creationists what science is and - just as important in such discussions - what is not science.

Do you have a source for the latter statement? I can’t recall coming across such a comforting assessment of the effect of climate change.

There are lots, for instance as shown here:https://www.sigmaaldrich.com/GB/en/products/analytical-chemistry/photometry-and-rapid-chemical-testing/test-strips-papers-and-readers I don’t know of one for discriminating between alcohols, but there is one for glucose, which uses a meter to detect the amount of gluconic acid generated by reaction with blood glucose.

Didn’t you start a thread on all this back in 2008? What’s different now?

Then you are poorly informed. Any sports trainer who is any good will be careful about red-lining it. Most training regimes are largely aerobic and do not proceed to exhaustion. I have rowed most of my life and even rowing training is a controlled mix of exercise that is mainly aerobic. We were also told never to train if we were not feeling well. It has been known for decades that training when you have flu can give you irreparable heart damage for example.

Not heat capacity but Latent Heat of Vaporisation of water, i.e. the heat absorbed in turning liquid water into vapour. This is very high for water, due mainly to the need to break hydrogen bonding between water molecules as they break away from the liquid.

Just leave it. Someone else may came along and add something. They don’t normally get closed unless a moderator considers the discussion has become pointless or objectionable.

Well, there are fuels that makes use of the bond energy of nitrogen, which seems to be the idea you are pursuing. Hydrazine for instance (H2N-NH2) has an enthalpy of combustion of 620kJ/mol, i.e. about 2/3 that of the N-N triple bond, and that is partly because it forms N2 as well as water when it burns, converting an N-N single bond into the triple bond. Hydrazine is also used as a rocket fuel without combustion, being decomposed by a catalyst into N2, NH3 and hydrogen. Again the formation of N2 is responsible for a large part of the heat output. But you can't separate nitrogen into atoms and store it in that form for fuel. That can't be done. By the way, I see you are still at school. I thought you might be. The reason I went into the thermodynamic equations was just in case you were starting to work with them in school - and because I thought it might be fun to apply them to this problem of yours. It is the kind of thing you will learn about if you study chemistry in the 6th form, at about the age of 16-17. If you have not got to that stage, don't worry.

It's not that simple, because it's to do with the statistical distribution of kinetic energy among the atoms and molecules and how that alters with temperature. But certainly there would be a lot more atomic N, if nothing else were going on at such enormously high temperatures (see later in the post). There are two formulae in chemical thermodynamics which enable us to work it out. The first I've already mentioned: ΔG = ΔH - TΔS. We now have the values for this reaction, so we can say ΔG = 945 x 1000 - (16400 +273)x115, which gives a value of ΔG = - 9.75 x 10⁵ . (I add 273 to the temperature as it needs to be absolute, i.e in K rather than deg C.) The second formula is ΔG = -RTlnK, where K is the equilibrium constant, in this case (p(2n))²/p(n2), p(2n) being the partial pressure of atomic N and p(n2) being that of molecular nitrogen. So lnK = - ΔG/RT = 9,75 x10⁵/(8.32 x (16400 +273)) ~ 7. So that makes K ~ 1000. So there will be approx √1000 times, i.e. about 30 x, as much atomic N as molecular N2 at that temperature........ ...or would be, if there were no other processes set in train by such an astronomically high temperature. However, at such a temperature you would no longer have entirely nitrogen atoms! The 1st ionisation energy of N is 1400kJ/mol. You would have a lot of N+ ions and free electrons, i.e. a plasma. This temperature is about 3 times that of the surface of the sun ( refer @chenbeier 's earlier post ). So there you have it. The bonding in nitrogen is so strong that to break it you need to start breaking up the atoms themselves and have to resort to stellar temperatures. Best to look elsewhere for a practical way to store energy.

There is no magic temperature at which the bonds suddenly break. At any given temperature, molecules have a statistical range of velocities. When you dissociate a substance by heating, what happens is some of the molecules get enough thermal kinetic energy for the bond holding the atoms together to break. If a dissociated atom later on encounters another dissociated atom then, unless the atoms have between them more energy than the bond energy, they will recombine. So what you have is a dynamic process, in which some molecules are splitting into their constituent atoms, and some atoms are recombining into molecules. This is a chemical equilibrium: N₂ <-> 2N. Whether most of the substance is in the form on the left hand side or the form on the right hand side depends on the bond energy, the entropy of the two states and the conditions (temperature and pressure). As you raise the temperature, the average velocity of the molecules increases. That means that the fraction of them with an energy greater than the bond energy increases. This results in a greater fraction of them being in the dissociated state at any given moment. A strong bond energy favours the left hand side, while a significant entropy of dissociation favours the right hand side, more so at higher temperatures. What I am pointing out is that - because the bond energy in this case is so high - it is not until you reach about 9000C that you will have equal amounts of molecular and atomic nitrogen. In fact, I've now got a more accurate value for the entropy of dissociation, 115J/K-mol, which leads to temperature of more like 8200C for a 50:50 mixture of atomic and molecular nitrogen. (This is in good agreement with @studiot's earlier post in this thread, in which he quoted a textbook saying that at 8000C, there would be 40% dissociation.) By the way, catalysts are irrelevant to this. A catalyst does not change the thermodynamics of a reaction, which is what determines the equilibrium state. It just accelerates the rate at which it gets to equilibrium from an initial set of reactants. In the Haber process, the catalyst accelerates the reaction by avoiding having to dissociate the nitrogen molecule into atoms before reacting with hydrogen. It does this by forming new bonds between nitrogen and the metal surface. The energy of the bound molecules and atoms stays lower, throughout the process, than if they had to be free atoms. As a result, a greater fraction of the molecules have enough energy to react - so it goes faster.