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Efficiency of a Carnot engine

Definition of efficiency for a heat engine. Efficiency of a Carnot Engine. Created by Sal Khan.

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  • leaf grey style avatar for user Shlomo Fingerer
    How similar are the engines we use in our cars to the "ideal" Carnot engine?
    (24 votes)
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  • leaf green style avatar for user NathanaelGreenaway
    what happened to the work done from points B to C, then D to A in the cycle? this seems to be totally neglected?
    (9 votes)
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    • leafers tree style avatar for user Sean Meadows
      The work that the system does from B to C is counterbalanced by the work that is done to the system from D to A.

      Visualize the Carnot cycle illustrated on a PV diagram. We know that internal energy is a state variable, which means the internal energy at point A will always be the same regardless of the path that leads to point A. Therefore, there is no change in internal energy when a Carnot cycle is completed, because we started at point A and finished at point A.

      We know that there is no change in internal energy from A to B, because there is no change in temperature from point A to B. The same is true from C to D. We know that there is a decrease in internal energy from B to C, because there is a decrease in temperature. However, this decrease in internal energy is counterbalanced by an increase in internal energy from D to A, because the internal energy at the end of the cycle must equal the internal energy at the start of the cycle.

      Finally, we know the thermodynamic processes from B to C and from D to A are adiabatic, which means the system did not gain or lose heat during these processes. We know that "change in internal energy" equals "heat gained" minus "work done by the system" (or plus "work done to the system"). Since no heat was gained or lost from B to C, the "change in internal energy" from B to C equals negative ( - ) "work done by the system" (the gas did work to increase the volume). Since no heat was gained or lost from D to A, the "change in internal energy" from D to A equals "work done to the system" (work was done to the gas to decrease the volume). We've already established that the change in internal energy from B to C counterbalances the change in internal energy from D to A. Therefore, the work done by the system from B to C counterbalances the work done to the system from D to A. This is why Sal neglected the work involved in these two adiabatic processes.
      (10 votes)
  • piceratops ultimate style avatar for user Bailan
    why isn't the Q2 integrated from Vc to Vd?
    (4 votes)
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    • aqualine ultimate style avatar for user Mengyuan Chen
      You see, when we are talking about Q2 we assume it is positive because that's better to understand. Its definition should be the heat given off by the system. If you integrate from Vc to Vd you derive a heat given to the system, which is negative. (Now I'm also a novice on thermodynamics I may be wrong... But the general idea is probably like that...)
      (1 vote)
  • piceratops ultimate style avatar for user Ashwin Nair
    how much will the engine's efficiency be if both the temperatures are kept at 0 degrees?
    (1 vote)
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    • leafers tree style avatar for user Ben Coomes
      0/0 does not simply equal 1. However, if we start at a temperature of 1 for the hot reservoir and for the cold reservoir, the efficiency will be 1 - 1/1 =0. If we halve the temperatures: 1 - .5/.5= 0. Halved again; 1 - .25/.25=0 and again: 1 - .125/.125=0... and so on. You'll never get to zero, but you can get infinity close, and as you approach zero, you will still observe that the efficiency is zero.
      Still unconvinced? 0 degrees Celsius is the same temperature as 273 Kelvin. Because you are dividing one temperature by another, units will cancel out. Therefore, they do not matter, as long as they are the same. 1 - 273/273 = 0, so after converting back to Celsius, we can safely say that the efficiency is 0 at 0 degrees Celsius.
      Hope this helped
      (5 votes)
  • male robot donald style avatar for user Zak
    I can't seem to figure out where this "net work" ends up going. Does the piston end up higher than when it started, even if the thermodynamic macrostates all go back to where they started? If someone could explain where the net positive work is actually doing I would greatly appreciate it.
    (3 votes)
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  • piceratops ultimate style avatar for user Dave Morgan
    Could a Carnot engine actually be built in real life?
    (2 votes)
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  • blobby green style avatar for user softport45
    How do you calculate the efficiency of a Carnot refrigerator? Is it Qc/(work input)?

    I tried it and came up with eff = 1/(1+(Th/Tc)) where Th is the hot side temp, and Tc the low side temperature. Is this correct?
    If so, then maximum efficiency would be 0.5 (when Th = Tc). That seems very low, I suspect that the efficiency should be 1 (for a Carnot refrigerator) when the hot and cold side temperatures are the same.
    (1 vote)
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    • leaf blue style avatar for user Ting Kuo
      Your first equation is correct, but your second one isn't. For the Carnot engine, the equation for efficiency with respect to temperature is [eff = 1+(Tc/Th)].

      To get the efficiency of a Carnot refrigerator, you would have to start from your first equation [eff = (Qc/W)], substitute W with (Qh-Qc), and then do the same procedure as Sal where [Qc = nRTc(ln(VA/VB))] and [Qh = nRTh(ln(VD/VC))].
      After substituting and simplifying, you should come to [eff = Tc/(Th-Tc)]
      (4 votes)
  • leaf green style avatar for user Haris Haziq
    How the efficiency of carnot engine varies with the variation in temperatue of source and sink?
    (1 vote)
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  • spunky sam blue style avatar for user Vishnu Gopalakrishnan
    Why exactly, for efficiency when we divide by energy added are we dividing only the energy input ? Shouldn't it also be the net energy that is Q1-Q2 ?
    (1 vote)
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  • blobby green style avatar for user James Davidson
    Can some please explain to me the difference between the efficiency of the Carnot cycle and a normal engine? I had a conceptual question and the correct answer was "The efficiency of this engine is greater than the ideal Carnot cycle efficiency." However, when i tried to calculate the efficiency of it i ended up with the same number for both the theoretical (carnot) and real engine efficiency. Used sals equation e = 1 - (Tc/Th) for the Carnot cycle and e = w/Qh for the real one. Any advice?
    (1 vote)
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Video transcript

I'm now going to introduce you to the notion of the efficiency of an engine. And it's represented by the Greek letter eta. Even though it sounds like an e, it looks like a kind of funny-looking n. So Greek letter eta, this is efficiency. And efficiency as applies to engines is kind of similar to the way we use the word in our everyday life. If I said, how efficient are you with your time, I care about, what are you doing with that hour that I gave you? Or if I said, how efficient are you with your money, I'd say, how much were you able to buy with that $100 I gave you? So efficiency with an engine is the same thing. What were you able to do with the stuff that I gave you? So in the engine world, we define efficiency to be, the work you did with the energy that I gave you to do the work, essentially. And in our Carnot world, up here, this is all the stuff I'd done before, what was the energy that I had given you? Well, the energy that I'd given you was the energy that came from this first reservoir up here. Remember, when we were moving the pebbles from A to B, we were moving the pebbles to keep it as a quasistatic process, so that we can go back, if we had to, and so that the system stayed in equilibrium the whole time. If we didn't have this reservoir here, the temperature would have gone down, because we're expanding the volume, and all of that. So we had to keep that reservoir there. The added heat to the system, we figured out this multiple times, was Q1. It was equivalent to the work we did over that time period, so it would have been the whole shaded region, not just what's inside the circle. But so the heat that we gave you was Q1. You later gave some heat back, when some work was done for you. But what we care about is the heat that we were given. So in this case, it would be Q1. And what was the work that you did? Well, the net work that you did was the shaded area. It was the area inside of our Carnot cycle. So that's our definition for efficiency. It's always going to be a fraction. Sometimes it's given as a percentage, where you just, you know, this is 0.56. You would call that 56% efficient. Which is essentially saying, you were able to transfer 56% of the heat energy that you were given, and turn that into useful work. And so it makes sense, at least in my head, that that would be the definition for efficiency. Now le'ts see if we can play around with this, and see how efficiency would play out with some of the variables we're dealing with in the Carnot cycle. So what was the work that we did? Well, we know our definition of internal energy. Our definition of internal energy has been more useful than you would have thought, for such a simple equation. Our change in internal energy is the net heat applied to the system minus the work done by the system. Right? Now, when you complete one Carnot cycle, when you go from a all the way around back to a, and actually, I'll make a little aside here. You could have actually gone the other way around the cycle. But when you go the way we went the first time, when you go clockwise, you're a Carnot engine. You're doing work, and you're transferring heat from T1 to T2. If you went the other way around the circle, essentially, you'd be a Carnot refrigerator, where you would have work being done to you, and I'll touch on this in a second, and you'd be transferring energy the other way. And this will be important to our proof of why the Carnot engine is the best engine, at least theoretically, from an efficiency point of view. If efficiency is all you cared about. But anyway. That's what I was talking about. So if I go a complete cycle in this PV diagram and I end up back at A, what's my change in internal energy? It's 0. My internal energy is a state variable, so my change in internal energy is 0. My change in entropy would also be 0. It's another state variable, when I get from A back to A. So over the course of this cycle, we know that my change in internal energy is 0. What's the net heat applied to the system? Well, we applied Q1 to the system, and then we took out Q2, right? We gave that to the second reservoir. We gave that down there to T2, to that second reservoir. And then minus work-- and all of this is equal to 0. This is the net, i just want to make it clear, this is the net heat applied to our system. So the work done by the system-- we just add W to both sides of this equation, you get W is equal to Q1 minus Q2. So there we have it. Let's just substitute that back here. And instead of W, we can write Q1 minus Q2 as the numerator in our in our efficiency definition, and then the denominator is still Q1. And we can do a little bit of math. This simplifies-- this is Q1. This is the heat we put into it. So it's the net heat we applied to the system divided by the heat we put into it. So this is equal to-- Q1 divided by Q1 is 1, minus Q2 over Q1. So once again, this is another interesting definition of efficiency. They're all algebraic manipulations using the definition of internal energy, and whatever else. Now let's see if we can somehow relate efficiency to our temperatures. Now, let me-- this is Q1 right there. So what were Q2 and Q1? What were they? What were their absolute values, not looking at the signs of them? I mean, we know that Q2 was transferred out of the system, so if we said Q2 in terms of the energy applied to the system, it would be a negative number. But if we just wanted to know the magnitude of Q2, what would it be, and the magnitude of Q1? Well, the magnitude of Q1-- let me draw a new Carnot cycle, just for cleanliness. I'll draw a little small one over here. That's my volume axis. That's my pressure axis. PV. I start here at some state, and then I go isothermally. What's a good color for an isothermal expansion? Maybe purple. It's kind of-- let's see. An isothermal expansion. I'm on an isotherm here. So I go down there, and I go to state, and I went down to state B. So this is A to B. And we know we are an isotherm. This is when Q1 was added. This is an isotherm. If your temperature didn't change, your internal energy didn't change. And like I said before, if your internal energy is 0, then the heat applied to the system is the same as the work you did. They cancel each other out. That's why you got to 0. So this Q1 that we apply to the system, it must be equal to the work we did. And the work we did is just the area under this curve. We did this multiple times. And why is it the area under the curve? Because it's a bunch of rectangles of pressure times volume. And then you just add up all the rectangles, an infinite number of infinitely narrow rectangles, and you get the area. And what is that? Just a review-- pressure times volume was the work, right? Because we're expanding the cylinder. We're moving up that piston. We're doing force times distance. So the amount Q1 is equal to that integral-- the amount of work we did as well-- is equal to the integral from V final-- I shouldn't say V final. from VB, from our volume at B-- oh sorry. From our volume at A to our volume at B. We're starting here, and we're going to our volume at B. And we're taking the integral of pressure-- and I've done this multiple times, but I'll do it again. So the pressure, our height, times our change in volume, dv. We go back to our ideal gas formula, PV is equal to nRT, divide both sides by V, and you get P is equal to nRT over V. And so you have Q1 is equal to the integral of VA from VA to VB of this little thing over here. nRT over V dv. All of this stuff up here, these are all constants. Remember, we're an isotherm. Our temperature isn't changing. We could write T1 there, because that's our temperature. We're at T1. We're on a T1 isotherm right there, because we're touching the T1 reservoir. But this is all a constant, so we can take it out of the equation. And then we've evaluated this multiple times, so I won't go into the mechanics of the integral. But so this is Q1 is equal to our constant terms, nRT1 times this definite integral. All we have left in the integral is 1 over V The antiderivative of that is natural log, and then you evaluated the two boundaries. So you get the natural log of VB minus VA, which is the same thing as the natural log of VB over VA. Fair enough. This right here is Q1. All right. Now what is Q2? Q2 was this part of the Carnot cycle. Q2 is when we went from C to D. So the magnitude of Q2 is the area under this curve, right here. Now this is the work done to of the system, so that's why we subtracted it out when we wanted to know the net work done by the system. And we ended up with this area over here, when you subtracted this as well, over here on this side. But if we just want to know the magnitude of Q2, we just take the integral under this curve. And what's the integral under that curve? This is the heat out of the system, the heat that had to be pushed out of the system as work was done to it. So that integral-- so we could just say, the magnitude of Q2-- I'll do it over here-- is equal to-- the same exact logic applies, just the boundaries are different. We're now going from-- and remember, when we cared about the direction, I would say, I'm going from VC to VD. But if I just want to know the absolute value of that area, because I just want to know the magnitude, I could go from VC to VD and just take the absolute value of it, or I could just go from VD to VC, and I'd get a positive area. So let me just do that. So it's the integral from VD to VC. Remember, the cycle we went from VC to VD, but I just want the absolute value. I want this to be positive. So I'm turning it the other way. Of P dv. And we do the exact same math there. You get Q2 is equal to nR-- this time the temperature is T2. We're on a T2 reservoir. Times the natural log-- and this time, what's it going to be? Times the natural log of-- instead of VB over VA, it's going to be VC divided by VD. Now let's use these two pieces of information and substitute them back into that results for efficiency we just got. We just learned that you could also write the efficiency of an engine to be equal to 1 minus Q-- let's look back at it. 1 minus Q2 over Q1. So let's substitute Q2 there and Q1 over here. And what do you get? You get the efficiency of your engine is equal to 1 minus-- Q2 is this expression over here-- nRT2 times the natural log of VC over VD, all of that divided by Q1, which is this one over here. nRT1 times the natural log of VB over VA. Now we can do a little bit of canceling. Obviously the n and the r is cancel. And we have these natural logs and all of that. But I had a whole video dedicated to show you that VC over dv is equal to VB over VA. Now, if we know that these are equal, then this is equal to this. So the natural logs of them are equal, so we can just divide. And what are we left with? We're left with the fact that efficiency can also be written as 1 minus T2 over T1 for a Carnot engine. Remember, this time, what we did over here, this applied to any engine. This was just a little bit of math and the definition of what work is, and-- well, I won't go too much into it right now. But this is for a Carnot engine, right? Because we did a little bit of work here that involved the Carnot cycle. But this is a pretty bit important outcome, because we're going to show that the Carnot engine, in the next video, is actually the most efficient engine that can ever be attained. Well, you have to be very careful about that, that when we say efficient, it means that between two temperature sources, you can't get a more efficient engine than the Carnot engine. Now, I'm not saying it's the best engine, or it's a practical engine, or you'd want it to power your lawn mower or your jet plane. I'm just saying that it's the most efficient engine between these two temperature reservoirs. And I'll show you that in the next video.