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PROFESSOR NELSON: Well, last
time we started in on a

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discussion of entropy, a new
topic, and I started out by

00:00:32.720 --> 00:00:37.670
writing out some somewhat
verbose written descriptions

00:00:37.670 --> 00:00:42.120
that had been formulated that
indicate certain limitations

00:00:42.120 --> 00:00:46.230
about things like the efficiency
of a heat engine,

00:00:46.230 --> 00:00:48.730
and what can be done reversibly
and so forth.

00:00:48.730 --> 00:00:54.970
And these verbal descriptions
lead to some pictures that I

00:00:54.970 --> 00:00:58.900
put up and I'll put up again
about how you might try to

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accomplish something like run an
engine or move heat from a

00:01:02.920 --> 00:01:05.960
colder to a warmer body.

00:01:05.960 --> 00:01:07.540
And what some of the
limitations are

00:01:07.540 --> 00:01:09.370
on things like that.

00:01:09.370 --> 00:01:12.550
I'll show those again, but what
I want to do mostly today

00:01:12.550 --> 00:01:17.070
is try to put a mathematical
statement of the second law in

00:01:17.070 --> 00:01:20.250
place that corresponds to
the verbal statements

00:01:20.250 --> 00:01:22.270
that we saw last time.

00:01:22.270 --> 00:01:31.340
So, just to review what we saw
before, we looked at a heat

00:01:31.340 --> 00:01:37.480
engine where there is some hot
reservoir at some temperature,

00:01:37.480 --> 00:01:51.640
T1, and it's connected to an
engine running in a cycle.

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We could, write the work that we
generate that comes out as

00:01:57.580 --> 00:02:14.560
negative w.

00:02:14.560 --> 00:02:22.870
And then there's a cold
reservoir at some lower

00:02:22.870 --> 00:02:25.610
temperature T2.

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So the way I've got things
written here, q1 is positive.

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Heat's flowing from the hot
reservoir to the engine that

00:02:36.140 --> 00:02:38.160
runs in a cycle.

00:02:38.160 --> 00:02:39.340
Work is coming out.

00:02:39.340 --> 00:02:42.350
A positive amount of work is
being generated and is coming

00:02:42.350 --> 00:02:50.980
out, so minus w is positive, and
this minus q2 is positive,

00:02:50.980 --> 00:03:01.360
that is heat is also flowing
into a cold reservoir.

00:03:01.360 --> 00:03:06.020
And what we saw last time is
that this was necessary to

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make things work, this part
of it in particular.

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You couldn't just run something
successfully in a

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cycle and get work out of it,
using the heat from the hot

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reservoir, without also
converting some of the heat

00:03:18.860 --> 00:03:21.130
that came in to heat that would

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flow into a cold reservoir.

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All the heat couldn't
get successfully

00:03:25.070 --> 00:03:27.690
converted into work.

00:03:27.690 --> 00:03:39.240
That would be desirable,
but it's not possible.

00:03:39.240 --> 00:03:43.990
And, similarly, if we try to run
things backward and build

00:03:43.990 --> 00:03:55.510
a refrigerator, so now we have
our cold reservoir, and we're

00:03:55.510 --> 00:04:03.870
going to remove heat from
it, and pump it

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up into a hot reservoir.

00:04:18.260 --> 00:04:22.070
And to do this, we saw that
you have to put work in,

00:04:22.070 --> 00:04:26.130
right, you can't just remove
heat from a cold reservoir,

00:04:26.130 --> 00:04:29.840
move it up to a hot reservoir,
without doing some work to

00:04:29.840 --> 00:04:34.690
accomplish that.

00:04:34.690 --> 00:04:37.750
So here, q2 is greater
than zero.

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The work is greater than zero,
and minus q1 is greater than

00:04:46.390 --> 00:04:51.790
zero, that is heat is
flowing this way.

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So these are just the pictures
that we saw last time.

00:04:58.220 --> 00:05:01.320
And then we had some statements
of the second law

00:05:01.320 --> 00:05:04.120
of thermodynamics that I won't
re-write up here, but, for

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example, Clausius gave us the
statement that it's impossible

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for a system to operate in a
cycle the takes heat from a

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cold reservoir, transfers it
to a hot reservoir, that is

00:05:14.120 --> 00:05:17.650
acts like a refrigerator,
without at some, at the same

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time, converting some
work into heat.

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Work has to come in to
make that happen.

00:05:22.040 --> 00:05:25.550
And we had the similar statement
by Kelvin about the

00:05:25.550 --> 00:05:29.060
heat engine that required that
some heat gets dumped into a

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cold reservoir in the process
of converting the heat from

00:05:31.770 --> 00:05:34.740
the hot reservoir into work.

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Fine.

00:05:36.800 --> 00:05:41.540
Now what I want to do is put up
a specific example of the

00:05:41.540 --> 00:05:48.440
cycle that can be undertaken
inside here in an engine, and

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we can just calculate from what
you've already seen of

00:05:50.590 --> 00:05:51.690
thermodynamics.

00:05:51.690 --> 00:05:54.760
What happens to the
thermodynamics parameters, and

00:05:54.760 --> 00:05:56.580
see the results in terms of the

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parameters including entropy.

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So let's try that.

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So, the engine that I'm going
to illustrate is called a

00:06:07.320 --> 00:06:12.100
Carnot engine.

00:06:12.100 --> 00:06:15.900
And the cycle it's going to
undertake is called a Carnot

00:06:15.900 --> 00:06:24.670
cycle, and it works the
following way: we're going to

00:06:24.670 --> 00:06:28.040
do pressure volume work.

00:06:28.040 --> 00:06:30.320
So this is something that,
by now, you're

00:06:30.320 --> 00:06:34.460
pretty familiar with.

00:06:34.460 --> 00:06:39.110
So we're going to start at one,
go to two and this is

00:06:39.110 --> 00:06:46.640
going to be in isotherm at
temperature T1, and all the

00:06:46.640 --> 00:06:58.590
paths here are going
to be reversible.

00:06:58.590 --> 00:07:00.370
So that's our first step.

00:07:00.370 --> 00:07:02.760
Then we're going to have
an adiabatic expansion.

00:07:02.760 --> 00:07:05.340
So this is an isothermal
expansion.

00:07:05.340 --> 00:07:17.320
Here comes an adiabatic
expansion, to point three.

00:07:17.320 --> 00:07:22.170
So at this point the temperature
will change.

00:07:22.170 --> 00:07:24.870
Then we're going to have another
isothermal step, a

00:07:24.870 --> 00:07:37.520
compression to some
point four.

00:07:37.520 --> 00:07:42.790
So this is an isotherm at some
different temperature T2, a

00:07:42.790 --> 00:07:45.260
cooler temperature, because
this was an expansion.

00:07:45.260 --> 00:07:47.390
We know in an adiabatic
expansion the

00:07:47.390 --> 00:07:49.910
system's going to cool.

00:07:49.910 --> 00:07:53.790
And now we're going to have
another adiabatic step, an

00:07:53.790 --> 00:08:01.560
adiabatic compression.

00:08:01.560 --> 00:08:02.140
And that's it.

00:08:02.140 --> 00:08:04.060
That's going to take us back
to our starting point.

00:08:04.060 --> 00:08:06.620
So that's our cycle.

00:08:06.620 --> 00:08:09.360
Of course there are lots of ways
we can execute the cycle,

00:08:09.360 --> 00:08:12.210
but this is a simple one, and
these are steps that we're all

00:08:12.210 --> 00:08:14.650
familiar with at this point.

00:08:14.650 --> 00:08:18.680
So this is the picture, and
this is a cycle that

00:08:18.680 --> 00:08:19.440
undertakes it.

00:08:19.440 --> 00:08:22.580
So let's just look step by
step at what happens.

00:08:22.580 --> 00:08:28.490
So going from one to two, it's
an isothermal expansion a T1,

00:08:28.490 --> 00:08:41.670
so delta u is q1, we'll
call it, plus w1,

00:08:41.670 --> 00:08:43.690
for the first step.

00:08:43.690 --> 00:08:51.550
Going from two to three, that's
an adiabatic expansion,

00:08:51.550 --> 00:08:55.510
so q is equal to zero
in that step.

00:08:55.510 --> 00:09:02.540
And delta u is equal to what
we'll call w1 prime.

00:09:02.540 --> 00:09:07.190
So we'll use w1 and w1 prime to
describe the work involved.

00:09:07.190 --> 00:09:15.590
The work input during
the expansion steps.

00:09:15.590 --> 00:09:23.365
Step three to four isothermal
compression, delta u is going

00:09:23.365 --> 00:09:27.320
to be q2 plus w2.

00:09:27.320 --> 00:09:29.710
And finally, we're going
to go back to one.

00:09:29.710 --> 00:09:32.950
We're going to close the cycle
with another adiabatic step.

00:09:32.950 --> 00:09:34.860
q is zero.

00:09:34.860 --> 00:09:46.590
Delta u is w2 prime.

00:09:46.590 --> 00:09:51.140
Now, the total amount of the
work that we can get out is

00:09:51.140 --> 00:10:03.970
just given by the area
inside this curve.

00:10:03.970 --> 00:10:12.220
We'll call it capital W. It's
just minus the sum of all

00:10:12.220 --> 00:10:14.880
these things, because of course
these are just defined

00:10:14.880 --> 00:10:18.330
as the work done by the
environment to the system.

00:10:18.330 --> 00:10:28.040
So it's minus w1 plus w1 prime
plus w2 plus w2 prime.

00:10:28.040 --> 00:10:42.390
The heat input is just q1, and
we'll define that as capital

00:10:42.390 --> 00:10:46.250
Q. And I just want to write
those because what I really

00:10:46.250 --> 00:10:49.020
want to get at is what's the
efficiency of the whole thing?

00:10:49.020 --> 00:10:52.280
And in practical terms, we can
define the efficiency as the

00:10:52.280 --> 00:10:55.490
ratio of the heat in
to the work out.

00:10:55.490 --> 00:11:17.450
We would like that to be
as high as possible.

00:11:17.450 --> 00:11:25.300
So it's just capital W over
capital Q, which is to say

00:11:25.300 --> 00:11:34.680
it's minus all that
stuff over q1.

00:11:39.160 --> 00:11:42.580
Now the first law is going to
hold in all of these steps,

00:11:42.580 --> 00:11:45.130
and we're going around
in a cycle.

00:11:45.130 --> 00:11:48.560
So what does that tell us about
a state function like u?

00:11:48.560 --> 00:11:53.310
What's delta u going around
the whole thing?

00:11:53.310 --> 00:11:54.990
I want a chorus in
the answer here.

00:11:54.990 --> 00:11:56.020
What's delta u?

00:11:56.020 --> 00:11:56.980
STUDENT: Zero.

00:11:56.980 --> 00:11:58.570
PROFESSOR NELSON: Excellent.

00:11:58.570 --> 00:12:01.390
MIT students, yes!

00:12:01.390 --> 00:12:08.550
OK, so the first law, we know
that the, going around the

00:12:08.550 --> 00:12:19.750
cycle the integral of
delta u is zero.

00:12:19.750 --> 00:12:23.510
And that has to equal
q plus w, summed

00:12:23.510 --> 00:12:27.520
up for all the steps.

00:12:27.520 --> 00:12:36.000
Which is to say q1 plus q2 is
equal to minus w1 plus w1

00:12:36.000 --> 00:12:48.200
prime plus w2 plus w2 prime.

00:12:48.200 --> 00:12:50.560
So that means we can rewrite
this efficiency.

00:12:50.560 --> 00:12:56.000
We can replace this sum by just
the some of q1 plus q2.

00:12:56.000 --> 00:13:03.400
So sufficiency is q1
plus q2 over q1.

00:13:06.760 --> 00:13:11.200
Or we can write it as
one plus q2 over q1.

00:13:18.780 --> 00:13:24.480
Now that already tells us what
we know, which is that the

00:13:24.480 --> 00:13:26.170
efficiency is going
to be something

00:13:26.170 --> 00:13:28.000
less than zero, right?

00:13:28.000 --> 00:13:32.410
Because we've got one and then
we're adding q2 over q1 to it,

00:13:32.410 --> 00:13:37.120
but we've got negative q2
is a positive number.

00:13:37.120 --> 00:13:39.110
Heat is flowing this way.

00:13:39.110 --> 00:13:43.880
So q2, the heat in this way
is negative. q2 over q1 is

00:13:43.880 --> 00:13:45.500
negative. q2 of course
is positive.

00:13:45.500 --> 00:13:50.220
That is heat flowing from the
hot reservoir to the engine.

00:13:50.220 --> 00:13:53.230
So that efficiency is something
less than one, and

00:13:53.230 --> 00:13:58.380
we'd like to figure
out what that is.

00:13:58.380 --> 00:14:05.370
So let's just state that.

00:14:05.370 --> 00:14:08.990
Well, now let's go back to each
of our individual steps

00:14:08.990 --> 00:14:14.350
and look, based on what we know
about how to evaluate the

00:14:14.350 --> 00:14:17.800
thermodynamic changes that take
place here, let's look at

00:14:17.800 --> 00:14:20.420
each one of the steps and
see what happens.

00:14:20.420 --> 00:14:36.730
So from one to two
it's isothermal.

00:14:36.730 --> 00:14:38.720
And now we're going to specify,
we're going to do a

00:14:38.720 --> 00:14:42.340
Carnot cycle for an ideal gas.

00:14:42.340 --> 00:14:43.650
It's an ideal gas.

00:14:43.650 --> 00:14:45.710
It's an isothermal change.

00:14:45.710 --> 00:14:49.870
What's delta u?

00:14:49.870 --> 00:14:53.040
You know, it's like lots of
courses and all sorts of

00:14:53.040 --> 00:14:57.240
things that you're seeing,
they're always the same.

00:14:57.240 --> 00:14:58.580
What delta u?

00:14:58.580 --> 00:15:00.000
STUDENT: Zero.

00:15:00.000 --> 00:15:04.330
PROFESSOR NELSON?

00:15:04.330 --> 00:15:04.540
Right.

00:15:04.540 --> 00:15:08.100
Delta u is zero, and it's
also equal to this.

00:15:08.100 --> 00:15:12.500
So that says the q1 is
the opposite of w1.

00:15:15.290 --> 00:15:24.430
It's an isothermal expansion,
so dw is just negative p dV.

00:15:24.430 --> 00:15:27.640
So it's, this is just
the integral from

00:15:27.640 --> 00:15:32.300
one to two of p dV.

00:15:32.300 --> 00:15:36.980
And it's an ideal gas,
isothermal, right.

00:15:36.980 --> 00:15:42.135
RT over V. Were going to make it
for a mole of gas, so it's

00:15:42.135 --> 00:15:48.650
R times T1, and then we'll have
dV over V. So that's just

00:15:48.650 --> 00:15:51.200
the log of V2 over V1.

00:15:51.200 --> 00:16:01.750
Let's have one mole,
n equals one.

00:16:01.750 --> 00:16:08.700
Okay two going to three that's
this, it's adiabatic, right.

00:16:08.700 --> 00:16:15.740
So delta u is just equal to the
work but we also know what

00:16:15.740 --> 00:16:17.100
happens because the
temperature is

00:16:17.100 --> 00:16:20.850
changing from T1 to T2.

00:16:20.850 --> 00:16:29.830
So delta you is just Cv
times T2 minus T1.

00:16:29.830 --> 00:16:31.350
du is Cv dT.

00:16:31.350 --> 00:16:44.250
It's an ideal gas, and that's
equal to w1 prime.

00:16:44.250 --> 00:16:50.450
Now, this is a reversible
adiabatic path, so there's a

00:16:50.450 --> 00:16:58.660
relationship that I'm sure
you'll remember.

00:16:58.660 --> 00:17:06.480
Namely, T2 over T1 is
equal to V2 over V3.

00:17:06.480 --> 00:17:08.930
Don't be confused by
the subscripts.

00:17:08.930 --> 00:17:12.750
We're talking about quantities
both starting at two

00:17:12.750 --> 00:17:15.210
and going to three.

00:17:15.210 --> 00:17:16.530
So it's V2 and V3.

00:17:16.530 --> 00:17:17.770
They're different volumes.

00:17:17.770 --> 00:17:20.190
The temperature though at two is
the same as it was at one,

00:17:20.190 --> 00:17:22.480
so it's still a temp
at T1, and this

00:17:22.480 --> 00:17:25.370
temperatures is at T2.

00:17:25.370 --> 00:17:31.570
Anyway T2 over T1 is equal
to V2 over V3 to the

00:17:31.570 --> 00:17:35.550
power gamma minus one.

00:17:35.550 --> 00:17:41.740
So we're going to kind of store
that for the moment.

00:17:41.740 --> 00:17:45.740
Now going from three to four
right, so we have another

00:17:45.740 --> 00:17:50.830
isothermal process for an ideal
gas, so I won't try to

00:17:50.830 --> 00:17:53.510
make you sing again so soon.

00:17:53.510 --> 00:17:56.040
Delta u is zero.

00:17:56.040 --> 00:18:02.750
And so just like here, now q2
is minus w2, that's integral

00:18:02.750 --> 00:18:06.870
going from three to four p dV.

00:18:10.220 --> 00:18:15.360
So it's R T2, right, now we're
at a lower temperature times

00:18:15.360 --> 00:18:23.230
the log of V4 over V3.

00:18:23.230 --> 00:18:36.740
So that's our work in
this path and heat.

00:18:36.740 --> 00:18:42.630
And finally going from
four back to one.

00:18:42.630 --> 00:18:47.860
So we've already seen
that q is zero.

00:18:47.860 --> 00:18:56.050
So we know that delta u is just
Cv times T1 minus T2.

00:18:56.050 --> 00:18:57.980
Now we're going from
T2 up to T1.

00:19:00.670 --> 00:19:08.600
And this is equal to w2 prime.

00:19:08.600 --> 00:19:11.870
And now, just like we had
before, again we've got a

00:19:11.870 --> 00:19:29.430
reversible adiabatic path, so
T1 over T2 has to equal V4

00:19:29.430 --> 00:19:38.950
over V1, to the gamma
minus one, okay.

00:19:38.950 --> 00:19:44.880
All right, now, if this is the
case, of course, this is just

00:19:44.880 --> 00:19:48.030
the inverse of this.

00:19:48.030 --> 00:19:50.520
So this just must be the
inverse of this.

00:19:50.520 --> 00:20:01.740
We can combine these two to see
that V4 over V1 must equal

00:20:01.740 --> 00:20:05.630
V3 over V2.

00:20:05.630 --> 00:20:08.890
So now we have a relationship
between the ratios of these

00:20:08.890 --> 00:20:21.510
volumes that are reached during
these adiabatic paths.

00:20:21.510 --> 00:20:41.300
Now, let's just look
at our efficiency.

00:20:41.300 --> 00:20:48.070
So we saw that efficiency
is one plus q2 over q1.

00:20:48.070 --> 00:20:52.430
Let's just use our expressions
that we've

00:20:52.430 --> 00:20:56.280
found for q2 and q1.

00:20:56.280 --> 00:20:58.130
We're going to have
q2 over q1.

00:20:58.130 --> 00:20:58.790
R is going to cancel.

00:20:58.790 --> 00:21:01.400
We're going to have the ratio of
temperatures and the ratio

00:21:01.400 --> 00:21:11.470
of these logs.

00:21:11.470 --> 00:21:28.120
So it's T2 over T1 times the
log of V4 over V3, over the

00:21:28.120 --> 00:21:41.030
log of V2 over V1, but we just
arrived at this relationship

00:21:41.030 --> 00:21:42.280
between these volumes.

00:21:42.280 --> 00:21:52.990
And this of course tells us that
V4 over V3 is V1 over V2.

00:21:52.990 --> 00:21:53.820
Right.

00:21:53.820 --> 00:21:56.800
I'm just inverting these.

00:21:56.800 --> 00:21:58.820
So here it is.

00:21:58.820 --> 00:22:03.960
Here's the V4 over
V3, oops, sorry.

00:22:03.960 --> 00:22:11.250
V2 over V1, this is equal
to the inverse of this.

00:22:11.250 --> 00:22:14.670
So the ratio of the logs
is just minus one.

00:22:14.670 --> 00:22:19.500
So this is just, sorry, I forgot
about the one here.

00:22:19.500 --> 00:22:30.210
One plus this, but this is the
same as one minus T2 over T1.

00:22:30.210 --> 00:22:34.820
That's our expression
for the efficiency.

00:22:34.820 --> 00:22:36.740
All right, isn't
that terrific?

00:22:36.740 --> 00:22:39.430
Now, we have an expression.

00:22:39.430 --> 00:22:42.300
We can figure out
the efficiency.

00:22:42.300 --> 00:22:44.540
All we need to know is
the temperatures.

00:22:44.540 --> 00:22:46.590
What's the temperature
of the hot reservoir?

00:22:46.590 --> 00:22:48.380
What's the temperature of
the cold reservoir?

00:22:48.380 --> 00:22:51.290
We're done.

00:22:51.290 --> 00:22:53.910
That's the efficiency.

00:22:53.910 --> 00:23:03.050
As we expected, it's less than
one, and of course if we build

00:23:03.050 --> 00:23:05.650
an engine, we want it to be as
high as possible, as close to

00:23:05.650 --> 00:23:06.730
one as possible.

00:23:06.730 --> 00:23:11.830
What that means is we want to
run the hot reservoir as hot

00:23:11.830 --> 00:23:16.620
as possible and the cold
reservoir as cold as possible.

00:23:16.620 --> 00:23:22.810
In principle, this value, this
efficiency, can approach 1 as

00:23:22.810 --> 00:23:26.250
the low temperature approaches
absolute zero.

00:23:26.250 --> 00:23:32.590
If this were to be an absolute
zero Kelvin, then we could we

00:23:32.590 --> 00:23:34.190
can have something,
wait a minute.

00:23:34.190 --> 00:23:36.100
Sorry, it's T2.

00:23:36.100 --> 00:23:41.000
As T2 goes to zero, the cold
reservoir, then this goes to

00:23:41.000 --> 00:23:44.970
zero and our efficiency
approaches one.

00:23:44.970 --> 00:23:47.180
So that would be the
best we can do.

00:23:47.180 --> 00:23:49.600
And that basically make
sense, right?

00:23:49.600 --> 00:23:54.090
The whole thing is being powered
by sending heat from

00:23:54.090 --> 00:23:57.180
the hot to the cold reservoir.

00:23:57.180 --> 00:24:00.380
The colder that cold reservoir
is, the hotter that hot

00:24:00.380 --> 00:24:01.970
reservoir is, the better
off we are.

00:24:01.970 --> 00:24:04.080
The more work we can
get out of it.

00:24:04.080 --> 00:24:10.120
The closer the efficiency
will get to one.

00:24:10.120 --> 00:24:15.550
So in some sense, the first law
would suggest you can sort

00:24:15.550 --> 00:24:17.430
of break even.

00:24:17.430 --> 00:24:20.180
That is, it gives us the
relationship between energy

00:24:20.180 --> 00:24:21.700
and work and heat.

00:24:21.700 --> 00:24:23.450
It might suggest that
you could convert

00:24:23.450 --> 00:24:25.270
all the work to heat.

00:24:25.270 --> 00:24:30.480
The second law says that really,
you can't do that.

00:24:30.480 --> 00:24:34.200
The only way you could possibly
contemplate it is to

00:24:34.200 --> 00:24:38.580
be working at absolute
zero Kelvin.

00:24:38.580 --> 00:24:42.510
Guess what the third law
is going to tell us?

00:24:42.510 --> 00:24:45.000
You can't get to zero Kelvin.

00:24:45.000 --> 00:24:48.140
We'll see that shortly.

00:24:48.140 --> 00:24:51.500
But this is those closest you
could come at least by trying

00:24:51.500 --> 00:25:28.820
to do that to having your
efficiency approach one.

00:25:28.820 --> 00:25:36.920
Now, we've seen that q2
over q1 is equal to

00:25:36.920 --> 00:25:40.880
negative T2 over T1.

00:25:40.880 --> 00:25:48.070
So let me just rewrite
that as q1 over T1 is

00:25:48.070 --> 00:25:54.190
minus q2 over T2.

00:25:54.190 --> 00:26:05.020
Or in other words, q1 over T1
plus q2 over T2 is zero.

00:26:05.020 --> 00:26:10.560
But q1 and q2, each one of those
is just the integrated

00:26:10.560 --> 00:26:14.390
amount of heat that was
transferred going along a

00:26:14.390 --> 00:26:18.550
reversible constant
temperature path.

00:26:18.550 --> 00:26:26.420
Which means that q1 over T1,
that's this delta S thing that

00:26:26.420 --> 00:26:28.690
we saw before.

00:26:28.690 --> 00:26:33.410
And what we can say about this
is it's saying that if we go

00:26:33.410 --> 00:26:41.780
around in a cycle and look at
dq reversible over T it's

00:26:41.780 --> 00:27:08.290
zero, because that's what
these quantities are.

00:27:08.290 --> 00:27:11.660
Now, of course, we can run the
engine backward and build a

00:27:11.660 --> 00:27:16.680
refrigerator, and if you've got
your lecture notes from

00:27:16.680 --> 00:27:20.100
last period, your 8-9, well,
they're labeled 8-9 lecture

00:27:20.100 --> 00:27:24.470
notes, I made an attempt to
define something which was a

00:27:24.470 --> 00:27:25.570
little bit misguided.

00:27:25.570 --> 00:27:28.380
And so instead of defining
efficiency the way you've got

00:27:28.380 --> 00:27:32.210
it written there, I'm going to
define what's called something

00:27:32.210 --> 00:27:34.720
different for a refrigerator
which is called the

00:27:34.720 --> 00:27:38.130
coefficient of performance.

00:27:38.130 --> 00:28:16.110
So it's the following: so we can
define the coefficient of

00:28:16.110 --> 00:28:29.300
performance written as eta
as q2 over minus w.

00:28:29.300 --> 00:28:34.530
In other words, you know, it's
how much heat can we pull out

00:28:34.530 --> 00:28:40.100
of the cold reservoir, for
whatever amount of work that

00:28:40.100 --> 00:28:42.680
we're going to put in
in order to do that?

00:28:42.680 --> 00:28:47.050
Well of course, we'd like that
to be as big as possible.

00:28:47.050 --> 00:28:59.300
But of course, this is just
q2 over minus q1 plus q2.

00:28:59.300 --> 00:29:17.090
Just to be clear, so it's heat
extracted over the work in.

00:29:17.090 --> 00:29:20.460
So let me just rewrite that as,
I just want to divide by

00:29:20.460 --> 00:29:21.990
q1 everywhere.

00:29:21.990 --> 00:29:31.430
So I'm going to write this as q2
over q1 over minus one plus

00:29:31.430 --> 00:29:35.490
q2 over q1.

00:29:35.490 --> 00:29:38.830
I want to do that because we
know that q2 over q1 is

00:29:38.830 --> 00:29:40.350
negative T2 over T1.

00:29:40.350 --> 00:29:46.360
So this is negative T2 over
T1 over negative one

00:29:46.360 --> 00:29:49.790
minus T2 over T1.

00:29:49.790 --> 00:29:52.770
I can cancel those, and
then I'm going to

00:29:52.770 --> 00:29:56.880
multiply through by T1.

00:29:56.880 --> 00:30:06.620
So this is just going to be T2
over T1 minus T2, that's our

00:30:06.620 --> 00:30:17.190
coefficient of performance.

00:30:17.190 --> 00:30:20.550
So it's a measure of how well
we can do to run the

00:30:20.550 --> 00:30:21.490
refrigerator.

00:30:21.490 --> 00:30:26.520
So you know, what the second law
is doing, in words, it's

00:30:26.520 --> 00:30:29.920
putting these restrictions on
how well or how effectively we

00:30:29.920 --> 00:30:33.970
can convert heat into work in
the case of the engine, or

00:30:33.970 --> 00:30:39.230
work into heat extracted in the
case of a refrigerator.

00:30:39.230 --> 00:30:45.000
And it follows qualitatively
from just your ordinary

00:30:45.000 --> 00:30:48.530
observations about the direction
in which things go

00:30:48.530 --> 00:30:49.700
spontaneously, right.

00:30:49.700 --> 00:30:52.430
You know the heat isn't going to
flow from a cold body to a

00:30:52.430 --> 00:30:57.770
hot body without putting some
work in to make that happen.

00:30:57.770 --> 00:31:01.810
And so, we'll be able of follow
that further and see

00:31:01.810 --> 00:31:07.120
really how to determine the
direction of spontaneity for a

00:31:07.120 --> 00:31:09.660
whole set of processes, really
in principle for any

00:31:09.660 --> 00:31:14.390
processes, went analyzed
properly.

00:31:14.390 --> 00:31:18.090
So the first step to doing
that is I want to just

00:31:18.090 --> 00:31:22.560
generalize our results so
far for a Carnot cycle.

00:31:22.560 --> 00:31:24.050
You might think well
okay, but this is a

00:31:24.050 --> 00:31:25.810
pretty specialized case.

00:31:25.810 --> 00:31:30.730
We've formulated one particular
kind of engine, and

00:31:30.730 --> 00:31:35.190
seen how we can analyze what
it does, come up with

00:31:35.190 --> 00:31:40.040
relations that seem of value
for efficiency and other

00:31:40.040 --> 00:31:41.480
quantities.

00:31:41.480 --> 00:31:45.180
How general is it?

00:31:45.180 --> 00:31:46.640
Well, let's take a look.

00:31:46.640 --> 00:31:56.880
So, what I want to do is make a
new engine which really just

00:31:56.880 --> 00:32:01.090
consists of two engines,
side by side.

00:32:01.090 --> 00:32:05.370
One is our Carnot engine as
we've seen it, and the other

00:32:05.370 --> 00:32:10.510
is just any other reversible
engine.

00:32:10.510 --> 00:32:23.770
So generalize our Carnot
engine results.

00:32:23.770 --> 00:32:31.790
So let's take our hot reservoir
and draw it bigger.

00:32:31.790 --> 00:32:35.030
We know it has to big anyway,
since we can extract heat from

00:32:35.030 --> 00:32:38.000
it without changing
the temperature.

00:32:38.000 --> 00:32:44.190
And on this side, we're going
to write out an engine, and

00:32:44.190 --> 00:32:47.740
we're going to say this is a
Carnot engine. so on the side

00:32:47.740 --> 00:32:54.220
of it we have q1 prime.

00:32:54.220 --> 00:32:57.380
We're going to have w prime.

00:32:57.380 --> 00:33:01.330
And we're going to have q2
prime, and we're going to draw

00:33:01.330 --> 00:33:06.980
the arrows in the positive
direction in these cases.

00:33:06.980 --> 00:33:10.730
Or in the defined directions
I should say.

00:33:10.730 --> 00:33:17.180
Here is our colder reservoir.

00:33:17.180 --> 00:33:20.580
OK, so here is just an engine
like what we've already seen,

00:33:20.580 --> 00:33:27.150
and I'm going to specify that
this is a Carnot engine which

00:33:27.150 --> 00:33:29.250
is to say all the results
that we just derived

00:33:29.250 --> 00:33:31.360
hold for this case.

00:33:31.360 --> 00:33:38.280
And side by side, we're going
to run another engine.

00:33:38.280 --> 00:33:45.270
So it's got q2, and it's
got a cycle w.

00:33:45.270 --> 00:33:47.450
And it's got q1.

00:33:51.920 --> 00:33:59.260
So this is some other
engine that runs

00:33:59.260 --> 00:34:04.280
using reversible processes.

00:34:04.280 --> 00:34:07.910
So I can define the efficiency
of each one of them.

00:34:07.910 --> 00:34:13.930
The efficiency for the one on
the left is minus w over q1,

00:34:13.930 --> 00:34:18.210
The efficiency prime for the
one on the right is minus w

00:34:18.210 --> 00:34:22.610
prime over q1 prime.

00:34:22.610 --> 00:34:30.150
Now I want to just assume that
in some way they're different.

00:34:30.150 --> 00:34:42.390
So let's assume that epsilon
prime is greater than epsilon.

00:34:42.390 --> 00:34:47.060
So in other words, this engine
is running less efficiently

00:34:47.060 --> 00:34:49.550
than my Carnot engine.

00:34:49.550 --> 00:34:50.660
It's also reversible.

00:34:50.660 --> 00:34:54.890
And now since it's reversible,
we can run

00:34:54.890 --> 00:34:58.240
it forward or backward.

00:34:58.240 --> 00:35:07.710
So we can run this one backward
and we can use the

00:35:07.710 --> 00:35:14.250
work that comes out as the
input, well sorry, use this

00:35:14.250 --> 00:35:17.710
work that comes out of the one
of the right to run this one

00:35:17.710 --> 00:35:20.330
backward, which is to
say we'll move heat

00:35:20.330 --> 00:35:23.540
from cold to hot.

00:35:23.540 --> 00:35:33.290
So use work out of right-hand
side to

00:35:33.290 --> 00:35:40.910
run left-hand backward.

00:35:40.910 --> 00:35:42.660
Why not?

00:35:42.660 --> 00:35:44.600
We need work to come
in, we might as

00:35:44.600 --> 00:36:00.870
well get it from here.

00:36:00.870 --> 00:36:10.980
So, the total work that we can
get out must be zero, out of

00:36:10.980 --> 00:36:12.630
the whole sum of them.

00:36:12.630 --> 00:36:15.120
After all, we're taking the
output work that we get from

00:36:15.120 --> 00:36:21.800
the right, and using it
all to drive the left.

00:36:21.800 --> 00:36:28.420
So that means that minus
w prime must equal w.

00:36:28.420 --> 00:36:33.300
And w is greater than zero. that
is in this one we have

00:36:33.300 --> 00:36:38.300
the environment doing
work on the system.

00:36:38.300 --> 00:36:44.400
OK, now we've assumed that
epsilon prime is

00:36:44.400 --> 00:36:46.210
greater than epsilon.

00:36:46.210 --> 00:36:51.670
So we can just write out what
those are, minus w prime over

00:36:51.670 --> 00:36:57.720
q1 prime is greater than
minus w over q1.

00:36:57.720 --> 00:37:04.300
But we know that minus w prime
is the same thing as w.

00:37:04.300 --> 00:37:11.250
So w over q1 prime is greater
than minus w over q1.

00:37:15.450 --> 00:37:19.410
Which is the same thing just
to be a little bit maybe

00:37:19.410 --> 00:37:25.420
pedantic because w over minus
q1, and the only reason I'm

00:37:25.420 --> 00:37:29.090
writing that is to illustrate
that this says that q1 must be

00:37:29.090 --> 00:37:34.040
less than q1 prime.

00:37:34.040 --> 00:37:42.710
So q1 is less than zero.

00:37:42.710 --> 00:37:45.810
Remember this one's running
backward, we're pumping heat

00:37:45.810 --> 00:37:52.130
up. q1 prime is greater
than zero, it's

00:37:52.130 --> 00:37:53.330
running as an engine.

00:37:53.330 --> 00:37:57.930
It's taking heat from
the hot reservoir.

00:37:57.930 --> 00:38:01.660
Well, that's interesting.

00:38:01.660 --> 00:38:11.090
That says that minus q1 prime
plus q1 is greater than zero.

00:38:11.090 --> 00:38:16.910
Well, it's a pretty interesting
result.

00:38:16.910 --> 00:38:20.950
There's no work being done on,
out of the whole thing.

00:38:20.950 --> 00:38:24.340
This is providing work that's
being used in here, but if you

00:38:24.340 --> 00:38:27.830
take the whole outside of the
surroundings and this whole

00:38:27.830 --> 00:38:32.450
thing is the system, no net
work, these things cancel each

00:38:32.450 --> 00:38:39.740
other, and yet heat's
going up.

00:38:39.740 --> 00:38:43.630
How did that happen?

00:38:43.630 --> 00:38:47.620
Well it happened because we
clearly must have a faulty

00:38:47.620 --> 00:38:50.310
assumption underlying what
we've just done.

00:38:50.310 --> 00:39:02.480
This can't possibly be true.

00:39:02.480 --> 00:39:08.900
This is impossible.

00:39:08.900 --> 00:39:12.700
So what this says is the
efficiency of any reversible

00:39:12.700 --> 00:39:16.190
engine has to be one
minus T2 over T1.

00:39:29.380 --> 00:39:32.830
It's not a result that's
specific to the one

00:39:32.830 --> 00:39:39.450
cycle that we put up.

00:39:39.450 --> 00:39:42.490
And you know, you could have a
reversible engine with lots

00:39:42.490 --> 00:39:47.390
and lots of steps, but you could
always break them down

00:39:47.390 --> 00:39:49.930
into some sequence of adiabatic

00:39:49.930 --> 00:39:51.400
and isothermal steps.

00:39:51.400 --> 00:39:53.860
So you know your cycle, you
know, you could have a whole

00:39:53.860 --> 00:40:04.230
complicated sequence on a p v
diagram of steps going back.

00:40:04.230 --> 00:40:07.740
As long as it's reversible, you
know what the efficiency

00:40:07.740 --> 00:40:11.550
has to be, and in principle, you
could break it down into a

00:40:11.550 --> 00:40:13.660
bunch of steps that you
could formulate as

00:40:13.660 --> 00:40:14.890
isothermal and adiabatic.

00:40:14.890 --> 00:40:17.170
They might have to be formulated
as very small

00:40:17.170 --> 00:40:24.560
steps, in order to do
that, but you could.

00:40:24.560 --> 00:40:39.270
What this says too, is this
result that we found, right,

00:40:39.270 --> 00:40:45.740
now of course that's our
integral dS is zero.

00:40:45.740 --> 00:40:55.660
It's general.

00:40:55.660 --> 00:41:11.330
Entropy S is a state function,
generally.

00:41:11.330 --> 00:41:13.710
Now remember, we went through
before how it's a state

00:41:13.710 --> 00:41:16.210
function but to calculate
it, you'd need to find a

00:41:16.210 --> 00:41:22.420
reversible path, along which
you can figure this out.

00:41:22.420 --> 00:41:31.970
But the fact is it's a state
function, in a general way.

00:41:31.970 --> 00:41:43.390
Now, if we go back to our Carnot
cycle which is a set of

00:41:43.390 --> 00:41:47.760
reversible paths, it's useful
to compare this to what

00:41:47.760 --> 00:41:51.080
happens in an irreversible
case.

00:41:51.080 --> 00:41:54.820
In a real engine, of course,
you can approach the

00:41:54.820 --> 00:41:59.030
reversible limit.

00:41:59.030 --> 00:42:02.790
Every step won't be perfectly
reversible.

00:42:02.790 --> 00:42:07.300
And of course it's not hard
to see what happens.

00:42:07.300 --> 00:42:11.310
Let's just take our reversible
engine and

00:42:11.310 --> 00:42:14.310
modify it a little bit.

00:42:14.310 --> 00:42:17.490
Let's imagine that instead
of all the steps being

00:42:17.490 --> 00:42:23.260
reversible, let's just put
in one irreversible step.

00:42:23.260 --> 00:42:24.720
Let's do this.

00:42:24.720 --> 00:42:29.540
Instead of this reversible
isothermal step, Let's make it

00:42:29.540 --> 00:42:35.370
an irreversible isothermal
step.

00:42:35.370 --> 00:42:41.960
We can have a different
isothermal step.

00:42:41.960 --> 00:42:46.440
Of course in the reversible
case, you're always pushing

00:42:46.440 --> 00:42:50.430
against an external pressure,
which is essentially equal to

00:42:50.430 --> 00:42:53.330
the internal pressure.

00:42:53.330 --> 00:42:54.950
Let's not do that.

00:42:54.950 --> 00:42:59.550
Let's imagine maybe instead we
just immediately dropped the

00:42:59.550 --> 00:43:01.910
pressure and let the
system expand

00:43:01.910 --> 00:43:04.100
against the lower pressure.

00:43:04.100 --> 00:43:06.080
Now we know we're going
to get less work out

00:43:06.080 --> 00:43:06.820
of it in that case.

00:43:06.820 --> 00:43:08.830
You've seen that before.

00:43:08.830 --> 00:43:12.760
And the work in this case
is the area inside here.

00:43:12.760 --> 00:43:17.830
The work in this step is just
the area under this curve.

00:43:17.830 --> 00:43:21.410
So in some way we're going to
have a difference here between

00:43:21.410 --> 00:43:27.580
the irreversible case and
the reversible one.

00:43:27.580 --> 00:43:53.060
So if we do that, then what I
want to do is just see what

00:43:53.060 --> 00:44:07.545
happens to dq over T, so in our
irreversible engine, one

00:44:07.545 --> 00:44:14.580
to two, what we know for sure
is that minus w in the

00:44:14.580 --> 00:44:19.840
irreversible step, that's the
work out, extracted in that

00:44:19.840 --> 00:44:24.070
step, is going to be smaller
than minus w in

00:44:24.070 --> 00:44:28.880
the reversible case.

00:44:28.880 --> 00:44:40.290
So w irreversible is bigger
than w reversible.

00:44:40.290 --> 00:44:47.030
And of course, in either case,
delta u is q plus w, so it's q

00:44:47.030 --> 00:44:52.540
irreversible plus w
irreversible, but of course, u

00:44:52.540 --> 00:44:55.330
being a state function it's
the same in either case.

00:44:55.330 --> 00:45:02.510
So it's the same as q reversible
plus w reversible.

00:45:02.510 --> 00:45:06.230
And we've just seen that this
is bigger than this, but the

00:45:06.230 --> 00:45:07.840
sums are equal.

00:45:07.840 --> 00:45:12.960
So this has to be
less than this.

00:45:12.960 --> 00:45:18.410
q irreversible is less
than q reversible.

00:45:18.410 --> 00:45:20.790
In other words, and irreversible
isothermal

00:45:20.790 --> 00:45:25.060
expansion takes less heat from
the hot reservoir than a

00:45:25.060 --> 00:45:27.130
reversible one does.

00:45:27.130 --> 00:45:29.830
It makes sense, right, because
you know we got less work out

00:45:29.830 --> 00:45:33.190
and delta u is the same right,
so it must be that less heat

00:45:33.190 --> 00:45:35.400
got transferred.

00:45:35.400 --> 00:45:39.480
So the expansion against lower
pressure draws less heat from

00:45:39.480 --> 00:45:41.870
the hot reservoir right.

00:45:41.870 --> 00:45:47.940
So now let's look at the
efficiency of our

00:45:47.940 --> 00:45:50.370
irreversible engine.

00:45:50.370 --> 00:45:55.260
So it's one plus q2.

00:45:55.260 --> 00:45:59.970
Now q2 was in this step, and
we're going to leave that

00:45:59.970 --> 00:46:21.270
reversible, right, but
q1 is irreversible.

00:46:21.270 --> 00:46:28.620
And this has got to be less than
one plus q2 reversible

00:46:28.620 --> 00:46:34.140
over q1 reversible, which is
to say it's less than our

00:46:34.140 --> 00:46:36.890
efficiency in the
reversible case.

00:46:36.890 --> 00:46:37.960
Why?

00:46:37.960 --> 00:46:41.710
This is smaller than this, but
it's a negative number.

00:46:41.710 --> 00:46:46.150
So this negative number has a
bigger magnitude than this

00:46:46.150 --> 00:46:48.160
negative number.

00:46:48.160 --> 00:46:51.120
We're subtracting them from one
and they're less than one,

00:46:51.120 --> 00:46:59.010
so this is bigger than this.

00:46:59.010 --> 00:47:04.760
And all we know about this is
that it's really for some

00:47:04.760 --> 00:47:06.970
irreversible reversible step.

00:47:06.970 --> 00:47:10.880
All we really needed to know
about it is that this is

00:47:10.880 --> 00:47:13.740
smaller right.

00:47:13.740 --> 00:47:19.080
So it's going to be the case for
any irreversible engine.

00:47:19.080 --> 00:47:24.410
And that's the point.

00:47:24.410 --> 00:47:38.060
So it's that the irreversible
efficiency is lower than the

00:47:38.060 --> 00:47:43.090
reversible one.

00:47:43.090 --> 00:47:47.690
But, of course, since we saw
that this is smaller than it

00:47:47.690 --> 00:48:17.420
was in the reversible case,
we can also write that dq

00:48:17.420 --> 00:48:28.900
irreversible over T is less
than dq reversible over T.

00:48:28.900 --> 00:48:33.800
Which is to say if we go
around a cycle for dq

00:48:33.800 --> 00:48:43.910
irreversible over T, that's
less than zero.

00:48:43.910 --> 00:48:50.130
It's only in the reversible case
that dq over T around a

00:48:50.130 --> 00:48:53.080
cycle is equal to zero.

00:48:53.080 --> 00:48:57.980
So to write that in a general
way, it's actually formulated

00:48:57.980 --> 00:49:04.780
by Clausius.

00:49:04.780 --> 00:49:10.160
Going around in a cycle the
integral of dq over T is less

00:49:10.160 --> 00:49:13.570
than or equal to zero.

00:49:13.570 --> 00:49:24.290
Never greater.

00:49:24.290 --> 00:49:29.750
OK, what we'll see shortly is
that this will allow us to see

00:49:29.750 --> 00:49:33.820
that for an isolated system the
entropy never decreases.

00:49:33.820 --> 00:49:36.460
It only can go up.

00:49:36.460 --> 00:49:42.350
And in fact any spontaneous
process will make it go up.

00:49:42.350 --> 00:49:45.470
Only in the case of reversible
processes.

00:49:45.470 --> 00:49:48.710
Of course, then you can see
that this will be zero.

00:49:48.710 --> 00:49:52.520
Anything else, which is to say
any spontaneous process, it'll

00:49:52.520 --> 00:49:54.970
be less than zero.

00:49:54.970 --> 00:50:00.030
We'll see that next time, and
then we'll generalize in a

00:50:00.030 --> 00:50:04.040
broader sense to look at the
direction in which spontaneous

00:50:04.040 --> 00:50:05.740
processes go.