Since one of the common complaints that creationists often use is to say that evolution violates the laws of thermodynamics, and since often they have difficulty articulating how thermodynamics works, I thought I'd write up what I hope is a reasonable summary about thermo, starting from basic principles. I'll try to keep the equations to a minimum, but they will necessarily come up.
While all of this is just my description of it, it is based upon my physics textbooks: Fundamentals of Physics, Third Edition by Halliday and Resnick; Physics Parts 1 and 2, Third Edition by Halliday and Resnick; and University Physics, Seventh Edition, by Sears, Zemansky, and Young.
The first thing to recognize is that thermodynamics is about how heat moves. To point out the obvious, that's what "thermodynamics" means: "heat-motion." In order to study thermodynamics, we have to understand heat, and the first thing about heat is temperature.
This is what the Zeroth Law of Thermodynamics is about. Specifically, the Zeroth Law states that if two objects are in thermal equilibrium with a third object, then they are in thermal equilibrium with each other. This may seem obvious: If A = B and B = C, then A = C, right? Well, not necessarily. Suppose I know you. And you know somebody else. Does that mean I know that somebody else? Not necessarily. It was only after the First and Second Laws of Thermodynamics were codified and experimented upon that the Zeroth Law was developed (thus the name). It seemed to be so obvious but as just seen, it isn't necessarily so.
It is because of the Zeroth Law that we can even really discuss temperature. It allows us to measure it reliably: By putting a calibrated device against an object, we can get a reading. If we get that same reading when putting the device against something else, then we know that the two objects have the same temperature because of the Zeroth Law.
Temperature is measured in Kelvin. Please note that it is not "degrees Kelvin" but simply "Kelvin." There are lots of ways to measure temperature such as by noting the shape of a piece of metal, the electrical resistance in a wire, the pressure of a gas, and so on. The accepted standard is to use that last one. By putting a gas in a constant volume and measuring the pressure exerted by the gas, we can come up with a scale of temperature.
This leads us to the question of heat. Heat is not temperature, but it is closely related to temperature. We must be very careful not to confuse heat with temperature, however. For example, when we say that we've made something hotter, that's a reference to temperature, not heat. Heat is energy and is measured in Joules. Temperature, as noted above, is measured in Kelvin. If two objects have different temperatures, then heat will flow between them if they are put in contact with each other. The heat causes the change in temperature of the two bodies by flowing out of one object and into the other.
Not every object changes the same amount in temperature when a given amount of heat is applied to it. This difference in reaction is called "heat capacity." It can be defined as such:
Q = cm(Tf - Ti)
Q is the amount of heat added in Joules
m is the mass of the object in kilograms
Tf is the final temperature of the body
Ti is the initial temperature of the body
c is the "specific heat capacity" of the material the object is made of.
For example, if I add 100 J of heat energy to a kilogram brass, it will have a temperature difference of about 0.26 K:
100 J = 380 J/(kg · K) · 1 kg · (Tf - Ti)
(Tf - Ti) = 100/380 = 0.26 K
But if I add 100 J of heat energy to a kilogram of ice, it will have a temperature difference of about 0.04K...about one-sixth as much as brass.
Too, just because heat is added to a substance doesn't mean its temperature will rise. When you hit the phase points (solid/liquid, liquid/gas, gas/plasma), the heat added does not cause an increase in temperature but rather a change of phase. Again, each material has a different way of changing phase and the amount of energy needed to make it do so. Water, for example, is more difficult to make melt than silver. The way the molecules of water bind together in its crystalline structure is much more difficult to break than the way the atoms of silver bind together in its crystalline structure. In this case:
Q = Lm
L is the "heat of fusion" (the amount of heat required to make it change phase from solid to liquid) measured in kJ/kg
m is the mass of the object measured in kg
Suppose we have a kilogram of lead at room temperature (293 K) and want to melt it. How much heat do we need to add?
Well, the melting point of lead is 601 K and the specific heat capacity of lead is 128 J/(kg · K). First, we want to raise the temperature of the lead to its melting point. Using the above formula, we get:
Q = 128 J/(kg · K) · 1 kg · (601 K - 293 K) = 39.4 kJ
Next, we need to actually melt the lead. Lead has a "heat of fusion" of 23.2 kJ/kg. Thus, we have:
Q = 23.2 kJ/kg · 1 kg = 23.2 kJ
Thus, we need to come up with 39.4 kJ + 23.2 kJ = 62.6 kJ of heat in order to melt a kilogram of lead starting at room temperature.
This point is to show how heat and temperature are related, but are not the same thing.
Heat and work are also related. One of the things we know from chemistry is that if you increase the temperature of a gas, you increase the amount of pressure it exerts. We can imagine a cylinder that has a piston on top of it that is supported by trapped gas. For the sake of the experiment, no gas can escape the confines of the cylinder/piston, but the piston is free to move up and down in the cylinder. If we put some weight on the piston, it will compress the gas and by varying the amount of weight on the piston, we can control the pressure and the volume of the gas.
Suppose we decrease the amount of weight on the piston. The pressure of the gas inside will cause the piston to rise up a little, ds. How much work is done by this change?
Well, the work is force through a distance so the change in work is:
dW = F · ds
The amount of force is the amount of pressure pressed against the area of the piston: F = pA. Thus:
dW = pA · ds
But notice that A · ds is the volume increase caused by the piston moving up. Thus, A · ds is the change in volume that is caused and therefore:
dW = p (A · ds) = p dV
Integrating this, we get:
W = Integral p dV from the initial volume to the final volume.
But notice, by increasing the temperature of the gas, we can increase the pressure if we keep the volume of the gas constant. But if we decide to let the volume increase so as to keep the pressure constant (by keeping the weight on the piston constant), we can do work by adjusting the temperature of the gas. And since the temperature of the gas is adjusted by adding heat, we learn that heat can be turned into work.
Notice that by controlling the position of the piston and the temperature, the amount of heat and the amount of work are not always equal. If we add weight to the piston proportionally to the amount of heat that we add to the gas to increase the temperature, we end up not doing any work: The piston doesn't move and yet we've spent a lot of heat.
Similarly, if we remove weight from the piston, it will rise, but we haven't added any heat to the system. Work is done from the internal energy that the gas has rather than from any external source.
Or, we can have a combination of the two where we add heat to the gas and remove weight from the piston and then we'll get a different amount of work being done.
This leads us to the First Law of Thermodynamics. Essentially, it states that everything has to go somewhere. In any process, all the heat and all the work has to be taken into account and all the energy of the system must balance.
One of the things we discover through experimentation is that it doesn't matter how we go about futzing with the system. The amount of heat change and the amount of work done, sum to the same amount no matter how we change the temperature and volume. The "internal energy" of a system is defined:
delta-U = Q - W
delta-U is the change in internal energy of the system
Q is the heat added/subtracted from the system
W is the work done by/to the system.
There are some special things about this:
If there is absolutely no transfer of heat, then the only change in internal energy is done by the work that is done by the system or to the system. Think of the piston where all we did was remove weight. We didn't add any heat to the system. Any change of energy is from the gas:
delta-U = -W
Such a change is called "adiabatic."
Notice that this does not mean the temperature of the gas does not change. It does: When a gas expands adiabatically, it cools down. But that change in temperature is not lost to the environment. It is the mere fact that the volume changed that caused the change in temperature.
If the gas is kept at the same temperature but we wish to have a change in work by either compressing it or letting it expand, we will have to add or remove heat from the gas in order to maintain equilibrium. If we compress the gas, it will increase in temperature so we will need to remove heat. If we expand the gas, it will decrease in temperature so we will need to add heat.
Such a process is called "isothermal."
Suppose that we do a whole bunch of things to the gas but at the end, we're back where we started. This is a cyclical process and at the end, the internal energy of the gas remains the same:
Q = W
It's this last part that we start getting into the Second Law of Thermodynamics. In a perfect engine, heat can be converted entirely into work. In reality, however, no engine can ever do this. There is always some heat lost in the process. The gas is touching the cylinder and thus there will be some heat exchange between the gas and the cylinder. If we add heat to the gas, not all of it can be used to move the piston.
The Second Law states that there is no process by which heat can be converted entirely into work. In an engine, we have a reservoir at a certain temperature and another reservoir at a lower temperature. We can extract heat energy from the high-temperature reservoir and transfer it to the low-temperature reservoir. This can be done simply by letting the two touch each other. But, we can insert a device between the two that can use that heat to do work such as the piston device described above.
From the First Law, we know that all the energy has to add up. Thus, we know that we can determine the amount of work done by an engine by examining the amount of heat taken out of the high-temperature reservoir and comparing it to the amount of heat delivered to the low-temperature reservoir. The difference must necessarily be the amount of work done:
|W| = |Qh| - |Qc|
|W| is the absolute value of the amount of work done (positive for an engine, negative for a refrigerator)
|Qh| is the absolute value of the amount of heat taken from the high-temperature reservoir (negative for an engine, positive for a refrigerator)
|Qc| is the absolute value of the amount of heat given to the low-temperature reservoir (positive for an engine, negative for a refrigerator)
Notice that I'm talking about a refrigerator. A refrigerator is essentially an engine that runs in reverse. That is, an engine takes heat from a high-temperature reservoir and sends it to a low-temperature reservoir, in the process doing work. A refrigerator, on the other hand, takes work and in the process transfers heat from a low-temperature reservoir to a high-temperature reservoir.
What the Second Law states is that you can't have that heat transfer from the low to the high without an input of work. While it is quite possible (and very simple) to have heat go from the high to the low, doing the reverse requires the input of work.
Thus, the Second Law can be stated in two, equivalent ways:
1) Heat can not be converted entirely into work: There are no perfect engines.
2) Work must be used to transfer heat from a lower temperature to a higher temperature: There are no perfect refrigerators.
Notice what this means. Suppose we have a regular refrigerator. Thus, it requires an input of work to move the heat. But if there were a perfect engine, then we could transfer all that heat into work to run the refrigerator. The end result would be the transfer of heat from the low to the high without any expenditure of work: The heat from the hot is converted entirely into work and is then given right back to the high by the refrigerator, along with some extra heat pulled from the low.
One of the things we get out of the Second Law is that some energy cannot be used for work. One way of visualizing this is that if we have two bodies at different temperature, we can pull heat from the high and send it to the low, doing some work in the process. But eventually, the high and the low will be at the same temperature and we won't be able to pull any more heat out of the high. Note, this does not mean that the high has been brought to absolute zero temperature. Instead, it means that the engine works off the temperature differential of the two reservoirs. If they are at the same temperature, no reaction can take place.
If, however, we were to take these bodies and put them into a system with a third body that is not at the same temperature, we can do work by pulling it out of the one that is at a higher temperature and sending it to the lower one until, once again, we are at thermal equilibrium.
One measure of this inability to do work is called "entropy." Contrary to what people may tell you, it has nothing to do with disorder. While the idea of disorder is a convenient metaphor for entropy, we must remember that it is a metaphor. Entropy has to do with heat and how you can't do any work at a certain temperature.
First, let us examine engines. Let's look at a specific type of engine called a Carnot engine. It consists of an isothermal expansion followed by an adiabatic expansion. Then, there is an isothermal compression followed by an adiabatic compression to the same state we were in at the start.
Notice that for the first step, we have a change in heat but no change in temperature. From our discussion above, this means that the internal energy of the system remains constant and the amount of heat is equal to the amount of work. From the kinetic theory of gases, we know that the amount of work done by an ideal gas is related to the change in volume:
|Qh| = |Wh| = nRThln(Vb/Va)
n is the number of moles of gas
R is the ideal gas constant
Th is the temperature at which the isothermal expansion takes place
Vb is the final volume
Va is the original volume
Notice that this will be the same for the isothermal compression:
|Qc| = |Wc| = nRTcln(Vc/Vd)
So this takes care of the heat changes in the isothermal steps.
If we divide these two, we get:
|Qh|/|Qc| = [Thln(Vb/Va)]/[Tcln(Vc/Vd)]
Again, from the kinetic theory of gases, we know that an adiabatic expansion between two temperatures has the same ratio of volume as an adiabatic compression between those same two temperatures. That is:
Vb/Va = Vc/Vd
Therefore, we can say:
|Qh|/|Qc| = Th/Tc
We can re-arrange this:
|Qh|/Th = |Qc|/Tc
Since this is a cyclic engine, Qh and Qc have opposite sign, so we can drop the absolute value signs and again re-arrange:
Qh/Th + Qc/Tc = 0
If we extend this process through a great number of adiabatic and isothermal steps, we get:
Sum Q/T = 0
And if we take the number of steps to infinity, creating a smooth process:
Integral dQ/T = 0
One of the things we have learned from the study of temperature and internal energy is that they are properties of the system. That is, a system has a set amount of internal energy and a temperature. Notice in this cyclic process, we have a variable that returns to zero when we come back to the original state. Therefore, this property must be akin to internal energy or temperature: A state variable.
We call this state variable "entropy":
dS = dQ/T
Notice the units on entropy: Joules/Kelvin. This makes sense given our previous understanding about the Second Law. The Second Law indicates that some energy is always lost in a process. Energy is measured in Joules. And the process takes place at a temperature. Thus, we are not surprised to find that entropy is measured in Joules/Kelvin. For every Kelvin of temperature change, some heat energy is lost.
Suppose we have a kilogram of ice melting to water. What is the change in entropy?
Swater - Sice = Integral dQ/T = 1/T · Integral dQ = Q/T
The heat required to melt the ice is:
Q = mL = 1 kg · 333 kJ/kg = 333 kJ
Thus:
Swater - Sice = Q/T = 333 kJ/273 K = 1220 J/K
And notice that this heat has to come from somewhere (First Law). Therefore, the entropy change of the environment is equal and opposite to what happened to the ice: -1220 J/K.
Given this definition of entropy, we can restate the Second Law as follows:
In any process that takes place in a closed system, the entropy of the entire system cannot decrease but must either increase or remain the same.
Notice that in the process of melting the ice, the total entropy remained the same for the entire system. We have not violated the Second Law. While it is true that this was a reversible process and real processes contain irreversible components, there is no violation. What we have described is the lower limit: The entropy remains the same. With irreversible processes, the entropy will increase, even if only slightly.
And finally, notice that at no point during this discussion did the concepts of "order" or "disorder" or "complexity" or "information" come up. Once again, that is because thermodynamics isn't about such things. It is about the movement of heat. It is about energy.
The second law of thermodynamics states that for a closed system, the change in entropy must always be non-negative. But since a great deal of reactions do not take place in a closed system, what do we do?
For any reaction, there is the system in which the reaction takes place and the surroundings of the system. Thus:
delta-Stotal = delta-Ssys + delta-Ssurr
This means that the change in entropy of the system might be negative so long as the change in entropy of the surroundings are sufficiently positive to have a non-negative result. Or conversely, the change in entropy of the surroundings may be negative so long as the change in entropy of the system is sufficiently positive to have a non-negative result:
delta-Stotal = delta-Ssys + delta-Ssurr > 0
There is another variable in thermodynamics called "enthalpy." Conceptually, enthalpy is the opposite of entropy. Whereas entropy is the amount of heat unavailable to do work, enthalpy is the heat that is available. Entropy is commonly given the variable S while enthalpy is commonly given the variable H.
Suppose our reaction is taking place at constant temperature and pressure. Then the energy absorbed by the system has to come from somewhere. And you guessed it, it comes from the surroundings:
qsurr = - delta-Hsys
As we'll recall, q is the variable associated with heat energy. The above is the first law of thermodynamics. Notice that the enthalpy change is opposite in sign from the heat used. If the surroundings gain heat, then the system loses it. If the surroundings lose heat, then the system gains it.
Now, the change in entropy of the surroundings is a measure of the heat change at that temperature:
delta-Ssurr = qsurr/T
Since qsurr is equal to - delta-Hsys, we can substitute:
delta-Ssurr = qsurr/T = - delta-Hsys/T
Substituting the above into the equation describing total entropy, we have:
delta-Ssys - delta-Hsys/T > 0
Now, all we have to do is some mathematical re-arranging:
T * delta-Ssys - delta-Hsys > 0 (multiply both sides by T)
-(delta-Hsys - T * delta-Ssys) > 0 (factor out the -1 and rearrange terms)
delta-Hsys - T * delta-Ssys < 0 (negation flips inequalities)
Now, since H, T, and S are all state functions, this equation of state functions must also be a state function. We call this result "free energy" and give it the variable G after Josiah Willard Gibbs who developed much of modern physical chemistry.
So long as G < 0, the reaction is spontaneous. If G > 0, then the reverse reaction is spontaneous. If G = 0, then the system is at equilibrium. This does not mean that no reaction is occurring. It means that no net reaction is occurring...for every forward reaction, a reverse reaction takes place.
For example, if you take a glass ball that is half-filled with liquid water, the empty space will fill with water vapor. The system is then at equilibrium: Molecules of water still break free from the liquid water to become gaseous, but in the process a molecule of gaseous water is captured by the liquid. The system is at equilibrium.
But notice what this means: What drives a reaction is a function of the energy available, the energy lost, and the temperature at which it takes place.
If there is a lot of energy available but the temperature is low, then even with a positive change in entropy, the reaction still won't happen if that entropy change isn't large enough.
Similarly, you can have a very large decrease in entropy and still have a spontaneous reaction so long as the change in enthalpy is sufficiently negative to counteract.
As an example: Is the reaction 2NO2(g) <-> N2O4(g) spontaneous at 298.15K?
First, calculate the enthalpy of the reaction:
delta-H0 = delta-H0f(N2O4, g) - 2delta-H0f(NO2, g) = 9.16 kJ - 2(33.2 kJ) = -57.2 kJ
Similarly:
delta-S0 = delta-S0N2O4 - 2S0NO2 = 304.2 - 2(239.9) = -175.6 J/K
Therefore:
delta-G0 = -57.2 kJ - (298.15 K)(-175.6e-3 kJ/K) = -4.8 kJ
Therefore, the reaction is spontaneous to the right. Notice that this happens even though there is a negative entropy change. That is because the entropy change is small enough that the negative enthalpy change overpowers it.
The idea that every reaction everywhere always has to have a positive entropy change in order to be spontaneous is simply false. You can have negative entropy changes and still be spontaneous. It depends upon the amount of energy available and the temperature at which the reaction takes place.
Now, that is the definition of "entropy." This idea that entropy is a measure of "disorder" is simply false. The concept of "disorder" is nothing more than a metaphor to help us understand the processes of what is going on.
quote:
You can't just blast sunlight on the earth and expect life to proliferate.
Why not? Is there some reason why sunlight cannot provide energy of reaction? It's how plants live, after all.
quote:
If that were the case, then all of the planets in the solar system would be inhabited with life. Clearly that is not the case.
And you can't think of a single reason why the earth is different from other planets other than god?
quote:
There first must exist some mechanism which converts raw energy in to useful energy.
Incorrect. Mere existence means that you can use energy. There does not need to be a "mechanism." Existence is mechanism enough.
Or are you saying that physics and chemistry need god? Ooh! It looks like we're back to the question that never gets answered!
Is there anything that happens on its own or is god required for everything?
quote:
For example: Sunlight blasts our roofs all of the time. The introduction of that energy certainly isn't improving the roof, now is it?
What is this word "improve" and what does it mean? I'm sorry, but science doesn't understand "improve." That's a perceptual, subjective determinant.
quote:
The big question is, why?
Not because of entropy. Nothing dies of entropy. Then entropy of an organism recently dead is pretty much the same as it was moments before when it was alive.
No, things die because the chemical reactions that are used to keep it alive have stopped. That's because no chemical reaction is perfect every time.
quote:
One side is arguing about thermodynamic entropy, while the other side is arguing about the other kind of entropy which stipulates that all systems tend towards disorder.
Incorrect.
One side is arguing about entropy while the other side is wondering why you're bring up entropy (and a fundamentally flawed, fantasy version of it at that) in the first place.
Things don't die because of entropy. Entropy gives life. You couldn't live if it weren't for entropy.
quote:
Why then is their not a specific name for it?
Because it doesn't exist...?
Nah, couldn't be that.
Rrhain
Thank you for your submission to
Science. Your paper was reviewed by a jury of seventh graders so that they could look for balance and to allow them to make up their own minds. We are sorry to say that they found your paper "bogus," specifically describing the section on the laboratory work "boring." We regret that we will be unable to publish your work at this time.
™ Version 4.2