Energy can be often converted from one form to the other, in theory, and with special lab equipment often only available to high school physics teachers such as friction-less surfaces and mass-less pulleys, with no losses, except when conversion from/to heat gets involved then energy is lost due to entropy.
So if you are going to heat up something you can expect losses in recoverable energy. What about phase changes at a constant temperature, is it the same loss, or is this energy pretty much fully recoverable?
Everything involves losses to heat. In some cases, we can make those losses small enough that we’re not too concerned about, and in some cases (not necessarily exactly the same cases), we’re working at a low enough level that we don’t worry about them because they’re too hard to figure out. In such cases, we often do the calculations assuming no losses, but they’re still there.
That said, in most phase changes, the energy transfers are all heat to begin with, so it doesn’t make sense to talk of “losses to heat”.
I’m not sure I fully understand your question, but I’ll chime in…
You mentioned entropy, so I’m assuming you’re familiar with “losses”.
When you heat something up, it’s “recoverable”, in the sense that when it cools down, you can “recover” the release of energy.
Just like when cars “recover” kinetic energy (typically lost during breaking) in the form of electrical energy.
So, when you heat something up and it changes phase, what happens? Let’s take water for example. You have a pot of room temperature water at sea level. It’s placed on a stovetop, where energy is transferred from the heating element to the water. The water increases in temperature. This transfer is called “sensible heat”. It is recoverable (minus entropy), as mentioned above. When it finally reaches 212 F, energy is still being transferred from heating element to the water, but, rather than increasing in temperature, the water begins to boil, producing steam. This transfer is called “latent heat” and it is also recoverable when the steam condenses. While producing steam (latent heat) doesn’t result in a temperature change, the energy required is huge compared to what it takes to raise the temperature of water (sensible heat). For example, it takes roughly 5-6x more energy to boil the hot 212 F water than it did to increase its temperature from just above freezing to 212 F.
Steam is actually an excellent “carrier” of energy. Since so much energy is required in producing the steam, it has “a lot to give” when it finally condenses. That is why, in industrial applications, steam (and not hot water) is the life blood for anything that requires heating.
Perhaps I am not stating this properly. In theory we can convert lets say electrical energy to mechanical at 100% efficiency (all energy input goes to mechanical) or kinetic to gravitational potential, and we can come close to 100%, but yes some small amounts go to heat mainly due to not having perfect theoretical materials. In theory we can not convert heat energy to mechanical with anywhere near that level of efficiency, but only as function of the difference between the absolute temperatures of the hot and cold sinks, which means instead of a ‘perfect’ goal of 100% to approach their is a hard limit of a much lower number (lets say 50% max even possible)
Does condition 1 or 2 apply to a phase change. Can we get all the energy out (minus small losses), or is it a function of the temperature differences, or perhaps it depends on the method used?
Can you give an example of when phase change and only phase change can be converted to mechanical energy?
Likewise, can you give an example of heat energy being converted to mechanical?
The common example of steam spinning a turbine involves using heat to illicit a phase change, which is then partially converted to mechanical energy. This incorporates both “heat” and “phase change”.
I was thinking, perhaps freezing a bucket of water and using it to raise the height of something very very heavy. And subsequently comparing the energy put into freezing the water versus the work done by raising the very heavy object.
Or, heating up a piece of metal and using the thermal expansion to push (or raise) a heavy object. In this case, you would need to know the coefficient of thermal expansion, strength of material, elastic modulus, specific heat and density of the material. You would then need to create an optimized shape for this material, which is the optimal point for strength and work (ie. a thin and long rod has the potential to expand more, and therefore do more work, however it will be weaker and more prone to failure). It would be an interesting, albeit iterative, process for someone with time on their hands.
Simple Phase change is 100% efficient. Unless you get a chemical change happening at the same time (oxidation is a problem in some applications), or if your final phase is not the same as your initial phase (which is almost always the case with steel), or quantum or electromagnetic or gravitational radiation (like the losses you would get if you counted the losses in your super-conductor when it had an unexpected phase change)
When you burn gasoline in an engine, most of the chemical energy is converted to heat energy, then you have the heat-engine problem getting the work out. In a phase-change engine, all of the phase energy is converted to heat energy, they you have the heat-engine problem of getting the work out.
The answer to your question is “yes.” The heat that causes a phase change is exactly the same as the heat that causes a temperature change when you are far away from a phase transition temperature, and can be treated exactly the same. Thermodynamics doesn’t know or care about changes in phase that happen in your working substance. The only thing that affects is whether the temperature changes you see are equal to heat divided by heat capacity, or that plus some combination of latent heats.
For example, let us suppose you have a heat engine that withdraws 1 MJ of heat energy from a hot reservoir and rejects some smaller amount into a cold reservoir, with the energy difference appearing as useful work. It does not matter whether either reservoir undergoes a phase change as a result of the heat withdrawn or added. Doesn’t change the thermodynamic analysis at all.
What may be causing puzzlement is this weird fact that heat can go into a material without raising its temperature. Isn’t heat just goosing atomic scale motion, and if the kinetic energy of particles goes up, shouldn’t the temperature, always?
The resolution is to realize that only in the gas phase are all of the degrees of freedom of a substance available for the transfer of energy. In the liquid or solid phases, some of them are “frozen out,” meaning no energy can go in or come out of them. They might as well not exist. So when heat is added, the temperature rises faster than it would if the all degrees of freedom were available – we record a lower heat capacity.
Now when we reach a phase transition, what happens is new degrees of freedom become “unstuck” and available for the deposit of kinetic energy. As we pour in (heat) energy, they start to get moving. But until all of the newly unstuck degrees of freedom are “brought up to speed,” so to speak, meaning they acquire the same average kinetic energy as all the other degrees of freedom, we are not raising the average kinetic energy (because the denominator of that “average” – the number of “active” degrees of freedom – is growing in perfect lockstep with the total amount of kinetic energy). So the temperature doesn’t rise. Nevertheless, the total amount of kinetic energy in all the microscopic degrees of freedom of the substance is increasing. That is, heat is flowing in.
Changes in temperature are only one way to induce a phase change. Pressure changes induce phase changes even in conditions of constant temperature/heat input. That’s still E[sub]k[/sub], but might not be easily recoverable if the phase change is to a more stable crystal form.
It was said (in one post above) that the phase change requires temperature (heat input), and thus heat and the heat engine limitations with that. But can not another method be used to induce a phase change, and avoid the heat (trying to come up with that). It was noted that pressure change could do this, but also something like electric input in a florescent bulb or laser resonance chamber (trouble there is you are creating EM radiation, heat - however that light emission not from the phase change from gas to plasma, it is the additional use of electrical energy once the plasma was created AFAIK, ), or the ion propulsion engine for spacecrafts where electrical power is used to phase change a gas into a plasma which then is used for kinetic energy propulsion.
OP: I think the above really answered the issue I think you’re struggling with. Which is not exactly the question(s) you’ve asked.
Heat is not really all that mysterious. Despite a lot of effort by textbook writers to make it so. Here’s a metaphor.
Imagine we had an engine that converted gravitational potential energy to kinetic energy. Something like this: Here on Earth mount an overhead pulley then loop a rope and a weight over the pulley. Now drop the weight from the top and use the spinning pulley to turn a generator. Now use a spherical cow for the weight so the system is friction-free and all the conversions run at 100% efficiency.
We now have an engine to extract 100% of the gravitational potential energy into electricity. So far so good, at least in principal.
The Earth is pretty big, so we ought to be able to extract truly vast amounts of energy from it, right? Right? Wrong.
You’re stuck with the fact you can only haul the cow up as high as the pulley is mounted, and the cow can only fall as far as to the ground below. That’s all the energy available. Sure, you can move the pulley higher, or dig a deeper pit below. But those two endpoints put hard limits on the outcome.
The metaphor is just that, a metaphor, and as such imperfect.
But the key issue isn’t that heat energy is somehow different than other forms. Carnot and the available temp delta isn’t an obstacle to efficiency so much as it’s an obstacle to quantity.
Very efficient in practice… There could be some losses directly resulting from phase change, the change in volume/structure must cause some movement … which must cause friction… which is half on the container/surface and that could lose the heat to the outside of the system… but for a “hold it in your hand” sized container its a small loss, because its only at the surface… If it was nanotechnology, it could be a significant loss.
There is a difference that I didn’t mention above, because I didn’t want to confuse the issue.
A heat engine, or a gravitational engine, works on the linear distance between the two states. A phase change engine works on an exponential distance between two states.
They are all 100% efficient, 100% reversable if you do them at equilibrium, not like the Otto Cycle or the Diesel cycle. But the phase change can be idealised as happening at a single point, which the gravitational engine and the heat engine can’t
That is, in theory, the phase change engine works with the ground at “almost” the same height as the pulley.
(Then as before you still need a cow and a hole in the ground to convert thermal to mechancal energy)
Yes. If we isolate a system in a perfect insulative box (and we can with today’s tech make a box that is within 0.0001% or something of being perfect using very thick insulation and many layers of vacuum), and we mix 100 C steam with something else, the equilibrium temperature of the box will reflect the energy we injected converting the 100 C water to steam.
Or, in other words, experiment one, near perfect box :
1 kg 100 C water + 1 kg iron at 0 C. Equilibrium temperature will reflect the heat contained in the water.
1 kg 100 C steam + 1 kg iron at 0 C. Equilibrium temperature will be higher than in trial (1). I don’t have time to calculate the equilibrium temp but it is a very simple linear equation where you add up the heats in all the inputs and determine the temperature shared in common by the water + iron that equals the heat at the state of the experiment.
No, we cannot convert electrical energy to mechanical energy at 100% efficiency, even in theory. To do so would violate the Second Law of Thermodynamics. The Second Law states that any spontaneous process (i.e. a process that proceeds in the forward direction) must increase the entropy of the universe. There are then two possibilities: 1. the entropy of the system increases, and 2. the entropy of the system decreases, but the entropy of the surroundings (the rest of the universe) increases by a larger amount.
In case 1, you have not converted all of the electrical energy to mechanical energy, because some of the energy went to increasing the disorder of the system. While the system still contains the same energy as it did before the transition (assuming no heat transfer between the system and surroundings), less of that energy is available to do work (because it has been used to increase the entropy of the system). Thus, the energy transfer could not have been 100% efficient.
In case 2, the system lost heat to the surroundings. The only way to increase the entropy of the surroundings is to have the surroundings absorb energy as heat from the system. Thus, the system has less energy than it started with, meaning that the energy conversion could not have been 100% efficient.
You got the law wrong, Basically entropy does not decrease, you have apparently take it as it must increase and yes generally it does increase but can remain the same. And it is the can remain the same that allows the in theory statement and 100% conversion from electrical to mechanical.
Yes, entropy of the universe could remain the same. However, there is then no thermodynamic driving force for the process to occur. The only way to carry out such a process is to have each step occur at equilibrium. But if the process is at equilibrium then there is no driving force pushing the process in one direction. Hence any process that actually proceeds cannot have the entropy of the universe remain constant.