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What I understand so far: in processes where the form of energy is replaced with another form and/or energy is transferred (and so work is done), there is always some degree of loss of energy as heat which is indicated by a change in entropy and which makes the process irreversible (getting the process back to where it began, will require an additional external energy expenditure to compensate for the loss). The Carnot cycle is a theoretical process which exhibits a zero entropy change and is therefore reversible.

When an atom absorbs a photon, its energy is completely transferred causing an increase in the potential energy of the system that is the atom. The photon energy equals this increase and the possible increases for the atom are quantized. So this is a zero-entropy process? There is no decay of some of the photon's energy? Since then it wouldn't be able to be absorbed anymore because that's not allowed by the quality of the system being a quantized one? When the atom returns to the ground state, the same photon is reproduced and emitted, so without any loss of energy as heat?

Related to thermal energy and work: why is it that thermal energy sometimes can do work(nuclear energy drives generators, pushes pistons to drive cars...) but in the case of entropy is considered as the forever loss of potential work. Is it because of 'density of the thermal energy' or because of the specific organization of the system (closed chamber with a movable piston) in which it is produced?

Help in elucidating and clearing this up is greatly appreciated

Filip VdB
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in processes where the form of energy is replaced with another form and/or energy is transferred (and so work is done), there is always some degree of loss of energy as heat

I would say that is true for bulk systems. Temperature (and heat transfer) simply don't make sense for single particle interactions. Temperature is a description of the statistical behavior of a region. It's a way of describing the interactions without tracking each individual particle. This makes talking about the entropy of one or two interacting particles more difficult (but also see Can a single classical particle have any entropy?)

why is it that thermal energy sometimes can do work...

Thermal energy (like just about every other form of energy) can only be exploited across a gradient. Thermal energy at a single temperature is not useful.

In the cases you list (nuclear power plant or internal combustion engine), these plants are exploiting the difference between two separate temperatures (a hot core and the colder environment). That makes them examples of heat engines.

But the process of exploiting this temperature gradient reduces the difference. It makes the hot side cooler and the cool side warmer. Over a sufficient period, this reduces the thermal gradient and the amount of work that can be extracted.

BowlOfRed
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The first thing to remember is that the second law of thermodynamics is probabilistic. If you have a system of many particles and many energy units, then the probability of entropy decreasing vanishes. But in the case of one photon and one atom, the probability of entropy decreasing may be quite high. The other thing to understand is that heat on such a small scale is basically meaningless when you are talking about one atom, since heat is basically the transfer of kinetic energy between particles.

Entropy differs from regular heat because it is "waste heat." It cannot be used due to the fact that entropy must increase (second law) and that fact is due to the fact that heat flows from hot to cold. How much entropy increases during the process depends on the engine's efficiency, but even the most efficient system must expel waste heat just to make sure the entropy of the universe stays constant. In the process of taking in heat from a heat resivoir, it takes in entropy as well.

Zulle
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