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

A heat engine converts heat energy into useful work, exemplified by car engines that burn fuel to generate motion. The efficiency of an engine, denoted as e, is calculated using the equation e = wqₕ or e = 1−qₗₕ. The second law of thermodynamics states that 100% efficiency is impossible, as engines must expel waste heat to a cold reservoir, preventing perpetual motion machines.

Introduction to Heat Engines

Hey, guys. So by now, we've talked a lot about the first law of thermodynamics. In 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 to 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 some of the equations and diagrams that you'll need. So let's go ahead and check it out here.

Basically, a heat engine is 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 what 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, alright? 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. Alright?

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 going to 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 in 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 does 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 in 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. Alright. 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. Alright, so that's the basics of how a heat engine works 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: ΔE=q-w. We also did in a previous video, took a look at cyclic processes, and cyclic processes had a special condition that the change in internal energy, ΔE=0. So directly from this equation, if ΔE=0, then that means that q=w. So what we saw is that for cycles, the w for the cycle equaled the q. 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. And 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 2 heats. We have one that's going in, one that's going out. So what is this qȈnet here? It's actually both of them. It's actually the heat that's going in, that's qȉh, and it's positive because it's getting added to your system. And then this qȉc 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. We want to calculate how much waste heat is expelled from the engine. So which variable is that? Well, if we want to solve for 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 qȉwaste. Alright. 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, but we're told in this problem is that 500 joules of heat is being taken into the heat engine. So which variable is that? Well, remember, your heat engine takes in heat from the hot reservoir. So that just means that this qȉh or qȉin is equal to the 500 joules. Now what your heat engine's doing is it's taking some of that heat energy and it's doing 300 joules of work. So that's pretty straightforward. Remember the work is always just gonna come out of your heat engine like this. So your w here is just gonna be the 300 joules. 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 deck in this diagram here, which is the heat that gets expelled out to the cold reservoir. So this is qȉc or in other words, qȉout. And that's basically what this waste heat represents. Alright? So the trick to these kinds of problems here is kind of translating which variable they're asking you to solve. So qȉwaste is equal to qȉc. In other words, it's equal to qȉout. I mean, any of these, they all mean the same thing.

An 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

Power Output of a Gasoline Engine

Hey, guys. So I've got a great sort of real-life example problem that's going to 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 are lots of parts to this problem, but I think it's a 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 3,700 joules of work per cycle. So whenever I get these sorts of numbers here, I like to draw the energy flow diagram. 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 1.6×104 joules of heat from combustion gasoline. So 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 the combustion of gas. Then it does some amount of work. Right? And this work here, we're told, is equal to 37,100 joules. The heat that it takes in is 1.6×104. 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? Well, remember, it's going to be heat, so it's going to be one of the cues, 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 going to 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 - Q_c and everything sort of absolute values, everything's positive numbers, just so you don't get confused. Alright? So what happens here is we're looking for the Q_c. We're going to rearrange this equation, and we're going to move the Q_c to the other side and then the work over to the other side, so that it becomes positive. Right? So what happens is the Q_c is just going to be Q_h - 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 or the cold reservoir as waste heat. So really what happens is you're going to take in 1.6×104, and then you're going to do 37,100 joules of work. So you're going to subtract that, and what you end up with is waste heat is 1.23×104 joules. Alright. So that is the first answer here. That's how much heat 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're asked to calculate what mass of fuel is burned each cycle. So what variable is that? Well for mass, we've always used m. So that's basically what we're looking for here. So where does m 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've looked at latent heats when we looked at calorimetry problems. We're told that this latent heat is this number over here. Remember what this means is that for every one kilogram of mass that you've burned, for every one kilogram of gasoline, you basically can extract this amount of energy from it. So whenever we used a sort of phase change here, combustion is sort of one of those phase changes, we used Q = mL. So really what happens here is that this Q_h, remember the heat that comes from combusting gasoline, is going to equal mass times however much hot times latent heat of combustion. Right? So what we're looking for here is this little m. So what happens here is I'm going to take my Q_h which is 1.6×104, and I'm going to divide it by the latent heat of combustion, 4.6×107. So if you look at the exponents here, we have 10⁴, 10⁷. We should expect to get a number that's pretty small here, right, because we've only generated 1.6 × 10⁴. When you work this out, what you're going to get is that the mass that gets combusted is 3.5×amp;10-4 kilograms. In other words, if you convert this to grams, that's about 0.35 grams of fuel. And that kind of 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 really 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 3,600 RPM, for those of you who are familiar with cars, then what is going to be the power output in kilowatts? So what essentially we're looking for here is what is p? Remember that is the equation for power. Well, we're going to have to use an old equation that relates power with the amount of energy and time. Right? So we're told this is, this is 60 cycles per second. What's its power output in kilowatts? So we're going to 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 or the work that's done is it's actually just going to be the 37,100 joules. Right? Every single cycle, you're producing 37,100 joules of work. So then what's the delta T? Well, the delta T is going to be a little bit trickier to solve because we have to relate it to the frequency. So I'm going to sort of go over here. So we're looking for delta T. Remember that frequency and time are related by inverses of each other, so frequency is equal to 1 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 going to be 60 over 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 over 60. So this is actually what we can now plug back into this equation here. Okay? So the power is going to be 37,100 divided by 1 over 60. And when you work this out, what you're going to get is you're going to get 2.22×105, 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 297 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.

A 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

Thermal Efficiency & The Second Law of Thermodynamics

Hey, guys. So in the last couple of videos, we were introduced to heat engines, and in some problems, you'll have to calculate something called the thermal efficiency of that engine. So that's what I want to introduce to 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 going to 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 a hot reservoir to 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 this efficiency is the lowercase letter e, and, basically, the equations are pretty straightforward. One of them you'll need to know is \( \frac{W}{Q_h} \). 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 \( Q_h - Q_c \). And if you go ahead and do that and simplify, you'll end up with this equation over here, \( 1 - \frac{Q_c}{Q_h} \). 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 going to do for these diagrams is calculate whatever variables are missing. In this first example here, we have 1,000 joules that get put in, 0 work that gets produced, and 1,000 joules that get put out. We want to calculate the efficiency here. So, the efficiency, remember, is just the amount of work you get out divided by the amount of heat that you put in. That's just gonna be 0 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 would get no useful work out of it. 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, we want to use the same equation here, heat efficiency equals \( \frac{W}{Q_h} \). 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 the equation that's over here because that has the 2 Q's, 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 going to be \( 1 - \frac{Q_c}{Q_h} \). And, basically, this is going to be \( 1 - \frac{600}{1000} \). It's always the smaller number over the bigger one. So if you work this out, what you're going to get is 0.4. Now, again, some professors might multiply by 100 or make you multiply by 100, and therefore, the 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 this engine, 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 \( W = Q_h - Q_c \). That's the first one that we learned. We actually can't use this because we have 2 unknown variables. We can't use this. Instead, we're going to use another equation, the efficiency equation, because we have \( e \) and \( W \). So this \( e \) here is equal to \( \frac{W}{Q_h} \), remember. So now, what we want to do is calculate this \( Q_h \) here, that's our missing variable. We're going to trade places with these two variables, and \( Q_h \) is going to be work divided by the efficiency. So this is going to be 600 divided by 0.3, and what you're going to get here are 2,000 joules. So 2,000 joules of heat energy were pumped into this heat engine. That's your \( Q_h \).

Now, basically, now that we have this \( Q_h \), we can actually go back and use this equation here and calculate the waste heat that gets expelled. This is going to be \( 2,000 - 600 \), and you're going to end up with 1,400 joules that leave 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 1,000 to 1,000. So the efficiency here is going to be \( e = \frac{W}{Q_h} \). That's just going to be \( \frac{1000}{1000} \), 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 one we're going to 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 “Drive" 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.

A 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. 1.695×10⁵ J
• C. 10,020 J
• D. 4.24×10⁵ J

Efficiency of a Nuclear Power Plant

Hey, guys. So let's get started with our example here. Now I want to mention right off the bat that you're going to see some units in here that are a little bit different than the ones that we're used to. For example, 250 megawatts of power or 550 megawatts out to the surrounding environments. But, ultimately, what we want to 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 equations for E. We've got the work over Q_H or the Q_C over Q_H. I'm just going to 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. In 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 use this equation when you're given units of power instead of joules or something like that. So, basically, what's going on is that you can take this work over q_H equation and you can turn it into, basically, a unit 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's ultimately are going to cancel. They don't really actually do anything to this equation, but this is sort of just how this is working out. You can use those variables. So basically, what's going on here is this power is 250 megawatts. And now all we have to do is figure out the amount of power or energy that flows into this nuclear power plant. So basically, we're going to have to figure out what is Q_H. Now what we're told here is that this expels 550 megawatts out to the surrounding environments. That's going to be the colder 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 = Q_H - 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 joules and expel 550 megajoules out to 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 Q_H is. So I'm going to move this Q_C over to the other side. We're going to 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; if you sort of draw the energy flow diagram like this, then we're going to see here that this Q_H is equal to 250 megawatts. This Q_C is 550 megawatts, 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 it 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 250 divided by the 800 megawatts. And what you're going to get here is an efficiency that's equal to, 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.

Here’s what students ask on this topic:

What is a heat engine and how does it work?

A heat engine is a device that converts heat energy into useful work. It operates by taking in heat from a high-temperature source (hot reservoir), using part of this energy to perform work, and then expelling the remaining heat to a low-temperature sink (cold reservoir). A common example is a car engine, which burns fuel to generate heat, converting some of this heat into mechanical work to move the car, while the rest is expelled as exhaust. The efficiency of a heat engine is never 100% due to the second law of thermodynamics, which mandates that some energy must always be lost as waste heat.

What is the second law of thermodynamics in relation to heat engines?

The second law of thermodynamics states that it is impossible for any heat engine to convert all the heat energy it absorbs into work. This means that no heat engine can be 100% efficient. Some of the input heat must always be expelled as waste heat to a cold reservoir. This principle prevents the creation of a perpetual motion machine, which would otherwise be able to sustain its own operation indefinitely without any external energy input. The second law is often expressed in terms of entropy, indicating that the total entropy of a closed system can never decrease over time.

How do you calculate the thermal efficiency of a heat engine?

The thermal efficiency (e) of a heat engine is a measure of how well it converts heat into work. It is calculated using the equation:

$e=\frac{W}{Q}$

where W is the work output and Q_H is the heat input from the hot reservoir. Alternatively, it can be expressed as:

$e=1-\frac{Q}{C}$

where Q_C is the heat expelled to the cold reservoir. This ratio indicates the fraction of heat converted to work, with the remainder being waste heat.

What is the significance of the hot and cold reservoirs in a heat engine?

The hot and cold reservoirs are crucial components of a heat engine. The hot reservoir provides the heat energy that the engine converts into work. This is typically a high-temperature source, such as burning fuel in a car engine. The cold reservoir, on the other hand, is where the waste heat is expelled. It is a low-temperature sink that absorbs the unused heat energy. The temperature difference between these two reservoirs drives the heat flow, enabling the engine to perform work. The efficiency of the engine depends on the temperatures of these reservoirs, with greater differences generally leading to higher efficiencies.

Why can't a heat engine achieve 100% efficiency?

A heat engine cannot achieve 100% efficiency due to the second law of thermodynamics. This law states that some energy must always be lost as waste heat to a cold reservoir. This unavoidable loss is due to the increase in entropy, which means that not all the input heat can be converted into work. As a result, every real heat engine must expel some heat to the environment, preventing it from being perfectly efficient. This principle ensures that perpetual motion machines, which would operate indefinitely without external energy, are impossible.