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I'm now going to introduce
you to the notion of the

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efficiency of an engine.

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And it's represented by
the Greek letter eta.

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Even though it sounds like an
e, it looks like a kind of

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funny-looking n.

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So Greek letter eta,
this is efficiency.

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And efficiency as applies to
engines is kind of similar to

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the way we use the word
in our everyday life.

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If I said, how efficient are
you with your time, I care

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about, what are you doing with
that hour that I gave you?

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Or if I said, how efficient are
you with your money, I'd

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say, how much were you able to
buy with that $100 I gave you?

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So efficiency with an engine
is the same thing.

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What were you able to do with
the stuff that I gave you?

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So in the engine world, we
define efficiency to be, the

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work you did with the energy
that I gave you to do the

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work, essentially.

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And in our Carnot world, up
here, this is all the stuff

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I'd done before, what was the
energy that I had given you?

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Well, the energy that I'd given
you was the energy that

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came from this first
reservoir up here.

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Remember, when we were moving
the pebbles from A to B, we

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were moving the pebbles to
keep it as a quasistatic

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process, so that we can go back,
if we had to, and so

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that the system stayed in
equilibrium the whole time.

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If we didn't have this
reservoir here, the

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temperature would have gone
down, because we're expanding

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the volume, and all of that.

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So we had to keep that
reservoir there.

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The added heat to the system,
we figured out this multiple

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times, was Q1.

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It was equivalent to the work we
did over that time period,

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so it would have been the whole
shaded region, not just

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what's inside the circle.

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But so the heat that
we gave you was Q1.

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You later gave some heat
back, when some work

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was done for you.

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But what we care about is the
heat that we were given.

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So in this case,
it would be Q1.

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And what was the work
that you did?

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Well, the net work that you
did was the shaded area.

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It was the area inside
of our Carnot cycle.

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So that's our definition
for efficiency.

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It's always going to
be a fraction.

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Sometimes it's given as a
percentage, where you just,

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you know, this is 0.56.

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You would call that
56% efficient.

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Which is essentially saying, you
were able to transfer 56%

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of the heat energy that you were
given, and turn that into

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useful work.

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And so it makes sense, at least
in my head, that that

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would be the definition
for efficiency.

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Now le'ts see if we can play
around with this, and see how

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efficiency would play out with
some of the variables we're

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dealing with in the
Carnot cycle.

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So what was the work
that we did?

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Well, we know our definition
of internal energy.

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Our definition of internal
energy has been more useful

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than you would have thought,
for such a simple equation.

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Our change in internal energy is
the net heat applied to the

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system minus the work
done by the system.

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Right?

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Now, when you complete one
Carnot cycle, when you go from

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a all the way around back to a,
and actually, I'll make a

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little aside here.

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You could have actually
gone the other

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way around the cycle.

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But when you go the way we went
the first time, when you

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go clockwise, you're
a Carnot engine.

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You're doing work, and
you're transferring

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heat from T1 to T2.

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If you went the other way
around the circle,

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essentially, you'd be a Carnot
refrigerator, where you would

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have work being done to you,
and I'll touch on this in a

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second, and you'd
be transferring

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energy the other way.

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And this will be important to
our proof of why the Carnot

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engine is the best engine, at
least theoretically, from an

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efficiency point of view.

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If efficiency is all
you cared about.

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But anyway.

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That's what I was
talking about.

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So if I go a complete cycle in
this PV diagram and I end up

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back at A, what's my change
in internal energy?

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It's 0.

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My internal energy is a state
variable, so my change in

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internal energy is 0.

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My change in entropy
would also be 0.

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It's another state variable,
when I get from A back to A.

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So over the course of this
cycle, we know that my change

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in internal energy is 0.

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What's the net heat applied
to the system?

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Well, we applied Q1 to the
system, and then we

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took out Q2, right?

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We gave that to the
second reservoir.

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We gave that down there to T2,
to that second reservoir.

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And then minus work-- and all
of this is equal to 0.

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This is the net, i just want to
make it clear, this is the

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net heat applied
to our system.

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So the work done by the system--
we just add W to both

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sides of this equation, you get
W is equal to Q1 minus Q2.

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So there we have it.

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Let's just substitute
that back here.

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And instead of W, we can write
Q1 minus Q2 as the numerator

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in our in our efficiency
definition, and then the

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denominator is still Q1.

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And we can do a little
bit of math.

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This simplifies-- this is Q1.

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This is the heat
we put into it.

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So it's the net heat we applied
to the system divided

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by the heat we put into it.

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So this is equal to-- Q1
divided by Q1 is 1,

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minus Q2 over Q1.

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So once again, this is another
interesting definition of

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efficiency.

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They're all algebraic
manipulations using the

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definition of internal energy,
and whatever else.

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Now let's see if we can somehow
relate efficiency to

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our temperatures.

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Now, let me-- this is
Q1 right there.

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So what were Q2 and Q1?

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What were they?

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What were their absolute values,
not looking at the

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signs of them?

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I mean, we know that Q2 was
transferred out of the system,

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so if we said Q2 in terms of
the energy applied to the

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system, it would be
a negative number.

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But if we just wanted to know
the magnitude of Q2, what

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would it be, and the
magnitude of Q1?

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Well, the magnitude of Q1--
let me draw a new Carnot

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cycle, just for cleanliness.

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I'll draw a little small
one over here.

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That's my volume axis.

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That's my pressure axis.

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PV.

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I start here at some state, and
then I go isothermally.

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What's a good color for an
isothermal expansion?

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Maybe purple.

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It's kind of-- let's see.

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An isothermal expansion.

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I'm on an isotherm here.

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So I go down there, and I
go to state, and I went

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down to state B.

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So this is A to B.

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And we know we are
an isotherm.

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This is when Q1 was added.

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This is an isotherm.

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If your temperature didn't
change, your internal energy

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didn't change.

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And like I said before, if your
internal energy is 0,

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then the heat applied to the
system is the same as

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the work you did.

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They cancel each other out.

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That's why you got to 0.

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So this Q1 that we apply to the
system, it must be equal

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to the work we did.

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And the work we did is just
the area under this curve.

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We did this multiple times.

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And why is it the area
under the curve?

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Because it's a bunch
of rectangles of

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pressure times volume.

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And then you just add up all
the rectangles, an infinite

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number of infinitely
narrow rectangles,

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and you get the area.

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And what is that?

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Just a review-- pressure times
volume was the work, right?

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Because we're expanding
the cylinder.

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We're moving up that piston.

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We're doing force
times distance.

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So the amount Q1 is equal to
that integral-- the amount of

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work we did as well-- is equal
to the integral from V final--

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I shouldn't say V final.

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from VB, from our volume
at B-- oh sorry.

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From our volume at A
to our volume at B.

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We're starting here, and we're
going to our volume at B.

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And we're taking the integral
of pressure-- and I've done

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this multiple times, but
I'll do it again.

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So the pressure, our
height, times our

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change in volume, dv.

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We go back to our ideal gas
formula, PV is equal to nRT,

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divide both sides by
V, and you get P is

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00:08:08,740 --> 00:08:13,050
equal to nRT over V.

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And so you have Q1 is equal to
the integral of VA from VA to

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VB of this little
thing over here.

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00:08:23,560 --> 00:08:27,750
nRT over V dv.

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All of this stuff up here,
these are all constants.

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Remember, we're an isotherm.

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Our temperature isn't
changing.

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We could write T1 there,
because that's our

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temperature.

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We're at T1.

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We're on a T1 isotherm right
there, because we're touching

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the T1 reservoir.

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But this is all a constant,
so we can take

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it out of the equation.

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00:08:42,870 --> 00:08:46,130
And then we've evaluated this
multiple times, so I won't go

200
00:08:46,130 --> 00:08:47,600
into the mechanics
of the integral.

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00:08:47,600 --> 00:08:55,970
But so this is Q1 is equal to
our constant terms, nRT1 times

202
00:08:55,970 --> 00:08:56,790
this definite integral.

203
00:08:56,790 --> 00:08:59,220
All we have left in the integral
is 1 over V The

204
00:08:59,220 --> 00:09:01,720
antiderivative of that is
natural log, and then you

205
00:09:01,720 --> 00:09:03,390
evaluated the two boundaries.

206
00:09:03,390 --> 00:09:11,040
So you get the natural log of VB
minus VA, which is the same

207
00:09:11,040 --> 00:09:15,450
thing as the natural
log of VB over VA.

208
00:09:15,450 --> 00:09:16,480
Fair enough.

209
00:09:16,480 --> 00:09:18,510
This right here is Q1.

210
00:09:18,510 --> 00:09:18,860
All right.

211
00:09:18,860 --> 00:09:20,500
Now what is Q2?

212
00:09:20,500 --> 00:09:23,220
Q2 was this part of
the Carnot cycle.

213
00:09:23,220 --> 00:09:25,720
Q2 is when we went
from C to D.

214
00:09:25,720 --> 00:09:28,540


215
00:09:28,540 --> 00:09:33,080
So the magnitude of Q2 is
the area under this

216
00:09:33,080 --> 00:09:34,330
curve, right here.

217
00:09:34,330 --> 00:09:36,680


218
00:09:36,680 --> 00:09:39,650
Now this is the work done to of
the system, so that's why

219
00:09:39,650 --> 00:09:41,770
we subtracted it out when we
wanted to know the net work

220
00:09:41,770 --> 00:09:43,050
done by the system.

221
00:09:43,050 --> 00:09:45,510
And we ended up with this area
over here, when you subtracted

222
00:09:45,510 --> 00:09:47,460
this as well, over here
on this side.

223
00:09:47,460 --> 00:09:50,180
But if we just want to know the
magnitude of Q2, we just

224
00:09:50,180 --> 00:09:55,150
take the integral under
this curve.

225
00:09:55,150 --> 00:09:58,350
And what's the integral
under that curve?

226
00:09:58,350 --> 00:10:01,370
This is the heat out of the
system, the heat that had to

227
00:10:01,370 --> 00:10:04,520
be pushed out of the system
as work was done to it.

228
00:10:04,520 --> 00:10:07,300
So that integral-- so we could
just say, the magnitude of

229
00:10:07,300 --> 00:10:16,980
Q2-- I'll do it over here-- is
equal to-- the same exact

230
00:10:16,980 --> 00:10:19,470
logic applies, just the
boundaries are different.

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00:10:19,470 --> 00:10:22,240
We're now going from-- and
remember, when we cared about

232
00:10:22,240 --> 00:10:25,090
the direction, I would say,
I'm going from VC to VD.

233
00:10:25,090 --> 00:10:28,140
But if I just want to know the
absolute value of that area,

234
00:10:28,140 --> 00:10:31,900
because I just want to know the
magnitude, I could go from

235
00:10:31,900 --> 00:10:34,900
VC to VD and just take the
absolute value of it, or I

236
00:10:34,900 --> 00:10:38,610
could just go from VD to VC, and
I'd get a positive area.

237
00:10:38,610 --> 00:10:41,350
So let me just do that.

238
00:10:41,350 --> 00:10:45,910
So it's the integral
from VD to VC.

239
00:10:45,910 --> 00:10:49,600
Remember, the cycle we went from
VC to VD, but I just want

240
00:10:49,600 --> 00:10:50,270
the absolute value.

241
00:10:50,270 --> 00:10:51,550
I want this to be positive.

242
00:10:51,550 --> 00:10:54,810
So I'm turning it
the other way.

243
00:10:54,810 --> 00:10:57,580
Of P dv.

244
00:10:57,580 --> 00:11:00,200
And we do the exact
same math there.

245
00:11:00,200 --> 00:11:06,050
You get Q2 is equal to
nR-- this time the

246
00:11:06,050 --> 00:11:07,070
temperature is T2.

247
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We're on a T2 reservoir.

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00:11:09,130 --> 00:11:14,260
Times the natural log--
and this time,

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00:11:14,260 --> 00:11:15,420
what's it going to be?

250
00:11:15,420 --> 00:11:21,150
Times the natural log of--
instead of VB over VA, it's

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00:11:21,150 --> 00:11:28,800
going to be VC divided by VD.

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00:11:28,800 --> 00:11:31,900


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00:11:31,900 --> 00:11:34,680
Now let's use these two pieces
of information and substitute

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00:11:34,680 --> 00:11:38,450
them back into that results for
efficiency we just got.

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00:11:38,450 --> 00:11:41,220
We just learned that you could
also write the efficiency of

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00:11:41,220 --> 00:11:46,170
an engine to be equal
to 1 minus Q-- let's

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00:11:46,170 --> 00:11:46,930
look back at it.

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00:11:46,930 --> 00:11:48,350
1 minus Q2 over Q1.

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00:11:48,350 --> 00:11:53,070


260
00:11:53,070 --> 00:11:56,390
So let's substitute Q2 there
and Q1 over here.

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00:11:56,390 --> 00:11:57,820
And what do you get?

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00:11:57,820 --> 00:12:01,590
You get the efficiency of your
engine is equal to 1 minus--

263
00:12:01,590 --> 00:12:08,090
Q2 is this expression over
here-- nRT2 times the natural

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00:12:08,090 --> 00:12:20,100
log of VC over VD, all of that
divided by Q1, which is this

265
00:12:20,100 --> 00:12:21,186
one over here.

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00:12:21,186 --> 00:12:29,720
nRT1 times the natural
log of VB over VA.

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00:12:29,720 --> 00:12:31,680
Now we can do a little
bit of canceling.

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00:12:31,680 --> 00:12:34,800
Obviously the n and
the r is cancel.

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00:12:34,800 --> 00:12:36,930
And we have these natural
logs and all of that.

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00:12:36,930 --> 00:12:41,300
But I had a whole video
dedicated to show you that VC

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00:12:41,300 --> 00:12:46,370
over dv is equal
to VB over VA.

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00:12:46,370 --> 00:12:48,450
Now, if we know that these
are equal, then

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00:12:48,450 --> 00:12:49,460
this is equal to this.

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00:12:49,460 --> 00:12:50,840
So the natural logs
of them are equal,

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00:12:50,840 --> 00:12:52,260
so we can just divide.

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00:12:52,260 --> 00:12:53,850
And what are we left with?

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00:12:53,850 --> 00:12:57,350
We're left with the fact that
efficiency can also be written

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00:12:57,350 --> 00:13:08,550
as 1 minus T2 over T1
for a Carnot engine.

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00:13:08,550 --> 00:13:11,980
Remember, this time, what we did
over here, this applied to

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00:13:11,980 --> 00:13:12,970
any engine.

281
00:13:12,970 --> 00:13:16,130
This was just a little bit of
math and the definition of

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00:13:16,130 --> 00:13:20,950
what work is, and-- well,
I won't go too much

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00:13:20,950 --> 00:13:21,850
into it right now.

284
00:13:21,850 --> 00:13:24,490
But this is for a Carnot
engine, right?

285
00:13:24,490 --> 00:13:26,580
Because we did a little bit of
work here that involved the

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00:13:26,580 --> 00:13:28,020
Carnot cycle.

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00:13:28,020 --> 00:13:30,850
But this is a pretty bit
important outcome, because

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00:13:30,850 --> 00:13:33,600
we're going to show that the
Carnot engine, in the next

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00:13:33,600 --> 00:13:36,800
video, is actually the most
efficient engine that can ever

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00:13:36,800 --> 00:13:37,870
be attained.

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00:13:37,870 --> 00:13:40,150
Well, you have to be very
careful about that, that when

292
00:13:40,150 --> 00:13:42,940
we say efficient, it means that
between two temperature

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00:13:42,940 --> 00:13:48,430
sources, you can't get a more
efficient engine than the

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00:13:48,430 --> 00:13:49,070
Carnot engine.

295
00:13:49,070 --> 00:13:50,620
Now, I'm not saying it's the
best engine, or it's a

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00:13:50,620 --> 00:13:53,630
practical engine, or you'd want
it to power your lawn

297
00:13:53,630 --> 00:13:56,050
mower or your jet plane.

298
00:13:56,050 --> 00:13:59,320
I'm just saying that it's the
most efficient engine between

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00:13:59,320 --> 00:14:01,340
these two temperature
reservoirs.

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00:14:01,340 --> 00:00:00,000
And I'll show you that
in the next video.