Entropy = Heat = confusion

[the.punk]

So the easiest conclusion from this is that when heat flows naturally from hot to cold, the entropy increases, but when we have to force heat to flow from cold to hot, the entropy decreases... but we have to pay for the decrease with work.

Thanks.
Before I understood it a bit. But now I understand it very good.
It's cool that someone is here who can explain it good. I understand it now more as before.
It's also cool that there is someone who is also interested in this. Because I'm also.

 

[RisingFury]

Good calculating.
Hot -> cold increases entropy and decreases enthalpy
cold -> hot decreases entropy and increases enthalpy (right?)
and because I can decrease the entropy in a own system by putting energy/work into.

Enthalpy is defined as
H = W + A
W is internal energy, A is work.

I'm not sure, but I think enthalpy wouldn't change... as temperature of the object drops, so does the internal energy, which is a function of temperature.
Now... in basic, the change of internal energy is Delta-W = m*c*Delta-T and so is the change of heat, which equals work, so A = m*c*Delta-T... if both of these are positive, then enthalpy would increase Delta-H = 2*m*c*Delta-T, but I think that change of internal energy in this case is negative, which means that the change of enthalpy is 0.

Though if I remember correctly this stuff only holds for constant volume...

 

[Artlav]

Now let's look at a simple fridge. We'll assume that the fridge is 100% efficient, converting all of the energy it receives from electricity into cooling and we'll assume the fridge is perfectly insulated so no heat is able to get inside it.

AKA, the spherical horse in a vacuum.

The energy is not converted into cooling, but into flow of some liquid, which happen to vaporise at the right time. The energy is flowing into the fridge at all times and flows out of the heat exchanger at all times, the heat taken is not shipped somewhere, but is dissipated into the machine, converted into various kinds of energy, etc, and only smal part of it is radiated at the proper place.
Also, does the 2nd heuristic of thermodynamics not apply to a working machine?
It's not forcing the heat to lower, it's the same flow of heat from hot to cold.

I fail to see what is increasing there, only re-distribution of heat/energy.

---------- Post added at 11:04 PM ---------- Previous post was at 11:02 PM ----------

TStill, how can one talk of increasing disarray within a system when it actually gets more complex?

What is complexity?

 

[Countdown84]

This is a conclusion I have reached on logic reasoning, but it seemed so contrary to what I learned about the concept of entropy: That increasing entropy, as you stated, means increasing disarray. It sounds somewhat strange to think of increasing disarray as increasing complexity. decreasing complexity sounds much more like it, but that's probably where I have my major misconception. Still, how can one talk of increasing disarray within a system when it actually gets more complex?

This is indeed confusing. I think many of us tend to think of "complex" in the context of technology - implying high levels of organization and structure. That context can be misleading here.

In the chemical sense, suppose you have a solution of pure water. This is very simple, also very organized, because everywhere you look, every molecule you examine, is the same.

Now imagine a solution of water that also contains a stew of dissolved ions, organic molecules, maybe even a little black pepper for good measure. Describing this system (mathematically, chemically, physically) is much more complicated, because each molecule you pick up might be different! And they're all moving around at random! It's a complex system because it contains a large number of components and physical/chemical interactions.

So in the sense we're talking about here, complexity = disorganization.

 

[RisingFury]

AKA, the spherical horse in a vacuum.

The energy is not converted into cooling, but into flow of some liquid, which happen to vaporise at the right time. The energy is flowing into the fridge at all times and flows out of the heat exchanger at all times, the heat taken is not shipped somewhere, but is dissipated into the machine, converted into various kinds of energy, etc, and only smal part of it is radiated at the proper place.
Also, does the 2nd heuristic of thermodynamics not apply to a working machine?
It's not forcing the heat to lower, it's the same flow of heat from hot to cold.

I fail to see what is increasing there, only re-distribution of heat/energy.

A fridge works by compressing a gas to the point where it turns into a liquid. That will cause the liquid to heat up, because it has to give off evaporation heat. To compress it to this point, you'll need at least as much work as the heat released from the liquid. The liquid then heats up and is let to cool down to room temperature. Then it's pumped still under pressure into the fridge where it is let to expand. Expanding will cause the liquid to evaporate, which needs evaporation heat. That means the liquid will cool down and then draw heat from the stuff inside the fridge.

Ok, let's imagine a fridge that's perfectly insulated, cooling down a mass of air. Flowing through the cooling pipes is something like freon.

Here is the proof that all the intermediate steps don't cause any change of entropy, only the initial and end state:


I'm gonna be using O as mass flow of freon and q as evaporation heat.

First, we compress freon:

A/t = P = Q/t = O * q

Compressing freon will heat it up:

O * q = O * c * Delta-T
Delta-T = q / c

Because q and c are positive numbers, Delta-T is positive, which means increase of temperature.

The change of entropy:

Delta-S/t = O * c * Ln((T+Delta-T)/T)
Since (T+Delta-T)/T is larger then one, the natural log is greater then one, which means that change of entropy per unit of time is positive.


Then we let the liquid freon cool off:

Delta-S/t = O*c*Ln(T/(T+Delta-T)) and now, T/(T+Delta-T) is negative.

These two changes are exactly the same, because the two logarithms are exactly opposite:

Ln((T+Delta-T)/T) = -Ln(T/(T+Delta-T))
Ln(T+Delta-T) - Ln(T) = -(Ln(T) - Ln(T+Delta-T))
Ln(T+Delta-T) - Ln(T) = Ln(T+Delta-T) - Ln(T)

^^ so you see, when first compressing and cooling down freon, that causes no change of entropy. Same goes with the second cycle of cooling it down and heating it up once in the fridge. Therefore, if all the states in between do not cause a change of entropy, the change is only dependent on the initial and end state.

---------- Post added at 08:06 PM ---------- Previous post was at 07:53 PM ----------

Delta-S = m*c*Ln[T1/T]
Delta-S = 845 J/K (again, positive)

I guess nobody noticed the error here...

It's supposed to be
Delta-S = m*c*Ln[T/T1]
Delta-S = -845 J/K, which makes it negative, however, it doesn't change the end result. The entropy is still positive.

The change from heating up the air was 963 J/K and the change from cooling water is -845. The only difference is that it brings the result to 118 J/K, instead of 1808.

But the point stands, the entropy of the system has increased.

 

[Linguofreak]

Basically, heat is the combination of entropy and energy.

A low energy, low entropy situation would be a delta glider sitting near another one. The temperature of the two is fairly low, and the kinetic energy of either as measured by the other is also low.

A high energy, low entropy situation would be those two gliders hurling towards each other at 30 km/s. he temperature of the two is fairly low, and the kinetic energy of either as measured by the other very high.

In this case the energy is higher than the previous situation, but the entropy (and heat) is pretty much the same.


A high energy, high entropy situation would be when they hit, creating a nice toasty fireball with bits of vaporized delta glider flying every which way. You have all the energy the two DG's had before they hit in a very small space (high energy), and instead of being in organized motion, the molecules of the two end up going in all directions (High entropy).

Notice that in this case the entropy and temperature is much higher than the previous situation, but the energy is the same. The entropy increase takes the form of the velocities of the molecules of the (former) DG's becoming much more spread out. (The molecules of each DG had been travelling in the same speed and direction, and now they are travelling at a whole bunch of speeds in a whole bunch of directions)

As the fireball expands, it cools, and entropy increases while temperature decreases (the molecules radiate off alot of their energy, and the expansion of a gas (much of the debris at 30 km/s is going to be gas) cools it (the relative motions of nearby molecules are slower)). The entropy is higher than the previous situation, and the energy is the same, while the temperature is lower. The entropy increase in this case comes from the positions of the molecules becoming much more spread out.

 

[RisingFury]

As the fireball expands, it cools

Actually, no... in space, there is no pressure acting on the gas, therefore there is no energy needed to expand it... the only cooling you'd get would come from radiating heat away.

You'll get the evidence for that from this experiment:

Take a container, split in half by a breakable wall. Fill one side with gas and empty the other side. Stick a thermometer in the gas and break the wall... you'll notice no change in temperature, therefore, no energy was released.

 

[jedidia]

This is indeed confusing. I think many of us tend to think of "complex" in the context of technology - implying high levels of organization and structure. That context can be misleading here.

Indeed I did. So, complexity in this context would be how difficult something is to describe. That makes sense. I think I start to get it now!

Plus the other misconception was that temperature does not equal heat, but is heat/volume, so increasing entropy does not necessarily mean that stuff gets warmer. I think I start to get it now.

Here is the proof that all the intermediate steps don't cause any change of entropy, only the initial and end state:

This, however, confused me a bit again... if there is no entropy change in all the intermediate steps, how can the end state be different than the initial state? (I probably didn't get it because I suck at math...)

 

[Urwumpe]

Take a container, split in half by a breakable wall. Fill one side with gas and empty the other side. Stick a thermometer in the gas and break the wall... you'll notice no change in temperature, therefore, no energy was released.

In other words, a gas cools when it has to work against another pressure during the expansion. It does not matter what provides the pressure. Also works with pistons. If you let the expanding gas push the piston, the gas cools. If you pull the piston faster, than the gas expands, it will keep it's temperature.

 

[Kozak]

Enthalpy is the specific internal energy of the substance. It's not specifically linked with heat. The energy of the reaction with enthalpy difference may be input or output by e.g. electromagnetic radiation.
Enthropy is measure of quantity of possible system states. If enthropy is increasing, the information does neither appear neither diasppear. The only thing that changes is amount of information needed to describe the system.
Heat in general has nothing to do with enthropy either.

 

[the.punk]

Enthalpy is defined as
H = W + A
W is internal energy, A is work.

I'm not sure, but I think enthalpy wouldn't change... as temperature of the object drops, so does the internal energy, which is a function of temperature.
Now... in basic, the change of internal energy is Delta-W = m*c*Delta-T and so is the change of heat, which equals work, so A = m*c*Delta-T... if both of these are positive, then enthalpy would increase Delta-H = 2*m*c*Delta-T, but I think that change of internal energy in this case is negative, which means that the change of enthalpy is 0.

Though if I remember correctly this stuff only holds for constant volume...

But I think enthalpy and entropy are directly linked together.
If entropy decreases enthalpy increases.
And if entropy increases enthalpy decreases.
It depends on the situation if one of those increases or decreases.
However in every situation enthalpy is linked with entropy. If it decreases the other increases I think.

 

[cjp]

In statistical physics, entropy is the logarithm of the number of microscopic states corresponding to a certain macroscopic state.

The logarithm is there just for convenience. Without the logarithm, the numbers would be astronomically high, even for small amounts of matter. Also, it gives the entropy some useful mathematical properties. But conceptually, you can do without the logarithm: whenever a certain value increases/decreases, its logarithm also increases/decreases.

A microscopic state is the state of the system at its lowest level: all positions, velocities of atoms and molecules, and all other state information, like the states of all atomic nuclei, electron energy levels & so on. So this has a very clear physical meaning.

A macroscopic state contains all all the things you can observe about the system on a macroscopic level. This includes things like its mass and its volume, and also statistical measures like its temperature and pressure.

Obviously, the macroscopic state is not as exact as the microscopic state description: you can't measure the complete microscopic state of large systems. So, each macroscopic state can correspond to a lot of different microscopic states. This is exactly what is described by the entropy value.

One example:
suppose you have a box containing a gas in equilibrium. Its volume, mass, temperature, pressure etc. don't change, so its macroscopic state doesn't change. Yet its microscopic state changes continuously, as its molecules fly around through the gas.

Other example:
suppose you look at the deserted city, and you take only a very quick look, to determine to what degree it has broken down. Here, the 'microscopic state' includes the positions of all pieces of walls, roofs etc., while the 'macroscopic state' doesn't include such detailed information. If you find the city completely intact, there is only one microscopic state possible: everything is still in place. So this state has a low entropy. The more the city is broken down, the higher the entropy: at 50% broken down, 50% of all pieces are where they don't belong, but there are millions of different ways of what their actual locations can be.

I think there is some kind of subjectivity in what state information actually belongs to the macroscopic state. I think the entropy kind of describes how much information is missing in the macroscopic state. When you inspect a system more closely, the actual entropy value might be lower. But as long as you are consistent in what you describe in the macroscopic state, the second law of thermodynamics should apply.

This second law of thermodynamics ('entropy never decreases') simply follows from statistics. When a microscopic state evolves randomly or semi-randomly, over the range of all physically possible states, it simply has the highest probability of ending up in the macroscopic state with the highest entropy. For large systems (more than just a few molecules), this probability is practically 100%: virtually all microscopic states belong to the macroscopic state with the highest entropy.

Looking at the deserted city again: if you randomly change something to it, chances are much higher that you will break it down more, than that you accidentally fix something.

PS.
This may be a different way of looking at entropy, but it is fully consistent with the 'thermodynamics' delta-S calculations mentioned earlier. The trick is that statistical physics also provides a very exact definition of temperature, which is such that the relations between energy, entropy and temperature are exactly as described in thermodynamics.

PS 2.
Enthalpy is not the 'opposite of entropy'. It's just a form of energy. They don't even have the same unit: enthalpy, as an energy form, has unit J (Joules), while entropy has unit J/K (Joules / Kelvin).

 

[Thunder Chicken]

Entropy used to drive me crazy, too. It's actually not so bad.

Heat and entropy are intimately linked. My intuitive non-math mental model is this:

1. Entropy is a measure of disorder.
2. The hotter (more energetic) a bit of substance is, the greater the disorder of the atoms and molecules in that bit.
3. Heat flow transfers energy from one bit of substance to another bit, and therefore entropy is transferred as well.

If you burn a bucket of gasoline and tally up all the entropy of the initial air and gasoline and the resulting combustion products, the entropy of the products is always more than the gas and air. This is the Second Law in action, and tells us that we can turn gas and air into combustion products, but we can't reverse the process without adding a lot of work.

 

[cjp]

Heat and entropy are intimately linked. My intuitive non-math mental model is this:

1. Entropy is a measure of disorder.
2. The hotter (more energetic) a bit of substance is, the greater the disorder of the atoms and molecules in that bit.

That's almost always true. In fact, the only exceptions I can think of are extremely exotic systems, with negative temperature (on the Kelvin scale!).

But entropy usually also increases when matter or energy is spread out over a larger area. As a result, you can have something expanding and cooling down simultaneously, with a total entropy increase.

All spontaneous reactions have a net entropy increase, or no entropy change. Entropy decrease is possible in 'non-spontaneous' reactions because work is added. This work comes from an external energy source (e.g. an electrical power plant), and in that energy source, a reaction is running with such a large entropy increase that it (more than) compensates the local entropy decrease of the non-spontaneous reaction. So the total entropy always increases.

When you have a high-pressure vessel, and you let gas escape from it through a valve, that gas will cool down as it escapes the valve. As soon as the valve is opened, you don't have to apply work anymore: it is a spontaneous process. So this is an example of a system that cools down while the entropy increases. Of course, the expansion of the high-pressure gas does the trick.

 

[Thunder Chicken]

But entropy usually also increases when matter or energy is spread out over a larger area. As a result, you can have something expanding and cooling down simultaneously, with a total entropy increase.

Yes, you are right - randomness is not just controlled by temperature (in gases, anyway). You can increase the order / decrease the entropy by pushing all of the wayward molecules into a smaller box (i.e. adding pressure work).