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## Physics library

### Course: Physics library > Unit 10

Lesson 3: Laws of thermodynamics- Macrostates and microstates
- Quasistatic and reversible processes
- First law of thermodynamics / internal energy
- More on internal energy
- What is the first law of thermodynamics?
- Work from expansion
- PV-diagrams and expansion work
- What are PV diagrams?
- Proof: U = (3/2)PV or U = (3/2)nRT
- Work done by isothermic process
- Carnot cycle and Carnot engine
- Proof: Volume ratios in a Carnot cycle
- Proof: S (or entropy) is a valid state variable
- Thermodynamic entropy definition clarification
- Reconciling thermodynamic and state definitions of entropy
- Entropy intuition
- Maxwell's demon
- More on entropy
- Efficiency of a Carnot engine
- Carnot efficiency 2: Reversing the cycle
- Carnot efficiency 3: Proving that it is the most efficient

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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.

## Want to join the conversation?

- How similar are the engines we use in our cars to the "ideal" Carnot engine?(25 votes)
- Usually 25% of the gasoline is converted to work. Cars with diesel engines however can convert up to 40% to work.(33 votes)

- 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)
- 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)

- why isn't the Q2 integrated from Vc to Vd?(4 votes)
- 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)

- how much will the engine's efficiency be if both the temperatures are kept at 0 degrees?(1 vote)
- 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)

- 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)
- Every time you see that the work done does not correspond to the change in internal energy, consider that heat is another way to change the internal energy.

delta U = Q+W (in)(1 vote)

- Could a Carnot engine actually be built in real life?(2 votes)
- No, Carnot engine only exists in ideal situation. That is we are assuming that there are no external interaction (e.g. heat transfer). This is impossible in real life(2 votes)

- 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)- 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)

- How the efficiency of carnot engine varies with the variation in temperatue of source and sink?(1 vote)
- Higher temperature difference means higher heat energy which will increase the efficiency.(2 votes)

- 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)
- 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)
- Carnot Engine is the reversible engine and it has the maximum efficiency than other engines if they are operating b/w the same temperature difference.!!(1 vote)

## 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.