23. The Second Law of Thermodynamics

Heat Engines and the Second Law of Thermodynamics

1

concept

## Introduction to Heat Engines

7m

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Hey guys. So by now we've talked a lot about the first law of thermodynamics and the next couple of videos, we'll start to talk about the second law of thermo and what we'll see is that it's not so much an equation as it is. A bunch of statements. And these statements have to do with these things called heat engines. So before we get there in this video, I want to introduce you what a heat engine is and hopefully I'll give you a really good analogy for sort of understanding these things. I'll also show you the equations and diagrams that you'll need. So let's go ahead and check it out here. Basically, a heat engine is it's kind of a machine or a device or something like that, and what it does is it converts heat energy into useful work. Now the analogy that I always love to use is that it's just like the engine that's in your car. A car engine is a perfect example of a heat engine. So, I always want you to think about it whenever we talk about these things or just in case you forget now without getting into the specifics, basically where your car works is it takes some gasoline, some fuel source and your car ignites that gas, it creates a ton of heat and thermal energy. And basically what your engine does is it takes that heat energy and then it uses it and converts it into useful work by turning the wheels of your car and that's what makes you go forward in your car. So it converts heat into useful work and it doesn't do this perfectly. So, what ends up happening is that some comes out the back end as exhaust. So, this is just the exhaust heat that comes out the back of your car. All right. So, that's really how any heat engine works. It takes in heat and then converts it into useful work and then it spits out whatever it doesn't use. All right. So, instead of having a bunch of different diagrams for a bunch of different types of machines, physicists a long time ago developed what's called an energy flow diagram. You're gonna see these things a lot in your books and classes. And basically what this thing is, is it represents the heat transfers that are happening and any kind of heat engine, regardless of whether it's a boat or a plane or a car, whatever. Alright, so there's a couple of important things you need to know about them. The first one is that this energy flow diagram has what's called a hot reservoir. Now, what's that mean reservoir? It's kind of like a weird word, but it's basically just a source of heat energy that's going into the engine. In my car engine analogy, the source of the heat energy was the burning gasoline that we ignited. Right? So the hot reservoir is just the gasoline that you're igniting. That's what the engine uses in order to produce work. So the second thing you need to know is the work is the usable energy that's produced by the engine that's turning the wheels of the car. And our analogy, right? So your car takes the gas, sets it on fire and then uses that to spin the wheels of your car and move you forward. All right now, lastly, we have what's called a cold reservoir and that's kind of like the opposite of a hot reservoir. It's basically where all the wasted heat energy goes once it's expelled out from the engine. And in my example here, that was basically just the exhaust pipe. That's where the rest of the not used. Heat energy goes. All right. So that's the basics of how a heat engine works, sort of conceptually. Let's take a look at some of the equations that you'll need. Now remember we talked about the first law of thermodynamics, which was this equation here, Delta E equals Q minus W. We also did in a previous video took a look at cyclic processes and cyclic process had a special condition that the change in internal energy, delta E was zero. So, directly from this equation, if delta is equal to zero and that means that Q has to equal W. So we saw, is that for cycles, the W for the cycle equaled the que the work done in the cycle equals the heat added during a cycle. Now, heat engines are always going to be cyclic. Now, this process here of burning gasoline and this and that doesn't just happen one time it happens over and over and over and over again. As long as you can feed your engine more gas. Right? So heat engines always happen in cyclic processes that repeat over and over again. We also have that heat is flowing in and out over a cycle. So basically just directly from this equation, the work that's done by the engine is equal to the heat that gets added over the cycle. Now, what happens is again, we have two heats, we have one that's going in one that's going out. So what is this Q. Net here? It's actually both of them. It's actually the heat that's going in, that's Q. H. And it's positive because it's getting added to your system and then this Q. C. Is leaving your system, so it gets a negative sign. So that's basically the equation that you're gonna see. One way you can kind of think about this is that the work that's done is equal to whatever heat that goes in minus whatever heat that goes out. But I just highlighted this one because this is the one that you're most likely going to see in your problems. So that's really all there is to it guys, let's go ahead and take a look at our example here. So we have a heat engine that's taking in 500 joules of heat, it's doing 300 joules of work, and we want to calculate how much waste heat is expelled from the engine. So, which variable is that? Well, if we want to sulfur some kind of a heat that's going to be a Q. We're solving for how much waste heat is expelled. So, I'm going to label this as Q waste. All right. So, before we actually get into any equations, let's sort of draw out in our energy flow Diagram, what's going on here? So, remember, we have a hot reservoir, we have a cold reservoir. What we're told in this problem is that 500 jewels of heat is being taken into the heat engine. So, which variable is that? Remember your heat engine takes in heat from the hot reservoir. So, that just means that this QH or Q in is equal to the 500 jules. Now, what your heat engine is doing is it's taking some of that heat energy and it's doing 300 jewels of work. So, that's pretty straightforward. The work is always just gonna come out of your heat engine like this. So, you're w here is just gonna be the 300 jewels. So, if we're asked to find out the heat, the waste heat, it's basically the one heat that we don't have in this debt in this diagram here, which is the heat that gets expelled out to the cold reservoir. So this is Q seed or in other words Q out. And that's basically what this waste heat represents. All right. So the trick to these kinds of problems here is kind of translating which variable they're asking you to solve? So Q waste is equal to Q. C. In other words it's equal to Q out. I need any of these. They all mean the same thing. All right. So that just means if we're looking for Q. C. Now we can just use this equation over here. So let's get started. The work done by the engine is equal to the nets. Heat transfer to the cycle which is gonna be qh minus Q. C. Alright, so, we have the work that's done by the system by the engine. That's the 300 jewels. And we also have the heat that gets added to the system. So we can solve this Q. C. Over here, just by rearranging. So basically what happens is you just flip you move this to the other side, this becomes Q. C. Equals Q. H minus the work done by the engine. So, it's just gonna be 500 jewels -300 jewels. And you end up with 200 jewels. All right. So, it's as simple as that basically what's going on here, is that there's 500 jewels that gets added to your system and then the engine sort of splits that up. Right? It does 300 joules of work and it spits out the rest which it doesn't use, which is the 200 jewels that's left over. So this w here there's 300 is always going to be the difference between what comes in and what goes out. Right. So, anyway, that's uh an introduction to heat engines. Let me know if you have any questions.

2

Problem

ProblemAn aircraft engine takes in 9 kJ of heat and expels 6.4 kJ of heat each cycle. How much mechanical work does the engine do each cycle?

A

15.4 kJ

B

2.6 kJ

C

2.6 J

D

6.4 kJ

3

example

## Power Output of a Gasoline Engine

6m

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Hey guys, so I've got a great sort of real life example problem that's gonna pull together a lot of equations that we've seen from heat and temperature. And even earlier stuff in physics and also with heat engines. So just kind of bear with me here, there's lots of parts to this problem, but I think it's a really, really great one. So let's get started. We have a gasoline engine that takes in this amount of heat and it does this by combusting gasoline and then it does 3700 jewels of work per cycle. So whenever I get these sort of numbers here, I'd like to draw the energy flow diagram. So I've got my energy flow diagram, this is my hot reservoir that pulls in heat to the cold reservoir and then it does some work and then it expels the rest out to the cold reservoir. Now, what we're told is that it takes in 1.6 times 10 to the four jewels of heat from combustion gasoline. Remember that's going to be Q. H. That's the heat that you draw in from the hot reservoir. And really this comes from combustion of gas, then it does some amount of work, right? And this work here, we're told is equal to 3700 jewels. The heat that it takes in is 1.6 times 10 to the fourth. And then what happens is it's going to expel some heat out to the cold reservoir. So this is just basically the waste heat, that's how much heat that doesn't get used by the engine. So let's get started with our first part here, we're asked to calculate the heat that is expelled each cycle. So which variable is that? We'll remember. It's gonna be heat. So it's gonna be one of the queues. And we're really looking for the waste heat. How much heat doesn't get used by the engine. So if we're looking for Q. C. Here, we're gonna start off with our equation that relates works and heats for heat engines. This equation over here. So we have the absolute value of the work is equal to Q. H. Minus Q. C. And everything sort of absolute values. Everything is positive numbers. Just so you don't get confused. All right. So what happens here is we're looking for the Q. C. We're gonna rearrange this equation and we're gonna move the QC to the other side and then the work over to the other side so that it becomes positive. Right? So what happens is that the QC is just gonna be Q. H minus W. And that should make some sense. Remember what happens here? Is that a heat engine takes in some heat from the heat engine from the hot reservoir? It does some work right? It sort of produces some mechanical work and then whatever it doesn't use gets sort of thrown out to the waste reservoir the cold reservoir as waste heat. So really what happens is you're gonna take in 1.6 times 10 to the fourth. And then you're gonna do 3700 jewels of work? So you're gonna subtract that and what you end up with is waste heat is 1.23 times to the fourth jules. Alright, so that is the first answer here. That's how much he gets expelled out. So now let's take a look at the second part of our problem. So, in the second part of our problem, we are asked to calculate what mass of fuel is burned each cycle. So what variable is that? Well, for mass, we've always use em. So that's basically what we're looking for here. So where does em come into play with our heat engine? Well, if you look at what happens, we're told that gasoline has a latent heat of combustion. Remember we looked at latent heat when we looked at kalorama tree problems, we're told that this latent heat is this number over here, Remember what this means? Is that for every one kg of mass that you've burned for every one kg of of gasoline, you basically can extract this amount of energy from it. So, whenever we used to sort of phase change here, combustion is sort of one of those phase changes, we used Q equals M. L. So, really, what happens here is that this qh remember the heat that comes from combusting gasoline is going to equal mass times however much sometimes latent heat of combustion. Right? So we're looking for here is this little m So what happens here is I'm gonna take my cue H which is 1.6 times 10 to the fourth, and I'm gonna divided by the latent heat of combustion 4.6 times 10 to the seventh. So, if you look at the exponents here, we have 10 of the 4, 10 of the seven, we should expect to get a number that's pretty small here, Right? Because We've only generated 1.6 times 10-4 when you work this out, What you're gonna get is that the mass that gets combusted is 3.5 times 10 to the minus four kg. In other words, uh, if you sort of convert this two g, that's about 0.35 g of fuel and that kinda makes sense. Right? Every time you have an explosion in the gasoline engine, it actually doesn't require a whole lot of gasoline because it mixes with air and the cylinders are small and things like that. So, it doesn't require a whole lot of gasoline and you can generate a large amount of work from it. Alright, so now let's move on to our last part over here. Our last part asks us to calculate, well, if this engine here completes 60 cycles per second, in other words, 3600 rpm. For those of you who are familiar with cars, then what is going to be the power output in kilowatts. So essentially we're looking for here is what is p remember that is the equation for power? Well, we're gonna have to use an old equation that relates power with the amount of energy and time. Right? So we're told this is the this is 60 cycles per second. What's its power output in kW? So we're gonna have to use this equation over here. That power is equal to work done, divided by Delta T. In other words, it's also equal to the amount of energy that's produced, divided by delta T. So what is the energy that gets produced here? Well, really what happens here, the work that's done is it's actually just gonna be the 3700 jewels, right? Every single cycle you're producing 3700 jewels of work. So then what's the delta T. While the delta T. Is going to be a little bit trickier to solve because we have to relate it to the frequencies. I'm going to sort of go over here. So we're looking for delta T. Remember that frequency and time are related by in verses of each other. So frequency is equal to one over T. We're told that the frequency is the 60 cycles per second. So remember this cycles per second here is going to be frequency. Alright, So this is gonna be 60/1. So what that means here, is that your delta t. The amount of time that it takes for every single second of this engine is actually 1/60. So this is actually where we can now plug back into this equation here. Okay, so the power is going to be 3700 divided by 1/60. And when you work this out, what you're gonna get is you're gonna get 2.22 times 10 to the fifth, and that's in watts. So this is also equal to about 222 kilowatts. And again, for those of you who are very familiar with cars, this is actually equal to about horsepower. Okay, so this is the correct answer. That's just sort of like a bonus here. So it's about 300 horsepower for this gasoline engine. So that's sort of like a real life example about how all of these numbers sort of work out. Alright, so that's it for this one. Thanks for bearing with me, and I'll see you in the next one.

4

Problem

ProblemA heat engine uses a tank of ice water as a cold reservoir. The engine takes in 8 kJ of heat from the hot reservoir, and the heat expelled melts 18g of ice in the tank. How much work does this engine do?

A

14,012 J

B

6012 J

C

1,988 J

D

8000 J

5

concept

## Thermal Efficiency & The Second Law of Thermodynamics

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Hey guys, so in the last couple videos, we were introduced to heat engines. And in some problems you'll have to calculate something called the thermal efficiency of the engine. So that's what I want to introduce you in this video and we're also going to see what the second law of thermodynamics is. Let's check this out here and we're gonna work a bunch of examples out together. So remember that heat engines, basically what they do is they produce work using the heat energy that is flowing through the engine from hot reservoir, too cold. So you have some heat that flows in right from hot to cold. And then basically the engine takes some of that and produces some work some useful energy. Now, an engine's thermal efficiency is basically just a measure of how good it is at doing that, how good it is from producing work from heat, different engines, like a gasoline engine versus a diesel engine operate on different cycles and they have different efficiencies. The letter that we'll use for the sufficiency is the lower case letter E. And basically the equations are pretty straightforward. One of them you'll need to know is W over Q. H. The way I like to think about this is this w is the work that gets put out by the engine. Q. Is what you put into the engine. So the efficiency is how much you get out divided by, how much you put in. Now, this is just a ratio. So sometimes your professors might want to multiply this by 100%. That's really up to you. Now, there's actually one other equation that you need to know and it's because we can rewrite this w here as Q H minus Q. C. And if you go ahead and do that and simplify, you'll end up with this equation over here, one minus Q. H. Sorry Qc over Q. H. It's always the smaller number. Over the bigger number. That's basically all there is to it guys, let's take a look at some examples here. So basically what we're gonna do for these diagrams is calculate whatever variables are missing. And this first example here we have 1000 jewels that gets put in zero work that gets produced and 1000 jewels that gets put out. We want to calculate the efficiency here. So the efficiency remember is just the amount of work that you get out divided by the amount of heat that you put in, that's just going to be zero, divided by 1000. And if you multiply this, you're just gonna get 0% which is actually a pretty terrible engine, you would put in a bunch of heat energy into this and you will get no useful work out of it kind of sucks. Let's move on to the second one here. Now we have the amount of heat that flows in the amount of heat that flows out, but now we actually don't know the amount of work that gets produced. So if we want to use the same equation here, heat efficiency equals W over Q. H. What happens is we don't know what this work is now, we could find it because remember this number here is always the difference between these two numbers or we can actually just take a more direct approach and actually use this equation that's over here, because that has the two cues which we do know. So it really just depends on which variables, you know which ones you're given. You can take a more direct approach here, so this is gonna be one minus Q. C over Q. H. And basically this is going to be one minus the 600 joules divided by 1000. It's always the small number over the bigger one. So if you work this out, you're gonna get is 0.4. Now again, some professors might multiply by 100 or make you multiplied by 100 and therefore the x efficiency will be expressed as 40%. And that's the answer. Let's move on to part C. Now, in part C we know the work and we know the efficiency of these engines but we actually don't know the heat that goes in and the heat that comes out. So we're going to calculate this now, you might be tempted to write down the equation W equals Q H minus Q C. That's the first one that we learned, we actually can't use this because we have two unknown variables. We can't use this instead. We're gonna use another equation. The efficiency equation because we have E. And W. So this E. Here is equal to W over Q. H. Remember. So now we want to do is we want to calculate this QH here, that's our missing variable. So we're gonna basically trade places with these two variables, and Q. H. is going to be the work divided by the efficiency. So this is gonna be 600 divided by uh and this is going to be 0.3 and what you're gonna get here is 2000 jewels. So 2000 jewels of heat energy was pumped into this heat engine. That's your cue. H. Now, basically, now that we have this qh we can actually go back and use this equation here and calculate the waste heat that gets expelled. This is going to be 2000 minus 600 you're gonna end up with 1400 jewels that leaves the engine and now last but not least in this engine over here, we have all of the heat that gets pumped in and then all of it gets converted into work. You have 1000 and 1000. So the efficiency here is going to be equals w over Q. H. That's just gonna be 1000 over 1000. And that's basically just going to equal 100%. Now, even though there is nothing in this equation that prevents this from happening. This kind of engine is actually impossible to make because doing so violates what's called the Second law of thermodynamics. So let's talk about that. We've worked a lot with the first law of thermodynamics. The second one unfortunately isn't really an equation so much as it is. A bunch of statements. Now, what's one statement? The first thing we're gonna talk about here is that it's actually impossible. It's impossible to take an engine and have it convert all of the heat into work with 100% efficiency. It's impossible to design an engine that basically spits out as much work as the heat that you plug into it. Engines must expel waste heat to the cold reservoir. This statement here is known as the kelvin or the engine statement of the Second law of thermodynamics. It's a really conceptual one. But basically the idea here is that if you could design an engine that takes all of the heat energy and then converts it into work, you could basically build an engine that actually propels or sustains itself using its own work. This is called a perpetual motion machine. And everything in physics tells us that you absolutely cannot do this. You can't just keep this process going forever. You have to have heat that goes to the cold reservoir and so they can start the cycle all over again. Anyway, so that's it for this one. Guys, let me know if you have any questions

6

Problem

ProblemA steam turbine takes in 75g of water and boils it as heat energy to run a 40% efficient engine. How much work does this engine do per cycle?

A

67,800 J

B

1.695×10

^{5}JC

10,020 J

D

4.24×10

^{5}J7

example

## Efficiency of a Nuclear Power Plant

3m

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Hey guys, so let's get started with our example here. Now I want to mention right off the bat, we're gonna see some units in here that are a little bit different than the ones that were used to, for example, 250 megawatts of power or 550 megawatts out to the surrounding environments. But ultimately what we wanna do is we want to calculate the thermal efficiency of this power plant. So let's just go ahead and get started here. What we want to calculate is E. So we have a couple of of equations for E. We've got the work over Qh or the Q. C over Q. H. I'm just gonna use the simplest one here. This work over Q. H. So this is E equals work divided by the heat that flows into this nuclear power plant here. However, what we have here is not the work, what we have is the amount of power that it produces. Now remember that power is really just work divided by delta T. So here's what's going on here. And some problems where you're given units of power instead of work, remember that the efficiency equation does not depend on time. So whenever you have these kinds of situations, you can still unit when you use this equation, when you're giving units of power instead of jewels or something like that. So basically what's going on is that you can take this work over Qh equation and you can turn it into basically a unit of, of power or in terms of the rate of energy. So for example, you can say this is going to be W over delta T divided by Q. H. Over delta T. So what's going to happen is that your delta T. Is ultimately are going to cancel. They don't really actually do anything to this equation. But this is sort of just how this how this is working out, you can use those variables. So basically what's going on here is this power is 250 MW and now all we have to do is figure out the amount of power or energy that flows into this nuclear power plants. So basically we're gonna have to figure out what is Q. H. Now we're told here is that this expels 550 megawatts out of the surrounding environment, that's going to be the cold reservoir. So this here really represents Q. C. So if I want to calculate Q. H, what I have to do is I have to go here and I have to use my W equals Q H minus Q. C. Equation. Now again you can still use this even though we're working with units of power. So one way you can kind of think about this is that in one cycle this nuclear power plant will produce 250 mega jewels and expel 550 mega jewels out of the environment. But if you just run it continuously then you can also express this in units of power. Alright, so that's kind of what's going on here. So my Q. I really need to figure out what this Qh is. So I'm gonna move this Q. C over to the other side, we're gonna get the W plus Q C. Is Q. H. So in other words, the 250 megawatts of power that I produce, that's the useful energy that you get out of it, plus the 550 that gets expelled out to the environment. And what you'll see here is that Q. H is equal to 800 megawatts. So what's going in here is if you sort of like if you, if you sort of draw the energy flow diagram like this, Then we're going to see here that this QH is equal to 250 MW, this QC. Is 550 MW. And the amount of work that you get out of it, this w here, I'm sorry, this is the 800 this is the 800 megawatts and the work that you get out of each cycle is 250 megawatts. Alright, so that's how you can kind of see what's going on here as the energy flow diagram. So now this is basically what we plug into this efficiency equation. So you have the 2 50 divided by the 800 megawatts and what you're gonna get here is an efficiency, that's equal to um It's going to be 31.3%, so that is your final answer. Alright? So hopefully that makes sense here guys, because you may see some questions like this later on in the future, so that's it for this one.

Additional resources for Heat Engines and the Second Law of Thermodynamics

PRACTICE PROBLEMS AND ACTIVITIES (4)

- A Carnot refrigerator is operated between two heat reservoirs at temperatures of 320 K and 270 K. (c) What is ...
- A Carnot refrigerator is operated between two heat reservoirs at temperatures of 320 K and 270 K. (b) If the r...
- A Carnot refrigerator is operated between two heat reservoirs at temperatures of 320 K and 270 K. (a) If in ea...
- A diesel engine performs 2200 J of mechanical work and discards 4300 J of heat each cycle. (a) How much heat m...