Energy transfers and entropy

Question

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

Answer 1

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.

Answer 2

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.

Answer 3

Imagine you have a bucket filled with water (energy), and you're pouring it into another bucket. If the process is perfectly controlled and there’s no leakage (analogous to a reversible process like the Carnot cycle), all the water gets transferred. However, in real-world scenarios, some water always leaks out as you're pouring (analogous to energy lost as heat, increasing entropy). To get back to the original state, you'd need to add more water externally (external energy expenditure). In irreversible processes (e.g., heat engines or biological processes), the "leakage" of energy as heat results in a net increase in entropy, making it impossible to return to the exact initial state without external input.

When an atom absorbs a photon, the photon's energy fits perfectly into one of the atom's quantized energy levels, like placing a piece into a jigsaw puzzle. The energy matches exactly, with no leftover pieces or "crumbs" (no loss of energy as heat). This precision means the process involves no increase in entropy. When the atom returns to the ground state, the same "piece" (photon with the exact energy) is emitted. Since there's no energy "leakage" or disorder introduced, the process remains reversible and entropy neutral. Guess Why? Because the photon's energy is precisely quantized. No intermediate chaotic states (like vibrations or heat) are involved in a pure absorption-emission process. In contrast, if energy was partially dissipated as heat, it would introduce disorder (entropy) and make the process irreversible. However, real-world systems (e.g., solids or liquids) often experience non-ideal conditions where some energy does leak as heat.

Thermal energy can do work only when there is a difference in temperature, akin to water being stored at a height in a dam. The water in the dam has the potential to flow downhill and drive turbines, converting its potential energy into work. In thermal systems, a temperature gradient (high temperature vs. low temperature) acts as the "height" that drives the conversion of heat into work.

On the other hand, if all the water is at the same level, like a stagnant pond, there’s no flow, and no work can be done. Similarly, when thermal energy is uniformly distributed (no temperature gradient), it is "spread out" and incapable of performing work—this is the state of maximum entropy.

Imagine a room full of inflated balloons. Each balloon represents a nucleus, tightly packed with energy. When a balloon pops (analogous to nuclear fission), the release of energy creates a localized high-pressure region (like heat or thermal energy). This sudden burst can push objects in its path, converting thermal energy into work (e.g., driving a piston or spinning a turbine). For work to happen, the thermal energy must create an organized motion (like a piston moving or a turbine spinning). The "density" or concentration of energy in one region (high-temperature gas) allows it to flow to a lower-energy region (cooler surroundings), making work possible.