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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

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00:00:32,720 --> 00:00:37,670
writing out some somewhat
verbose written descriptions

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00:00:37,670 --> 00:00:42,120
that had been formulated that
indicate certain limitations

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00:00:42,120 --> 00:00:46,230
about things like the efficiency
of a heat engine,

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00:00:46,230 --> 00:00:48,730
and what can be done reversibly
and so forth.

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00:00:48,730 --> 00:00:54,970
And these verbal descriptions
lead to some pictures that I

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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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00:00:58,900 --> 00:01:02,920
accomplish something like run an
engine or move heat from a

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00:01:02,920 --> 00:01:05,960
colder to a warmer body.

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00:01:05,960 --> 00:01:07,540
And what some of the
limitations are

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00:01:07,540 --> 00:01:09,370
on things like that.

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00:01:09,370 --> 00:01:12,550
I'll show those again, but what
I want to do mostly today

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00:01:12,550 --> 00:01:17,070
is try to put a mathematical
statement of the second law in

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place that corresponds to
the verbal statements

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00:01:20,250 --> 00:01:22,270
that we saw last time.

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00:01:22,270 --> 00:01:31,340
So, just to review what we saw
before, we looked at a heat

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engine where there is some hot
reservoir at some temperature,

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00:01:37,480 --> 00:01:51,640
T1, and it's connected to an
engine running in a cycle.

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00:01:51,640 --> 00:01:57,580
We could, write the work that we
generate that comes out as

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00:01:57,580 --> 00:02:14,560
negative w.

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00:02:14,560 --> 00:02:22,870
And then there's a cold
reservoir at some lower

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00:02:22,870 --> 00:02:25,610
temperature T2.

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00:02:25,610 --> 00:02:32,830
So the way I've got things
written here, q1 is positive.

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00:02:32,830 --> 00:02:36,140
Heat's flowing from the hot
reservoir to the engine that

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00:02:36,140 --> 00:02:38,160
runs in a cycle.

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00:02:38,160 --> 00:02:39,340
Work is coming out.

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00:02:39,340 --> 00:02:42,350
A positive amount of work is
being generated and is coming

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00:02:42,350 --> 00:02:50,980
out, so minus w is positive, and
this minus q2 is positive,

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00:02:50,980 --> 00:03:01,360
that is heat is also flowing
into a cold reservoir.

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00:03:01,360 --> 00:03:06,020
And what we saw last time is
that this was necessary to

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00:03:06,020 --> 00:03:08,860
make things work, this part
of it in particular.

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00:03:08,860 --> 00:03:11,260
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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00:03:15,140 --> 00:03:18,860
reservoir, without also
converting some of the heat

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00:03:18,860 --> 00:03:21,130
that came in to heat that would

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00:03:21,130 --> 00:03:22,580
flow into a cold reservoir.

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00:03:22,580 --> 00:03:25,070
All the heat couldn't
get successfully

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00:03:25,070 --> 00:03:27,690
converted into work.

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00:03:27,690 --> 00:03:39,240
That would be desirable,
but it's not possible.

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00:03:39,240 --> 00:03:43,990
And, similarly, if we try to run
things backward and build

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00:03:43,990 --> 00:03:55,510
a refrigerator, so now we have
our cold reservoir, and we're

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00:03:55,510 --> 00:04:03,870
going to remove heat from
it, and pump it

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00:04:03,870 --> 00:04:07,660
up into a hot reservoir.

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00:04:18,260 --> 00:04:22,070
And to do this, we saw that
you have to put work in,

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00:04:22,070 --> 00:04:26,130
right, you can't just remove
heat from a cold reservoir,

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00:04:26,130 --> 00:04:29,840
move it up to a hot reservoir,
without doing some work to

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00:04:29,840 --> 00:04:34,690
accomplish that.

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00:04:34,690 --> 00:04:37,750
So here, q2 is greater
than zero.

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00:04:37,750 --> 00:04:46,390
The work is greater than zero,
and minus q1 is greater than

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00:04:46,390 --> 00:04:51,790
zero, that is heat is
flowing this way.

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00:04:51,790 --> 00:04:58,220
So these are just the pictures
that we saw last time.

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00:04:58,220 --> 00:05:01,320
And then we had some statements
of the second law

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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

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00:05:14,120 --> 00:05:17,650
acts like a refrigerator,
without at some, at the same

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00:05:17,650 --> 00:05:19,890
time, converting some
work into heat.

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00:05:19,890 --> 00:05:22,040
Work has to come in to
make that happen.

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00:05:22,040 --> 00:05:25,550
And we had the similar statement
by Kelvin about the

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00:05:25,550 --> 00:05:29,060
heat engine that required that
some heat gets dumped into a

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00:05:29,060 --> 00:05:31,770
cold reservoir in the process
of converting the heat from

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00:05:31,770 --> 00:05:34,740
the hot reservoir into work.

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00:05:34,740 --> 00:05:36,800
Fine.

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00:05:36,800 --> 00:05:41,540
Now what I want to do is put up
a specific example of the

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cycle that can be undertaken
inside here in an engine, and

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00:05:48,440 --> 00:05:50,590
we can just calculate from what
you've already seen of

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00:05:50,590 --> 00:05:51,690
thermodynamics.

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What happens to the
thermodynamics parameters, and

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see the results in terms of the

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00:05:56,580 --> 00:05:57,860
parameters including entropy.

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00:05:57,860 --> 00:05:59,390
So let's try that.

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00:05:59,390 --> 00:06:07,320
So, the engine that I'm going
to illustrate is called a

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00:06:07,320 --> 00:06:12,100
Carnot engine.

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00:06:12,100 --> 00:06:15,900
And the cycle it's going to
undertake is called a Carnot

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00:06:15,900 --> 00:06:24,670
cycle, and it works the
following way: we're going to

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00:06:24,670 --> 00:06:28,040
do pressure volume work.

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00:06:28,040 --> 00:06:30,320
So this is something that,
by now, you're

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00:06:30,320 --> 00:06:34,460
pretty familiar with.

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00:06:34,460 --> 00:06:39,110
So we're going to start at one,
go to two and this is

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00:06:39,110 --> 00:06:46,640
going to be in isotherm at
temperature T1, and all the

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00:06:46,640 --> 00:06:58,590
paths here are going
to be reversible.

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00:06:58,590 --> 00:07:00,370
So that's our first step.

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00:07:00,370 --> 00:07:02,760
Then we're going to have
an adiabatic expansion.

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00:07:02,760 --> 00:07:05,340
So this is an isothermal
expansion.

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00:07:05,340 --> 00:07:17,320
Here comes an adiabatic
expansion, to point three.

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00:07:17,320 --> 00:07:22,170
So at this point the temperature
will change.

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00:07:22,170 --> 00:07:24,870
Then we're going to have another
isothermal step, a

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00:07:24,870 --> 00:07:37,520
compression to some
point four.

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00:07:37,520 --> 00:07:42,790
So this is an isotherm at some
different temperature T2, a

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00:07:42,790 --> 00:07:45,260
cooler temperature, because
this was an expansion.

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00:07:45,260 --> 00:07:47,390
We know in an adiabatic
expansion the

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00:07:47,390 --> 00:07:49,910
system's going to cool.

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00:07:49,910 --> 00:07:53,790
And now we're going to have
another adiabatic step, an

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00:07:53,790 --> 00:08:01,560
adiabatic compression.

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00:08:01,560 --> 00:08:02,140
And that's it.

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00:08:02,140 --> 00:08:04,060
That's going to take us back
to our starting point.

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00:08:04,060 --> 00:08:06,620
So that's our cycle.

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00:08:06,620 --> 00:08:09,360
Of course there are lots of ways
we can execute the cycle,

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00:08:09,360 --> 00:08:12,210
but this is a simple one, and
these are steps that we're all

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00:08:12,210 --> 00:08:14,650
familiar with at this point.

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00:08:14,650 --> 00:08:18,680
So this is the picture, and
this is a cycle that

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00:08:18,680 --> 00:08:19,440
undertakes it.

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00:08:19,440 --> 00:08:22,580
So let's just look step by
step at what happens.

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00:08:22,580 --> 00:08:28,490
So going from one to two, it's
an isothermal expansion a T1,

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00:08:28,490 --> 00:08:41,670
so delta u is q1, we'll
call it, plus w1,

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00:08:41,670 --> 00:08:43,690
for the first step.

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00:08:43,690 --> 00:08:51,550
Going from two to three, that's
an adiabatic expansion,

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00:08:51,550 --> 00:08:55,510
so q is equal to zero
in that step.

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00:08:55,510 --> 00:09:02,540
And delta u is equal to what
we'll call w1 prime.

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00:09:02,540 --> 00:09:07,190
So we'll use w1 and w1 prime to
describe the work involved.

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00:09:07,190 --> 00:09:15,590
The work input during
the expansion steps.

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00:09:15,590 --> 00:09:23,365
Step three to four isothermal
compression, delta u is going

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00:09:23,365 --> 00:09:27,320
to be q2 plus w2.

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00:09:27,320 --> 00:09:29,710
And finally, we're going
to go back to one.

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00:09:29,710 --> 00:09:32,950
We're going to close the cycle
with another adiabatic step.

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00:09:32,950 --> 00:09:34,860
q is zero.

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00:09:34,860 --> 00:09:46,590
Delta u is w2 prime.

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00:09:46,590 --> 00:09:51,140
Now, the total amount of the
work that we can get out is

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00:09:51,140 --> 00:10:03,970
just given by the area
inside this curve.

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00:10:03,970 --> 00:10:12,220
We'll call it capital W. It's
just minus the sum of all

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00:10:12,220 --> 00:10:14,880
these things, because of course
these are just defined

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00:10:14,880 --> 00:10:18,330
as the work done by the
environment to the system.

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00:10:18,330 --> 00:10:28,040
So it's minus w1 plus w1 prime
plus w2 plus w2 prime.

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00:10:28,040 --> 00:10:42,390
The heat input is just q1, and
we'll define that as capital

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00:10:42,390 --> 00:10:46,250
Q. And I just want to write
those because what I really

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00:10:46,250 --> 00:10:49,020
want to get at is what's the
efficiency of the whole thing?

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00:10:49,020 --> 00:10:52,280
And in practical terms, we can
define the efficiency as the

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00:10:52,280 --> 00:10:55,490
ratio of the heat in
to the work out.

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00:10:55,490 --> 00:11:17,450
We would like that to be
as high as possible.

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00:11:17,450 --> 00:11:25,300
So it's just capital W over
capital Q, which is to say

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00:11:25,300 --> 00:11:34,680
it's minus all that
stuff over q1.

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00:11:39,160 --> 00:11:42,580
Now the first law is going to
hold in all of these steps,

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00:11:42,580 --> 00:11:45,130
and we're going around
in a cycle.

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00:11:45,130 --> 00:11:48,560
So what does that tell us about
a state function like u?

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00:11:48,560 --> 00:11:53,310
What's delta u going around
the whole thing?

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00:11:53,310 --> 00:11:54,990
I want a chorus in
the answer here.

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00:11:54,990 --> 00:11:56,020
What's delta u?

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00:11:56,020 --> 00:11:56,980
STUDENT: Zero.

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00:11:56,980 --> 00:11:58,570
PROFESSOR NELSON: Excellent.

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00:11:58,570 --> 00:12:01,390
MIT students, yes!

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00:12:01,390 --> 00:12:08,550
OK, so the first law, we know
that the, going around the

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00:12:08,550 --> 00:12:19,750
cycle the integral of
delta u is zero.

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00:12:19,750 --> 00:12:23,510
And that has to equal
q plus w, summed

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00:12:23,510 --> 00:12:27,520
up for all the steps.

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00:12:27,520 --> 00:12:36,000
Which is to say q1 plus q2 is
equal to minus w1 plus w1

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00:12:36,000 --> 00:12:48,200
prime plus w2 plus w2 prime.

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00:12:48,200 --> 00:12:50,560
So that means we can rewrite
this efficiency.

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00:12:50,560 --> 00:12:56,000
We can replace this sum by just
the some of q1 plus q2.

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00:12:56,000 --> 00:13:03,400
So sufficiency is q1
plus q2 over q1.

159
00:13:06,760 --> 00:13:11,200
Or we can write it as
one plus q2 over q1.

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00:13:18,780 --> 00:13:24,480
Now that already tells us what
we know, which is that the

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00:13:24,480 --> 00:13:26,170
efficiency is going
to be something

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00:13:26,170 --> 00:13:28,000
less than zero, right?

163
00:13:28,000 --> 00:13:32,410
Because we've got one and then
we're adding q2 over q1 to it,

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00:13:32,410 --> 00:13:37,120
but we've got negative q2
is a positive number.

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00:13:37,120 --> 00:13:39,110
Heat is flowing this way.

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00:13:39,110 --> 00:13:43,880
So q2, the heat in this way
is negative. q2 over q1 is

167
00:13:43,880 --> 00:13:45,500
negative. q2 of course
is positive.

168
00:13:45,500 --> 00:13:50,220
That is heat flowing from the
hot reservoir to the engine.

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00:13:50,220 --> 00:13:53,230
So that efficiency is something
less than one, and

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00:13:53,230 --> 00:13:58,380
we'd like to figure
out what that is.

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00:13:58,380 --> 00:14:05,370
So let's just state that.

172
00:14:05,370 --> 00:14:08,990
Well, now let's go back to each
of our individual steps

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00:14:08,990 --> 00:14:14,350
and look, based on what we know
about how to evaluate the

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00:14:14,350 --> 00:14:17,800
thermodynamic changes that take
place here, let's look at

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00:14:17,800 --> 00:14:20,420
each one of the steps and
see what happens.

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00:14:20,420 --> 00:14:36,730
So from one to two
it's isothermal.

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00:14:36,730 --> 00:14:38,720
And now we're going to specify,
we're going to do a

178
00:14:38,720 --> 00:14:42,340
Carnot cycle for an ideal gas.

179
00:14:42,340 --> 00:14:43,650
It's an ideal gas.

180
00:14:43,650 --> 00:14:45,710
It's an isothermal change.

181
00:14:45,710 --> 00:14:49,870
What's delta u?

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00:14:49,870 --> 00:14:53,040
You know, it's like lots of
courses and all sorts of

183
00:14:53,040 --> 00:14:57,240
things that you're seeing,
they're always the same.

184
00:14:57,240 --> 00:14:58,580
What delta u?

185
00:14:58,580 --> 00:15:00,000
STUDENT: Zero.

186
00:15:00,000 --> 00:15:04,330
PROFESSOR NELSON?

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00:15:04,330 --> 00:15:04,540
Right.

188
00:15:04,540 --> 00:15:08,100
Delta u is zero, and it's
also equal to this.

189
00:15:08,100 --> 00:15:12,500
So that says the q1 is
the opposite of w1.

190
00:15:15,290 --> 00:15:24,430
It's an isothermal expansion,
so dw is just negative p dV.

191
00:15:24,430 --> 00:15:27,640
So it's, this is just
the integral from

192
00:15:27,640 --> 00:15:32,300
one to two of p dV.

193
00:15:32,300 --> 00:15:36,980
And it's an ideal gas,
isothermal, right.

194
00:15:36,980 --> 00:15:42,135
RT over V. Were going to make it
for a mole of gas, so it's

195
00:15:42,135 --> 00:15:48,650
R times T1, and then we'll have
dV over V. So that's just

196
00:15:48,650 --> 00:15:51,200
the log of V2 over V1.

197
00:15:51,200 --> 00:16:01,750
Let's have one mole,
n equals one.

198
00:16:01,750 --> 00:16:08,700
Okay two going to three that's
this, it's adiabatic, right.

199
00:16:08,700 --> 00:16:15,740
So delta u is just equal to the
work but we also know what

200
00:16:15,740 --> 00:16:17,100
happens because the
temperature is

201
00:16:17,100 --> 00:16:20,850
changing from T1 to T2.

202
00:16:20,850 --> 00:16:29,830
So delta you is just Cv
times T2 minus T1.

203
00:16:29,830 --> 00:16:31,350
du is Cv dT.

204
00:16:31,350 --> 00:16:44,250
It's an ideal gas, and that's
equal to w1 prime.

205
00:16:44,250 --> 00:16:50,450
Now, this is a reversible
adiabatic path, so there's a

206
00:16:50,450 --> 00:16:58,660
relationship that I'm sure
you'll remember.

207
00:16:58,660 --> 00:17:06,480
Namely, T2 over T1 is
equal to V2 over V3.

208
00:17:06,480 --> 00:17:08,930
Don't be confused by
the subscripts.

209
00:17:08,930 --> 00:17:12,750
We're talking about quantities
both starting at two

210
00:17:12,750 --> 00:17:15,210
and going to three.

211
00:17:15,210 --> 00:17:16,530
So it's V2 and V3.

212
00:17:16,530 --> 00:17:17,770
They're different volumes.

213
00:17:17,770 --> 00:17:20,190
The temperature though at two is
the same as it was at one,

214
00:17:20,190 --> 00:17:22,480
so it's still a temp
at T1, and this

215
00:17:22,480 --> 00:17:25,370
temperatures is at T2.

216
00:17:25,370 --> 00:17:31,570
Anyway T2 over T1 is equal
to V2 over V3 to the

217
00:17:31,570 --> 00:17:35,550
power gamma minus one.

218
00:17:35,550 --> 00:17:41,740
So we're going to kind of store
that for the moment.

219
00:17:41,740 --> 00:17:45,740
Now going from three to four
right, so we have another

220
00:17:45,740 --> 00:17:50,830
isothermal process for an ideal
gas, so I won't try to

221
00:17:50,830 --> 00:17:53,510
make you sing again so soon.

222
00:17:53,510 --> 00:17:56,040
Delta u is zero.

223
00:17:56,040 --> 00:18:02,750
And so just like here, now q2
is minus w2, that's integral

224
00:18:02,750 --> 00:18:06,870
going from three to four p dV.

225
00:18:10,220 --> 00:18:15,360
So it's R T2, right, now we're
at a lower temperature times

226
00:18:15,360 --> 00:18:23,230
the log of V4 over V3.

227
00:18:23,230 --> 00:18:36,740
So that's our work in
this path and heat.

228
00:18:36,740 --> 00:18:42,630
And finally going from
four back to one.

229
00:18:42,630 --> 00:18:47,860
So we've already seen
that q is zero.

230
00:18:47,860 --> 00:18:56,050
So we know that delta u is just
Cv times T1 minus T2.

231
00:18:56,050 --> 00:18:57,980
Now we're going from
T2 up to T1.

232
00:19:00,670 --> 00:19:08,600
And this is equal to w2 prime.

233
00:19:08,600 --> 00:19:11,870
And now, just like we had
before, again we've got a

234
00:19:11,870 --> 00:19:29,430
reversible adiabatic path, so
T1 over T2 has to equal V4

235
00:19:29,430 --> 00:19:38,950
over V1, to the gamma
minus one, okay.

236
00:19:38,950 --> 00:19:44,880
All right, now, if this is the
case, of course, this is just

237
00:19:44,880 --> 00:19:48,030
the inverse of this.

238
00:19:48,030 --> 00:19:50,520
So this just must be the
inverse of this.

239
00:19:50,520 --> 00:20:01,740
We can combine these two to see
that V4 over V1 must equal

240
00:20:01,740 --> 00:20:05,630
V3 over V2.

241
00:20:05,630 --> 00:20:08,890
So now we have a relationship
between the ratios of these

242
00:20:08,890 --> 00:20:21,510
volumes that are reached during
these adiabatic paths.

243
00:20:21,510 --> 00:20:41,300
Now, let's just look
at our efficiency.

244
00:20:41,300 --> 00:20:48,070
So we saw that efficiency
is one plus q2 over q1.

245
00:20:48,070 --> 00:20:52,430
Let's just use our expressions
that we've

246
00:20:52,430 --> 00:20:56,280
found for q2 and q1.

247
00:20:56,280 --> 00:20:58,130
We're going to have
q2 over q1.

248
00:20:58,130 --> 00:20:58,790
R is going to cancel.

249
00:20:58,790 --> 00:21:01,400
We're going to have the ratio of
temperatures and the ratio

250
00:21:01,400 --> 00:21:11,470
of these logs.

251
00:21:11,470 --> 00:21:28,120
So it's T2 over T1 times the
log of V4 over V3, over the

252
00:21:28,120 --> 00:21:41,030
log of V2 over V1, but we just
arrived at this relationship

253
00:21:41,030 --> 00:21:42,280
between these volumes.

254
00:21:42,280 --> 00:21:52,990
And this of course tells us that
V4 over V3 is V1 over V2.

255
00:21:52,990 --> 00:21:53,820
Right.

256
00:21:53,820 --> 00:21:56,800
I'm just inverting these.

257
00:21:56,800 --> 00:21:58,820
So here it is.

258
00:21:58,820 --> 00:22:03,960
Here's the V4 over
V3, oops, sorry.

259
00:22:03,960 --> 00:22:11,250
V2 over V1, this is equal
to the inverse of this.

260
00:22:11,250 --> 00:22:14,670
So the ratio of the logs
is just minus one.

261
00:22:14,670 --> 00:22:19,500
So this is just, sorry, I forgot
about the one here.

262
00:22:19,500 --> 00:22:30,210
One plus this, but this is the
same as one minus T2 over T1.

263
00:22:30,210 --> 00:22:34,820
That's our expression
for the efficiency.

264
00:22:34,820 --> 00:22:36,740
All right, isn't
that terrific?

265
00:22:36,740 --> 00:22:39,430
Now, we have an expression.

266
00:22:39,430 --> 00:22:42,300
We can figure out
the efficiency.

267
00:22:42,300 --> 00:22:44,540
All we need to know is
the temperatures.

268
00:22:44,540 --> 00:22:46,590
What's the temperature
of the hot reservoir?

269
00:22:46,590 --> 00:22:48,380
What's the temperature of
the cold reservoir?

270
00:22:48,380 --> 00:22:51,290
We're done.

271
00:22:51,290 --> 00:22:53,910
That's the efficiency.

272
00:22:53,910 --> 00:23:03,050
As we expected, it's less than
one, and of course if we build

273
00:23:03,050 --> 00:23:05,650
an engine, we want it to be as
high as possible, as close to

274
00:23:05,650 --> 00:23:06,730
one as possible.

275
00:23:06,730 --> 00:23:11,830
What that means is we want to
run the hot reservoir as hot

276
00:23:11,830 --> 00:23:16,620
as possible and the cold
reservoir as cold as possible.

277
00:23:16,620 --> 00:23:22,810
In principle, this value, this
efficiency, can approach 1 as

278
00:23:22,810 --> 00:23:26,250
the low temperature approaches
absolute zero.

279
00:23:26,250 --> 00:23:32,590
If this were to be an absolute
zero Kelvin, then we could we

280
00:23:32,590 --> 00:23:34,190
can have something,
wait a minute.

281
00:23:34,190 --> 00:23:36,100
Sorry, it's T2.

282
00:23:36,100 --> 00:23:41,000
As T2 goes to zero, the cold
reservoir, then this goes to

283
00:23:41,000 --> 00:23:44,970
zero and our efficiency
approaches one.

284
00:23:44,970 --> 00:23:47,180
So that would be the
best we can do.

285
00:23:47,180 --> 00:23:49,600
And that basically make
sense, right?

286
00:23:49,600 --> 00:23:54,090
The whole thing is being powered
by sending heat from

287
00:23:54,090 --> 00:23:57,180
the hot to the cold reservoir.

288
00:23:57,180 --> 00:24:00,380
The colder that cold reservoir
is, the hotter that hot

289
00:24:00,380 --> 00:24:01,970
reservoir is, the better
off we are.

290
00:24:01,970 --> 00:24:04,080
The more work we can
get out of it.

291
00:24:04,080 --> 00:24:10,120
The closer the efficiency
will get to one.

292
00:24:10,120 --> 00:24:15,550
So in some sense, the first law
would suggest you can sort

293
00:24:15,550 --> 00:24:17,430
of break even.

294
00:24:17,430 --> 00:24:20,180
That is, it gives us the
relationship between energy

295
00:24:20,180 --> 00:24:21,700
and work and heat.

296
00:24:21,700 --> 00:24:23,450
It might suggest that
you could convert

297
00:24:23,450 --> 00:24:25,270
all the work to heat.

298
00:24:25,270 --> 00:24:30,480
The second law says that really,
you can't do that.

299
00:24:30,480 --> 00:24:34,200
The only way you could possibly
contemplate it is to

300
00:24:34,200 --> 00:24:38,580
be working at absolute
zero Kelvin.

301
00:24:38,580 --> 00:24:42,510
Guess what the third law
is going to tell us?

302
00:24:42,510 --> 00:24:45,000
You can't get to zero Kelvin.

303
00:24:45,000 --> 00:24:48,140
We'll see that shortly.

304
00:24:48,140 --> 00:24:51,500
But this is those closest you
could come at least by trying

305
00:24:51,500 --> 00:25:28,820
to do that to having your
efficiency approach one.

306
00:25:28,820 --> 00:25:36,920
Now, we've seen that q2
over q1 is equal to

307
00:25:36,920 --> 00:25:40,880
negative T2 over T1.

308
00:25:40,880 --> 00:25:48,070
So let me just rewrite
that as q1 over T1 is

309
00:25:48,070 --> 00:25:54,190
minus q2 over T2.

310
00:25:54,190 --> 00:26:05,020
Or in other words, q1 over T1
plus q2 over T2 is zero.

311
00:26:05,020 --> 00:26:10,560
But q1 and q2, each one of those
is just the integrated

312
00:26:10,560 --> 00:26:14,390
amount of heat that was
transferred going along a

313
00:26:14,390 --> 00:26:18,550
reversible constant
temperature path.

314
00:26:18,550 --> 00:26:26,420
Which means that q1 over T1,
that's this delta S thing that

315
00:26:26,420 --> 00:26:28,690
we saw before.

316
00:26:28,690 --> 00:26:33,410
And what we can say about this
is it's saying that if we go

317
00:26:33,410 --> 00:26:41,780
around in a cycle and look at
dq reversible over T it's

318
00:26:41,780 --> 00:27:08,290
zero, because that's what
these quantities are.

319
00:27:08,290 --> 00:27:11,660
Now, of course, we can run the
engine backward and build a

320
00:27:11,660 --> 00:27:16,680
refrigerator, and if you've got
your lecture notes from

321
00:27:16,680 --> 00:27:20,100
last period, your 8-9, well,
they're labeled 8-9 lecture

322
00:27:20,100 --> 00:27:24,470
notes, I made an attempt to
define something which was a

323
00:27:24,470 --> 00:27:25,570
little bit misguided.

324
00:27:25,570 --> 00:27:28,380
And so instead of defining
efficiency the way you've got

325
00:27:28,380 --> 00:27:32,210
it written there, I'm going to
define what's called something

326
00:27:32,210 --> 00:27:34,720
different for a refrigerator
which is called the

327
00:27:34,720 --> 00:27:38,130
coefficient of performance.

328
00:27:38,130 --> 00:28:16,110
So it's the following: so we can
define the coefficient of

329
00:28:16,110 --> 00:28:29,300
performance written as eta
as q2 over minus w.

330
00:28:29,300 --> 00:28:34,530
In other words, you know, it's
how much heat can we pull out

331
00:28:34,530 --> 00:28:40,100
of the cold reservoir, for
whatever amount of work that

332
00:28:40,100 --> 00:28:42,680
we're going to put in
in order to do that?

333
00:28:42,680 --> 00:28:47,050
Well of course, we'd like that
to be as big as possible.

334
00:28:47,050 --> 00:28:59,300
But of course, this is just
q2 over minus q1 plus q2.

335
00:28:59,300 --> 00:29:17,090
Just to be clear, so it's heat
extracted over the work in.

336
00:29:17,090 --> 00:29:20,460
So let me just rewrite that as,
I just want to divide by

337
00:29:20,460 --> 00:29:21,990
q1 everywhere.

338
00:29:21,990 --> 00:29:31,430
So I'm going to write this as q2
over q1 over minus one plus

339
00:29:31,430 --> 00:29:35,490
q2 over q1.

340
00:29:35,490 --> 00:29:38,830
I want to do that because we
know that q2 over q1 is

341
00:29:38,830 --> 00:29:40,350
negative T2 over T1.

342
00:29:40,350 --> 00:29:46,360
So this is negative T2 over
T1 over negative one

343
00:29:46,360 --> 00:29:49,790
minus T2 over T1.

344
00:29:49,790 --> 00:29:52,770
I can cancel those, and
then I'm going to

345
00:29:52,770 --> 00:29:56,880
multiply through by T1.

346
00:29:56,880 --> 00:30:06,620
So this is just going to be T2
over T1 minus T2, that's our

347
00:30:06,620 --> 00:30:17,190
coefficient of performance.

348
00:30:17,190 --> 00:30:20,550
So it's a measure of how well
we can do to run the

349
00:30:20,550 --> 00:30:21,490
refrigerator.

350
00:30:21,490 --> 00:30:26,520
So you know, what the second law
is doing, in words, it's

351
00:30:26,520 --> 00:30:29,920
putting these restrictions on
how well or how effectively we

352
00:30:29,920 --> 00:30:33,970
can convert heat into work in
the case of the engine, or

353
00:30:33,970 --> 00:30:39,230
work into heat extracted in the
case of a refrigerator.

354
00:30:39,230 --> 00:30:45,000
And it follows qualitatively
from just your ordinary

355
00:30:45,000 --> 00:30:48,530
observations about the direction
in which things go

356
00:30:48,530 --> 00:30:49,700
spontaneously, right.

357
00:30:49,700 --> 00:30:52,430
You know the heat isn't going to
flow from a cold body to a

358
00:30:52,430 --> 00:30:57,770
hot body without putting some
work in to make that happen.

359
00:30:57,770 --> 00:31:01,810
And so, we'll be able of follow
that further and see

360
00:31:01,810 --> 00:31:07,120
really how to determine the
direction of spontaneity for a

361
00:31:07,120 --> 00:31:09,660
whole set of processes, really
in principle for any

362
00:31:09,660 --> 00:31:14,390
processes, went analyzed
properly.

363
00:31:14,390 --> 00:31:18,090
So the first step to doing
that is I want to just

364
00:31:18,090 --> 00:31:22,560
generalize our results so
far for a Carnot cycle.

365
00:31:22,560 --> 00:31:24,050
You might think well
okay, but this is a

366
00:31:24,050 --> 00:31:25,810
pretty specialized case.

367
00:31:25,810 --> 00:31:30,730
We've formulated one particular
kind of engine, and

368
00:31:30,730 --> 00:31:35,190
seen how we can analyze what
it does, come up with

369
00:31:35,190 --> 00:31:40,040
relations that seem of value
for efficiency and other

370
00:31:40,040 --> 00:31:41,480
quantities.

371
00:31:41,480 --> 00:31:45,180
How general is it?

372
00:31:45,180 --> 00:31:46,640
Well, let's take a look.

373
00:31:46,640 --> 00:31:56,880
So, what I want to do is make a
new engine which really just

374
00:31:56,880 --> 00:32:01,090
consists of two engines,
side by side.

375
00:32:01,090 --> 00:32:05,370
One is our Carnot engine as
we've seen it, and the other

376
00:32:05,370 --> 00:32:10,510
is just any other reversible
engine.

377
00:32:10,510 --> 00:32:23,770
So generalize our Carnot
engine results.

378
00:32:23,770 --> 00:32:31,790
So let's take our hot reservoir
and draw it bigger.

379
00:32:31,790 --> 00:32:35,030
We know it has to big anyway,
since we can extract heat from

380
00:32:35,030 --> 00:32:38,000
it without changing
the temperature.

381
00:32:38,000 --> 00:32:44,190
And on this side, we're going
to write out an engine, and

382
00:32:44,190 --> 00:32:47,740
we're going to say this is a
Carnot engine. so on the side

383
00:32:47,740 --> 00:32:54,220
of it we have q1 prime.

384
00:32:54,220 --> 00:32:57,380
We're going to have w prime.

385
00:32:57,380 --> 00:33:01,330
And we're going to have q2
prime, and we're going to draw

386
00:33:01,330 --> 00:33:06,980
the arrows in the positive
direction in these cases.

387
00:33:06,980 --> 00:33:10,730
Or in the defined directions
I should say.

388
00:33:10,730 --> 00:33:17,180
Here is our colder reservoir.

389
00:33:17,180 --> 00:33:20,580
OK, so here is just an engine
like what we've already seen,

390
00:33:20,580 --> 00:33:27,150
and I'm going to specify that
this is a Carnot engine which

391
00:33:27,150 --> 00:33:29,250
is to say all the results
that we just derived

392
00:33:29,250 --> 00:33:31,360
hold for this case.

393
00:33:31,360 --> 00:33:38,280
And side by side, we're going
to run another engine.

394
00:33:38,280 --> 00:33:45,270
So it's got q2, and it's
got a cycle w.

395
00:33:45,270 --> 00:33:47,450
And it's got q1.

396
00:33:51,920 --> 00:33:59,260
So this is some other
engine that runs

397
00:33:59,260 --> 00:34:04,280
using reversible processes.

398
00:34:04,280 --> 00:34:07,910
So I can define the efficiency
of each one of them.

399
00:34:07,910 --> 00:34:13,930
The efficiency for the one on
the left is minus w over q1,

400
00:34:13,930 --> 00:34:18,210
The efficiency prime for the
one on the right is minus w

401
00:34:18,210 --> 00:34:22,610
prime over q1 prime.

402
00:34:22,610 --> 00:34:30,150
Now I want to just assume that
in some way they're different.

403
00:34:30,150 --> 00:34:42,390
So let's assume that epsilon
prime is greater than epsilon.

404
00:34:42,390 --> 00:34:47,060
So in other words, this engine
is running less efficiently

405
00:34:47,060 --> 00:34:49,550
than my Carnot engine.

406
00:34:49,550 --> 00:34:50,660
It's also reversible.

407
00:34:50,660 --> 00:34:54,890
And now since it's reversible,
we can run

408
00:34:54,890 --> 00:34:58,240
it forward or backward.

409
00:34:58,240 --> 00:35:07,710
So we can run this one backward
and we can use the

410
00:35:07,710 --> 00:35:14,250
work that comes out as the
input, well sorry, use this

411
00:35:14,250 --> 00:35:17,710
work that comes out of the one
of the right to run this one

412
00:35:17,710 --> 00:35:20,330
backward, which is to
say we'll move heat

413
00:35:20,330 --> 00:35:23,540
from cold to hot.

414
00:35:23,540 --> 00:35:33,290
So use work out of right-hand
side to

415
00:35:33,290 --> 00:35:40,910
run left-hand backward.

416
00:35:40,910 --> 00:35:42,660
Why not?

417
00:35:42,660 --> 00:35:44,600
We need work to come
in, we might as

418
00:35:44,600 --> 00:36:00,870
well get it from here.

419
00:36:00,870 --> 00:36:10,980
So, the total work that we can
get out must be zero, out of

420
00:36:10,980 --> 00:36:12,630
the whole sum of them.

421
00:36:12,630 --> 00:36:15,120
After all, we're taking the
output work that we get from

422
00:36:15,120 --> 00:36:21,800
the right, and using it
all to drive the left.

423
00:36:21,800 --> 00:36:28,420
So that means that minus
w prime must equal w.

424
00:36:28,420 --> 00:36:33,300
And w is greater than zero. that
is in this one we have

425
00:36:33,300 --> 00:36:38,300
the environment doing
work on the system.

426
00:36:38,300 --> 00:36:44,400
OK, now we've assumed that
epsilon prime is

427
00:36:44,400 --> 00:36:46,210
greater than epsilon.

428
00:36:46,210 --> 00:36:51,670
So we can just write out what
those are, minus w prime over

429
00:36:51,670 --> 00:36:57,720
q1 prime is greater than
minus w over q1.

430
00:36:57,720 --> 00:37:04,300
But we know that minus w prime
is the same thing as w.

431
00:37:04,300 --> 00:37:11,250
So w over q1 prime is greater
than minus w over q1.

432
00:37:15,450 --> 00:37:19,410
Which is the same thing just
to be a little bit maybe

433
00:37:19,410 --> 00:37:25,420
pedantic because w over minus
q1, and the only reason I'm

434
00:37:25,420 --> 00:37:29,090
writing that is to illustrate
that this says that q1 must be

435
00:37:29,090 --> 00:37:34,040
less than q1 prime.

436
00:37:34,040 --> 00:37:42,710
So q1 is less than zero.

437
00:37:42,710 --> 00:37:45,810
Remember this one's running
backward, we're pumping heat

438
00:37:45,810 --> 00:37:52,130
up. q1 prime is greater
than zero, it's

439
00:37:52,130 --> 00:37:53,330
running as an engine.

440
00:37:53,330 --> 00:37:57,930
It's taking heat from
the hot reservoir.

441
00:37:57,930 --> 00:38:01,660
Well, that's interesting.

442
00:38:01,660 --> 00:38:11,090
That says that minus q1 prime
plus q1 is greater than zero.

443
00:38:11,090 --> 00:38:16,910
Well, it's a pretty interesting
result.

444
00:38:16,910 --> 00:38:20,950
There's no work being done on,
out of the whole thing.

445
00:38:20,950 --> 00:38:24,340
This is providing work that's
being used in here, but if you

446
00:38:24,340 --> 00:38:27,830
take the whole outside of the
surroundings and this whole

447
00:38:27,830 --> 00:38:32,450
thing is the system, no net
work, these things cancel each

448
00:38:32,450 --> 00:38:39,740
other, and yet heat's
going up.

449
00:38:39,740 --> 00:38:43,630
How did that happen?

450
00:38:43,630 --> 00:38:47,620
Well it happened because we
clearly must have a faulty

451
00:38:47,620 --> 00:38:50,310
assumption underlying what
we've just done.

452
00:38:50,310 --> 00:39:02,480
This can't possibly be true.

453
00:39:02,480 --> 00:39:08,900
This is impossible.

454
00:39:08,900 --> 00:39:12,700
So what this says is the
efficiency of any reversible

455
00:39:12,700 --> 00:39:16,190
engine has to be one
minus T2 over T1.

456
00:39:29,380 --> 00:39:32,830
It's not a result that's
specific to the one

457
00:39:32,830 --> 00:39:39,450
cycle that we put up.

458
00:39:39,450 --> 00:39:42,490
And you know, you could have a
reversible engine with lots

459
00:39:42,490 --> 00:39:47,390
and lots of steps, but you could
always break them down

460
00:39:47,390 --> 00:39:49,930
into some sequence of adiabatic

461
00:39:49,930 --> 00:39:51,400
and isothermal steps.

462
00:39:51,400 --> 00:39:53,860
So you know your cycle, you
know, you could have a whole

463
00:39:53,860 --> 00:40:04,230
complicated sequence on a p v
diagram of steps going back.

464
00:40:04,230 --> 00:40:07,740
As long as it's reversible, you
know what the efficiency

465
00:40:07,740 --> 00:40:11,550
has to be, and in principle, you
could break it down into a

466
00:40:11,550 --> 00:40:13,660
bunch of steps that you
could formulate as

467
00:40:13,660 --> 00:40:14,890
isothermal and adiabatic.

468
00:40:14,890 --> 00:40:17,170
They might have to be formulated
as very small

469
00:40:17,170 --> 00:40:24,560
steps, in order to do
that, but you could.

470
00:40:24,560 --> 00:40:39,270
What this says too, is this
result that we found, right,

471
00:40:39,270 --> 00:40:45,740
now of course that's our
integral dS is zero.

472
00:40:45,740 --> 00:40:55,660
It's general.

473
00:40:55,660 --> 00:41:11,330
Entropy S is a state function,
generally.

474
00:41:11,330 --> 00:41:13,710
Now remember, we went through
before how it's a state

475
00:41:13,710 --> 00:41:16,210
function but to calculate
it, you'd need to find a

476
00:41:16,210 --> 00:41:22,420
reversible path, along which
you can figure this out.

477
00:41:22,420 --> 00:41:31,970
But the fact is it's a state
function, in a general way.

478
00:41:31,970 --> 00:41:43,390
Now, if we go back to our Carnot
cycle which is a set of

479
00:41:43,390 --> 00:41:47,760
reversible paths, it's useful
to compare this to what

480
00:41:47,760 --> 00:41:51,080
happens in an irreversible
case.

481
00:41:51,080 --> 00:41:54,820
In a real engine, of course,
you can approach the

482
00:41:54,820 --> 00:41:59,030
reversible limit.

483
00:41:59,030 --> 00:42:02,790
Every step won't be perfectly
reversible.

484
00:42:02,790 --> 00:42:07,300
And of course it's not hard
to see what happens.

485
00:42:07,300 --> 00:42:11,310
Let's just take our reversible
engine and

486
00:42:11,310 --> 00:42:14,310
modify it a little bit.

487
00:42:14,310 --> 00:42:17,490
Let's imagine that instead
of all the steps being

488
00:42:17,490 --> 00:42:23,260
reversible, let's just put
in one irreversible step.

489
00:42:23,260 --> 00:42:24,720
Let's do this.

490
00:42:24,720 --> 00:42:29,540
Instead of this reversible
isothermal step, Let's make it

491
00:42:29,540 --> 00:42:35,370
an irreversible isothermal
step.

492
00:42:35,370 --> 00:42:41,960
We can have a different
isothermal step.

493
00:42:41,960 --> 00:42:46,440
Of course in the reversible
case, you're always pushing

494
00:42:46,440 --> 00:42:50,430
against an external pressure,
which is essentially equal to

495
00:42:50,430 --> 00:42:53,330
the internal pressure.

496
00:42:53,330 --> 00:42:54,950
Let's not do that.

497
00:42:54,950 --> 00:42:59,550
Let's imagine maybe instead we
just immediately dropped the

498
00:42:59,550 --> 00:43:01,910
pressure and let the
system expand

499
00:43:01,910 --> 00:43:04,100
against the lower pressure.

500
00:43:04,100 --> 00:43:06,080
Now we know we're going
to get less work out

501
00:43:06,080 --> 00:43:06,820
of it in that case.

502
00:43:06,820 --> 00:43:08,830
You've seen that before.

503
00:43:08,830 --> 00:43:12,760
And the work in this case
is the area inside here.

504
00:43:12,760 --> 00:43:17,830
The work in this step is just
the area under this curve.

505
00:43:17,830 --> 00:43:21,410
So in some way we're going to
have a difference here between

506
00:43:21,410 --> 00:43:27,580
the irreversible case and
the reversible one.

507
00:43:27,580 --> 00:43:53,060
So if we do that, then what I
want to do is just see what

508
00:43:53,060 --> 00:44:07,545
happens to dq over T, so in our
irreversible engine, one

509
00:44:07,545 --> 00:44:14,580
to two, what we know for sure
is that minus w in the

510
00:44:14,580 --> 00:44:19,840
irreversible step, that's the
work out, extracted in that

511
00:44:19,840 --> 00:44:24,070
step, is going to be smaller
than minus w in

512
00:44:24,070 --> 00:44:28,880
the reversible case.

513
00:44:28,880 --> 00:44:40,290
So w irreversible is bigger
than w reversible.

514
00:44:40,290 --> 00:44:47,030
And of course, in either case,
delta u is q plus w, so it's q

515
00:44:47,030 --> 00:44:52,540
irreversible plus w
irreversible, but of course, u

516
00:44:52,540 --> 00:44:55,330
being a state function it's
the same in either case.

517
00:44:55,330 --> 00:45:02,510
So it's the same as q reversible
plus w reversible.

518
00:45:02,510 --> 00:45:06,230
And we've just seen that this
is bigger than this, but the

519
00:45:06,230 --> 00:45:07,840
sums are equal.

520
00:45:07,840 --> 00:45:12,960
So this has to be
less than this.

521
00:45:12,960 --> 00:45:18,410
q irreversible is less
than q reversible.

522
00:45:18,410 --> 00:45:20,790
In other words, and irreversible
isothermal

523
00:45:20,790 --> 00:45:25,060
expansion takes less heat from
the hot reservoir than a

524
00:45:25,060 --> 00:45:27,130
reversible one does.

525
00:45:27,130 --> 00:45:29,830
It makes sense, right, because
you know we got less work out

526
00:45:29,830 --> 00:45:33,190
and delta u is the same right,
so it must be that less heat

527
00:45:33,190 --> 00:45:35,400
got transferred.

528
00:45:35,400 --> 00:45:39,480
So the expansion against lower
pressure draws less heat from

529
00:45:39,480 --> 00:45:41,870
the hot reservoir right.

530
00:45:41,870 --> 00:45:47,940
So now let's look at the
efficiency of our

531
00:45:47,940 --> 00:45:50,370
irreversible engine.

532
00:45:50,370 --> 00:45:55,260
So it's one plus q2.

533
00:45:55,260 --> 00:45:59,970
Now q2 was in this step, and
we're going to leave that

534
00:45:59,970 --> 00:46:21,270
reversible, right, but
q1 is irreversible.

535
00:46:21,270 --> 00:46:28,620
And this has got to be less than
one plus q2 reversible

536
00:46:28,620 --> 00:46:34,140
over q1 reversible, which is
to say it's less than our

537
00:46:34,140 --> 00:46:36,890
efficiency in the
reversible case.

538
00:46:36,890 --> 00:46:37,960
Why?

539
00:46:37,960 --> 00:46:41,710
This is smaller than this, but
it's a negative number.

540
00:46:41,710 --> 00:46:46,150
So this negative number has a
bigger magnitude than this

541
00:46:46,150 --> 00:46:48,160
negative number.

542
00:46:48,160 --> 00:46:51,120
We're subtracting them from one
and they're less than one,

543
00:46:51,120 --> 00:46:59,010
so this is bigger than this.

544
00:46:59,010 --> 00:47:04,760
And all we know about this is
that it's really for some

545
00:47:04,760 --> 00:47:06,970
irreversible reversible step.

546
00:47:06,970 --> 00:47:10,880
All we really needed to know
about it is that this is

547
00:47:10,880 --> 00:47:13,740
smaller right.

548
00:47:13,740 --> 00:47:19,080
So it's going to be the case for
any irreversible engine.

549
00:47:19,080 --> 00:47:24,410
And that's the point.

550
00:47:24,410 --> 00:47:38,060
So it's that the irreversible
efficiency is lower than the

551
00:47:38,060 --> 00:47:43,090
reversible one.

552
00:47:43,090 --> 00:47:47,690
But, of course, since we saw
that this is smaller than it

553
00:47:47,690 --> 00:48:17,420
was in the reversible case,
we can also write that dq

554
00:48:17,420 --> 00:48:28,900
irreversible over T is less
than dq reversible over T.

555
00:48:28,900 --> 00:48:33,800
Which is to say if we go
around a cycle for dq

556
00:48:33,800 --> 00:48:43,910
irreversible over T, that's
less than zero.

557
00:48:43,910 --> 00:48:50,130
It's only in the reversible case
that dq over T around a

558
00:48:50,130 --> 00:48:53,080
cycle is equal to zero.

559
00:48:53,080 --> 00:48:57,980
So to write that in a general
way, it's actually formulated

560
00:48:57,980 --> 00:49:04,780
by Clausius.

561
00:49:04,780 --> 00:49:10,160
Going around in a cycle the
integral of dq over T is less

562
00:49:10,160 --> 00:49:13,570
than or equal to zero.

563
00:49:13,570 --> 00:49:24,290
Never greater.

564
00:49:24,290 --> 00:49:29,750
OK, what we'll see shortly is
that this will allow us to see

565
00:49:29,750 --> 00:49:33,820
that for an isolated system the
entropy never decreases.

566
00:49:33,820 --> 00:49:36,460
It only can go up.

567
00:49:36,460 --> 00:49:42,350
And in fact any spontaneous
process will make it go up.

568
00:49:42,350 --> 00:49:45,470
Only in the case of reversible
processes.

569
00:49:45,470 --> 00:49:48,710
Of course, then you can see
that this will be zero.

570
00:49:48,710 --> 00:49:52,520
Anything else, which is to say
any spontaneous process, it'll

571
00:49:52,520 --> 00:49:54,970
be less than zero.

572
00:49:54,970 --> 00:50:00,030
We'll see that next time, and
then we'll generalize in a

573
00:50:00,030 --> 00:50:04,040
broader sense to look at the
direction in which spontaneous

574
00:50:04,040 --> 00:50:05,740
processes go.