Entropy and the Second Law of Thermodynamics - Online Tutor, Practice Problems & Exam Prep

Entropy is a measure of a system's randomness, reflecting how energy is distributed within it. In thermodynamics, higher temperatures correlate with higher entropy due to increased energy spread. The change in entropy (ΔS) can be calculated using the equation ΔS = qT, where q is heat and T is temperature in Kelvin. The second law of thermodynamics states that the total entropy of a system and its surroundings always increases, emphasizing the irreversible nature of energy transfer processes.

Intro to Entropy

Hey, everyone. So in this video, I'm gonna introduce a concept called entropy. I'm super excited for this video because entropy is one of those tricky, sort of hard-to-understand conceptual things. But hopefully, by the end of this, you're gonna have a great solid understanding of what entropy is all about. And I'm also gonna show you the equation that you need to solve problems. So let's go ahead and get started here. What is entropy all about? Well, almost certainly the definition you're gonna get is that it's a measure of something called disorder, but I never really like this definition because disorder is kind of vague. What does it mean to be disordered? How do you measure or calculate something's disorderedness? So I think a better definition is that it's a measure of an object or a system's randomness. So it kinda has to do with statistics and probability.

Let me show you an example here. Imagine I had these 2 jars of different colored balls and I was gonna mix them together in a big jar and shake the whole thing up. Now before I've mixed anything, the 2 sort of colors are separated in their own containers and this system doesn't have a whole lot of randomness. But when you combine them and you shake the whole thing up, you're gonna end up with something that could look like this in which the balls sort of mix around and get more randomized. And if you keep shaking this, you're just gonna end up with more and more random configurations. The balls will never separate back into their own colors again. So what happens here is because there's less randomness in the before case before I mix them up, there's lower entropy. But once I mix everything up, there's more randomness and therefore higher entropy. Alright? So what does this have to do with thermodynamics? Well, you can imagine that instead of mixing these containers of balls together, I'm mixing jars of different gas particles together. So that's sort of the connection there.

So in thermodynamics specifically, entropy measures how randomly a system's energy is spread out. So let me show you another example of this. Imagine I had 2 jars, but now one's full of ice and one is full of water. And I ask you which one is more disordered. Now you might be tempted to say it's the ice. Notice how the ice cubes are sort of in their random positions, whereas the water is sort of relatively flat. It's not really doing anything. But, actually, this would be the wrong answer. The ice is actually the one that has less randomness. And so what happens here is you have to look down at the atomic level to see how the energy is spread out. Remember that for ice, which is a solid, all of the atoms are sort of locked together like this in their positions and they can't move around, which means that the energy can't really spread out enough. Whereas the water which is a liquid, remember these liquid particles, they're actually able to freely move around because the bonds have broken. So because these things are able to move around freely and therefore more randomly, the energy is more spread out. So the whole idea here is that the water has more randomness even though it looks like it is less disordered and therefore higher entropy, whereas the ice has lower entropy. So in general, because things with higher temperatures have more energy, the energy can get spread around a little bit more. They have higher randomness and therefore higher entropy. Alright. So hopefully that's a great conceptual understanding of what entropy is.

Now let's go ahead and talk about the equation. So in all problems, you're gonna need to calculate the change in entropy, or in most problems rather than the entropy itself. The symbol we use for the change in entropy is Δs (Delta s). Remember, Delta means change and s is for entropy. The equation is actually really straightforward. It's actually just qt where q is just the heat. That's a variable we've seen over and over again, and T is the temperature, which must be in Kelvin. Now what's really important about this equation is that it only works for isothermal processes. This only works when there is no change in temperature. One way to remember this is that this equation doesn't have a Delta T. It just has one T. So you don't need multiple T's. Usually in a problem, you'll just be given 1 and that's what you plug into this problem here. Now just, again, the units are also just gonna be in joules per joules per Kelvin and it's something you can just get from this equation here. Alright. So let's get started with our first example.

In our first example, we're gonna add 24100 joules of energy to a body of water that's so large that its temperature remains constant, And that's really important because remember our entropy equation only works for constant temperature. And we wanna calculate the change in entropy and that's just Delta s. So we wanna calculate Delta s and in order to do that, we're just gonna use our new equation. It's gonna be q divided by T. So we need to figure out how much heat gets transferred divided by the temperature. Now we're told in this problem that we're gonna add 24100 joules of heat, so that's gonna be our q. It's 24100. Now we just have to plug in the temperature. Do we plug in 27 or something else? Well, remember, there's only one temperature. It is gonna be this 27, but we actually have to convert it to Kelvin because this T has to be in Kelvin. So we're gonna have to add 273 to this. So we're gonna do 27 +273 and this just becomes 24100 divided by 300. When you work this out here, you're gonna have 80 joules per Kelvin, and that is the answer. So one way you can kind of think about this is that you added energy to a system, a body of water. The temperature didn't change. What happens is you basically spread out a little bit more energy throughout that body of water. So the entropy did increase by 80 joules per Kelvin. Alright? Let's move on to our second example here.

We're gonna calculate the change in entropy when 2 kilograms of water freezes completely to ice. Now what does that mean? Well, we've seen these kinds of problems before where we have a substance that's changing phases. We have water that's freezing to ice. And we wanna calculate the change in entropy, that's just gonna be Delta s, and we're just gonna use q over T. Now what happens here is we have to figure out how much heat is transferred into this, is transferred when you have water freezing to ice. What happens is for phase changes, we had this old equation here that q equals mL. We used this a lot in calorimetry. So you're just gonna replace this q with mL and we're gonna use the appropriate latent heats. Remember there were two of them for water. We had the latent heat of fusion and the latent heat of vaporization when something turned to steam. Now we're only actually gonna use this LF because that's when water is changing to ice and vice versa. So we're gonna use MLF and then we're gonna divide it by the temperature. Now, what is this temperature here? Well, it's the temperature at which water freezes to ice. So remember that the delta T for a phase change is equal to 0, and that's really important here because that means that we can use this equation here q over T. If delta T wasn't 0 we couldn't use this equation. So there's no change in temperature when you have a phase change, and this temperature for freezing is just 0 Celsius, which is just 273 Kelvin. So what happens here is this becomes t freeze and now we can just plug everything in. So we're gonna have the mass which is 2, the latent heat of fusion which is 3.34×105, and now you're gonna divide this by 273. Now the one thing that we forgot though is a minus sign. Remember that we have water that's turning and changing to ice, so we actually have to remove heat in order to do that. So this q here actually has a negative sign. So this is gonna be negative MLF and we're gonna have a little negative sign here. Now, when you work this out here, let me just move this down, What you're gonna end up with is that Delta s is equal to negative 2,447 joules per Kelvin. So we can see here is that in some cases, the change entropy will be positive and in other cases, it will be negative. And the sign of Delta s is always gonna be the same as q. And, basically, the rule is that if you add heat, which means that q is a positive number, then the entropy increases, and that's what we saw in the first example. If you remove heat, then q is a negative and the change in entropy or the entropy decreases. Alright. So that's it for this one, guys. Let me know if you have any questions.

3 moles of an ideal gas are compressed isothermally at 20°C. During this compression, 1850 J of work is done on the gas. What is the change of entropy of the gas?

• A. –24.9 J/K
• B. 6.31 J/K
• C. –6.31 J/K
• D. –9158 J/K

You have a block of ice at 0°C. Heat is added to the ice, causing an increase in entropy of 120J/K. How much ice melts into water in this process?

• A. 0.098 kg
• B. 0 kg
• C. 1.32×10^-6 kg
• D. 1.47×10⁵ kg

Entropy Increase When Braking

Hey, everybody. So let's take a look at our problems. We have a car that's driving along the road. It's going to suddenly slam on the brakes, and we want to calculate what's the change in entropy of the air. So in other words, what we want to calculate here is delta S, and I want to say delta S for the air. But before we get started, I want to sort of draw out what's going on here. Right? So we've got this little road, and I'm going to draw this little car like this. Don't make fun of my car drawings. Like that. And then what happens is it's chugging along initially at some initial speed, which is 20 meters per second. Then what happens is over some time, it's going to slam to a stop. So eventually what happens is that V final is going to be 0. Alright? So what's going on here? If you want to calculate the change in the entropy, we're going to have to use the change entropy equation. Now remember, we can only use this equation Q over T as long as the temperature remains constant, but it does because we're told in this problem that when you slam on the brakes, the air, which is, like, so massive and large, remains at a constant temperature of 20 degrees Celsius. So what happens is if you add a little bit of heat to it, the air is obviously so massive that it doesn't really change temperature. So we have delta S for the air is the heat that's added to the air divided by that temperature. Right? So we have what that temperature is. So the question here is, well, how much heat is actually added to the air? We don't have an equation for this. Right? We can't use like Q equals MCΔT or something like that because we don't have the mass of the air, we don't have the specific heat. So what's really going on? Well, what happens here is that we have to realize that this initial velocity for the car means that this car has some kinetic energy. So in other words, this kinetic energy, I'm going to say K_initial is equal to 12mv02. Squared. Actually, I'll call this K_zero. So I'll call this K_zero. And eventually what happens is that when you slow to a stop, right, you slam on the brakes, eventually when you get to this point here your K_final is equal to 0. So in other words, you had some kinetic energy and then you lost it because you slammed on the brakes. So then where did all of this energy go? Remember that energy can only change forms. It can't be created or destroyed. It doesn't go to potential energy because it's still staying at the same height. So basically what happens here is that all of the K that gets transferred basically goes into friction. It goes into the friction of the brakes and that gets dissipated as heat. Remember, friction, when you rub something, it's really just heat transfer. So what happens here is that the delta K, the change in the kinetic energy, really just becomes Q. It becomes the Q that gets given off to the air. Alright? So that's what's going on. So basically, what we can do here is we can say, well, the Q_air really comes from the change in the kinetic energy. Now really what happens is this change is really just the 12mv02 because the K_final is 0. So we don't have to do final minus initial. So this really just becomes one half, and this is going to be mv02. We have the mass of the car and the initial speed, so then we can go ahead and do that. So we're going to have one half of the mass, which is 1500 times the initial velocity, which is 20 squared. So now you divide that by the initial temperature. Now the temperature is 20 degrees Celsius, but we have to convert that to Kelvin. So this just becomes 293 Kelvin, and that's what we plug into our denominator here. So if you work this out, what you're going to get is that the change in the entropy for the air is equal to, 1.02 times 10 to the third, and this is going to be joules per Kelvin. Alright? So that's how much entropy gets sort of dissipated. That's how much energy you've now dissipated. The energy is a little bit more random and therefore the entropy of the universe increases. Alright. So that's it for this one guys. Let me know if you have any questions.

Calculating Entropy Changes for Systems of Objects

Hey, everyone. So in the last couple of videos, we saw how to calculate the change in entropy, but that was just for one object. In a lot of problems, you'll have to calculate the entropy change for a system of multiple objects. It could be 2 or more, or in some cases, the change in entropy of the universe. For example, the problem we're going to work out in this video together has a hot reservoir that's connected to a cold reservoir. It could be something like a heat engine or something like that, and what we want to do is calculate the net change in entropy of the system. So what we're going to do is work out this problem together because I want to show you a step by step process to solve these problems so you get the right answer every single time. So let's just go ahead and get to it. The very first step that we want to do is actually just draw a diagram. So, again, we have a hot reservoir connected to a cold reservoir. I'm just going to draw a sketch out real quick. So we're told that this hot reservoir has a temperature of 200 Celsius, but we want to convert that to Kelvin, and that's just going to be 473 Kelvin. And we know it's connected, we don't know exactly how, to a cold reservoir. So it's connected to this cold reservoir here, and this cold reservoir has a temperature of 100 Celsius, which is 373 Kelvin. Now, again, we don't know how they're connected. All we know is that 400 joules of heat gets transferred from the hot to the cold. Remember heat flows from hot to cold like this, so we have q=400J.

Alright. So that's the first step here. We've drawn the diagram. The next thing we want to do is we want to write our Δs equation, but because this is a system, we're going to write Δs total . So this is going to be Δs for the total system here. Now the only equation that we've seen for Δs is this one down here. That's q/t, and remember this only works for isothermal processes. However, if we look at the problem, it is isothermal because the reservoirs are so big that the heat exchange has no effect on their temperature, so we absolutely can use this equation. The problem is that this equation only worked when we had one object like the hot reservoir or the cold reservoir. What we want to do is compute the change in entropy of the system, so we have to consider both of those objects. So, in general, what happens is whenever you're writing the Δs total equation, that's just going to be Δs1+Δs2+Δs3. Basically, you just add up whatever objects in your system are exchanging heats and changing entropies, and that will be the total change in entropy.

So we're going to come back to this last part in just a second here. So, basically, what happens is this hot reservoir is losing heat and so that's going to change entropy. I'm going to call this ΔsHR, and this cold reservoir is absorbing heat and so it's going to change entropy as well. I'm going to call this ΔsCR. So the total is just going to be ΔsHR+ΔsCR. So now what I can do is just replace these equations in the next step. So I have that Δs total is going to equal, well, this is going to be q/t plus, and this is going to be q/t. However, these things are at different temperatures, so we don't want to confuse them. So I'm going to call this one TH, and this one is going to be TC. We have those values over here. So now we just start replacing all these numbers with the values that we know. What's the heat that's added or what's the heat that is flowing out of the hot reservoir? It's 400 joules. So because of that, what happens is heat is being removed and we add a negative sign. So this is going to be negative 400/473. Now what happens with the cold reservoir? Well, the hot reservoir is transferring 400 to it, so if the hot reservoir loses 400, the cold reservoir gains 400. So this is going to be positive 400/373. Right? So those are your numbers. Basically, what you're going to end up with here is that Δs total is equal to when you plug this in, you're going to get 0.85 plus this is going to be 1.07, and then when you add the whole thing up, you're going to get 0.22 joules per Kelvin, and that is the answer. Alright?

So what happens is that some part of our system here lost entropy or decreased in entropy. Another part of our system increased in entropy, and the overall effect was that the overall change in of entropy in the system was positive 0.22 joules per Kelvin. And in fact, this is always going to happen. This is actually one of the statements of the second law of thermodynamics, called the entropy statement. Remember, the second law was just a bunch of statements. This one's called the entropy statement, and what it says is that it's impossible for any process to cause an overall decrease in the entropy of a system or, in some cases, the universe. And that's actually the most important part of that sentence there, the entropy of the universe, because what we saw here is that one part of our system did, in fact, decrease in entropy, so it's perfectly fine. If any part of your system decreases in entropy, all that happens is that another part, like the cold reservoir in our problem, must increase in entropy at least the same amount, if not more.

Alright? So that's how to solve these kinds of problems, guys. So, basically, when you go back to this Δs total equation here, the last piece is that when you add up all the changes in entropy for a system, it will always be greater than or equal to 0. Another way of saying this is that the entropy of the universe is always increasing. The best you could possibly hope for is that the change in entropy for a system is 0. That's the best case scenario.

Alright? So, really, that's all there is to it, guys. The last point I want to make is that sometimes for this reason, entropy is called time's arrow, which is basically just a useful way to explain why processes happen the way they do, why they happen irreversibly. For instance, like heat transfers, like we just saw in our problem here. So heat transfer happens only from hot to cold, and it never flows the other way around, and that's because this system here increases, or this process increases in entropy. You can't get that energy back. The energy is more spread out. Another example is friction. When you rub your hands together, you're generating heat, you're spreading out more energy. That's a process that increases the entropy of the universe, so it never will happen backward. If you put your hand on a coffee cup or something, and it's hot, that heat will flow from coffee to your hand and never the other way around. That's just another example. Alright, guys. So, basically, time flows or moves in the direction of increasing entropy. So that's it for this one, guys. Let me know if you have any questions.

A non-Carnot heat engine operates between a hot reservoir at 610K and a cold reservoir at 320K. In a cycle, it takes in 6400 J of heat and does 2200 J of work. What is the total change in entropy of the universe over the cycle?

• A. 2.6 J/K
• B. 9.5 J/K
• C. 16.3 J/K
• D. –2.6 J/K

Entropy of Carnot Engine

Hey everybody. So let's check out our example problem. We have a Carnot heat engine that's taking in some heat from a hot reservoir, expelling it to a cold, and we want to calculate the total change in entropy of the universe. Now we already know this is going to be an entropy problem that involves multiple objects, because the hot and the cold reservoirs are both exchanging heats at different temperatures. So let's go ahead and draw our quick little diagram here. That's my hot. That's my engine, that's my cold. We've got the temperature of the hot is 500. The temperature of the cold is 350. All the stuff is in Kelvin. And so we have the heat that's taken in from the hot reservoir is 2,000 Joules. So that's my QH. Right? That's QH = 2,000 Joules. So then what happens is, it expels some heat to the cold reservoir, but we actually don't know what that Qc is. So the next thing we want to do is just set up our delta S total equation. The total change in entropy for the universe is going to be adding up the two entropy changes for the hot and cold reservoirs. So in other words, delta SH + delta SC.

So what happens here is that this delta SH, because we can assume that this happens at constant temperature, is going to be Qh over Th. But remember, we have to add a negative sign because the hot reservoir loses some heat to the engine, whereas this delta SC here is going to be plus QC over TC. And that's because this cold reservoir is absorbing heat from the engine. Right? So one's positive, one's negative. Okay? So, basically, let's go ahead and figure this out. Right? So we've got delta S, or sorry, negative QH, which is going to be negative 2,000 divided by the 500. Now we have the TC here. We have this 350. But we don't actually have what the QC is. So we're going to have to go figure that out. Now, the way that we did this in another problem was that we actually had the work done. And so if we use the work equation, right, this W equation, we calculate QC. But we actually don't have that here. So in order to figure out what this QC here is, I'm going to need another equation. So let's go ahead and do that. So which equation can we use for the Carnot cycle? We actually have a couple of them. If we knew the efficiency we would use this equation, but we don't have the efficiency. We can't use this equation here. So the only one that I can use that's sort of special to a Carnot engine is going to be this one over here.

This the ratio of the heats is equal to the ratio of the temperatures. Remember, we can only use this for a Carnot cycle. So let's set this up here. So we've got QC over QH equal to TC over TH. Alright. So if you look at this, I have 3 of 4 variables. Right? I've got all the heats and the temperatures. So I can calculate this real quick here. This QC is really just going to be well, TC over TH is 350 over 500. So we're going to multiply that times the QH, which was 2,000. If you go ahead and work this out, what you're going to get here is you're going to get, let's see, I get 1400. So this is 1400 Joules. That's what we plug into this equation now. Alright? So basically, this just becomes, negative 2,000 over 500 plus 1400 over 350. Now when you work this out, what you're going to get is that this delta S total here is actually equal to 0. So essentially what happens is these two things will cancel each other out. So what's going on here is that because Carnot heat engines are ideal and they are perfectly reversible, then essentially what that means is that the change in entropy for a Carnot cycle is always equal to 0. So in other words, Carnot engines are perfect and perfectly reversible, then basically what happens is that the best outcome happens. You have no total change in entropy of the universe.

Alright, guys. So that's it for this one.

Here’s what students ask on this topic:

What is entropy in thermodynamics?

Entropy in thermodynamics is a measure of a system's randomness or disorder. It quantifies how energy is distributed within a system. Higher entropy indicates a more random distribution of energy. For example, in a gas, particles move freely and randomly, resulting in high entropy. In contrast, in a solid, particles are more ordered, leading to lower entropy. The concept is crucial in understanding the second law of thermodynamics, which states that the total entropy of a system and its surroundings always increases, emphasizing the irreversible nature of energy transfer processes.

How do you calculate the change in entropy?

The change in entropy (ΔS) can be calculated using the equation:

$\mathrm{\Delta S}=\frac{q}{T}$

where $q$ is the heat added to or removed from the system, and $T$ is the temperature in Kelvin. This equation is valid for isothermal processes, where the temperature remains constant. For phase changes, the heat $q$ can be calculated using $q=mL$, where $m$ is the mass and $L$ is the latent heat.

What is the second law of thermodynamics?

The second law of thermodynamics states that the total entropy of a system and its surroundings always increases over time. This law emphasizes the irreversible nature of energy transfer processes. It implies that natural processes tend to move towards a state of maximum entropy or disorder. For example, heat will always flow from a hotter object to a cooler one, never the reverse, leading to an increase in the overall entropy of the system. This law is fundamental in understanding why certain processes occur spontaneously and why others do not.

Why is entropy called 'time's arrow'?

Entropy is often referred to as 'time's arrow' because it provides a direction to the flow of time. In thermodynamics, processes that increase entropy are irreversible, meaning they can only proceed in one direction. For instance, heat naturally flows from hot to cold, and this process increases the entropy of the system. This unidirectional flow of energy and increase in entropy gives a sense of time moving forward. The concept helps explain why certain processes, like the mixing of gases or the melting of ice, occur spontaneously and cannot be reversed without external intervention.

What is the relationship between entropy and temperature?

Entropy and temperature are directly related in thermodynamics. Higher temperatures generally correspond to higher entropy because increased thermal energy allows particles to move more freely and randomly. This increased randomness leads to a greater distribution of energy within the system, thus increasing its entropy. For example, water at a higher temperature has more entropy than ice at a lower temperature because the water molecules move more freely. The relationship is quantitatively described by the equation:

$\mathrm{\Delta S}=\frac{q}{T}$

where $q$ is the heat added and $T$ is the temperature in Kelvin.