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Transcript
Thermodynamics
Thermodynamics
Heat and Forces
Thermodynamics
Dynamics is to do
with forces and
motion
Thermal
Thermometer
Thermo means
heat
Internal Energy, Heat and Work



Internal energy is the total
PE and KE of all the
molecules in an object
Heat is energy that is
transferred due to
temperature difference
(conduction, convection or
radiation)
Work is done when a force
changes its point of
application
F
First Law
ΔQ = heat added
ΔU =increase in
internal energy
ΔW work done by
the system
ΔQ = ΔU + ΔW
100 = 20 + 80
ΔQ = 100J
ΔW = +80J
ΔU = 20 J
First Law
ΔQ = heat added
ΔU =increase in
internal energy
ΔW work done by
the system
ΔQ = -60J
ΔQ = ΔU + ΔW
-60 = 40 - 100
Negative because work
done ON the gas
ΔW = -100J
ΔU= 40J
ΔW = PΔv



ΔW = F x d
ΔW = PxA xd
ΔW = PΔv
d x A = ΔV
p
ΔW = pΔV
v1
v2
A
F
d
v
P = F/A
F = PxA
Isothermal






p
Constant Temperature
ΔU = 0
ΔQ = ΔU + ΔW
ΔQ = ΔW
Boyles’ Law is an
example of isothermal
change
ΔQ
Compression must
happen slowly and walls
should be thin so that
heat can escape
ΔQ
ΔW
v
ΔW
ΔW
Adiabatic






No heat enters or leaves
the system
ΔQ = 0
ΔQ =ΔU + ΔW
ΔU = -ΔW
The compression must
happen quickly and the
walls should be well
insulated
Diesel engines use
adiabatic compression to
ignite the fuel
Once again the
adiabatic line is
The line is steeper
steeper than the
than the isothermal
isothermal. The
line. The pressure
pressure goes down
goes up because
because the volume
the volume gets
gets bigger and
smaller and
because the
because the goes
gas
temperature
getsdown
warmer
p
v
-ΔU
ΔU
-ΔW
ΔW
Isobaric


Isobaric means
constant
pressure
As in Charles
law
p
ΔW = pΔv
v
Isovolumetric




Δv = 0
Like in the
pressure law
What is ΔW?
ΔW = 0
p
v
Heat Engine




A device that changes heat
energy to useful work is called
a heat engine.
A steam engine and an
internal combustion engine
are both examples of a heat
engine
You do not need to know the
details of how these work.
We just draw a very simple
diagram.
Efficiency = ΔW/Qhot
Efficiency = Qhot-Qcold
Qhot
hot
Q hot
Qhot = Qcold + ΔW
ΔW
Q cold
cold
Remember this is the best an
engine can possibly be at
these temperatures. A real
engine will be less efficient
than this
We all know that efficiency cannot be
Carnot Efficiency



more than 100% so ΔW has to be less
than QH
However with heat engines it has to be
a lot less than this or it breaks the
second law of thermodynamic
A clever French dude called Carnot
worked out that the best efficiency a
heat engine can ever have is
600 – 300
300
Efficiency = Thot – Tcold
= 50%
Thot
What is the Carnot efficiency of this engine?
600K
Q hot
ΔW
Q cold
300K
Carnot Cycle





Carnot also worked out
that to get this
maximum efficiency
you would need
Isothermal
compression
Adiabatic compression
Isothermal expansion
Adiabatic expansion
p
ΔW
v
Coin toss




If you toss two
coins what is the
most likely
outcome?
2 heads,
2 tails,
1 head and 1 tail
H
H
H
T
25%
50%
T
H
T
T
25%
Drop balls in the box
3
41
2


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
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If we drop 4 balls what is the most
likely outcome?
4-0 1 way
3-1 4 ways
2-2 6 ways
1-3 4 ways
0-4 1 way
3 4
2 4
2 23 31
4
1 13 42
4
1 12 33
4
1 1212324
34
It doesn’t have to end
like this but this is the
most common way
1 2
1 3
12 13 44
21 23 34
31 22 44
1412323434
100 balls





If you drop 100 balls the chances of all landing
on 1 side is 2100
This is 1.3 x 1030
The universe is 4.7 x 1017 seconds old
If all 6 billion people on Earth dropped balls
every second for the entire age of the universe
they would be very unlikely to get 100 on one
side
You are very likely to get about half on each
side
Air molecules move at random


Gasp
!

What’s to stop all the air
molecules moving to the
other side of the room?
Nothing, except the laws of
probability
When we deal with a billion
billion air molecules it is as
near certain as anything
can be that this won’t
happen.
Entropy
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
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
With twelve balls
there is a 1 in 212
chance that they
will unmix.
This is 1 in 4096
It might happen
With a hundred
balls it wont
With a billions and
billions of
molecules it wont
This is why heat engines cannot be
100% efficient
Kinetic energy
is organised.
All the
molecules
move in the same direction
KE changes
to internal
energy
in a collision
there
are so many
molecules
that the
chances
of this
IfBut
all the
molecules
vibrate
in the same
direction
at the
same time then
Internal
energy
is
random.
All
the
molecules
vibrate
independently
happening
are
rediculous.
warm
objects
would
suddenly start to move
The second Law of Thermodynamics





The entropy of a system and its
environment cannot decrese
Entropy means disorder
Smashed cups cannot mend themselves
Warm water cannot be unmixed to give hot
and cold water
Heat energy cannot be completely turned
into work
ΔS = ΔQ/T







The above equation can be used
to calculate the change in
entropy (ΔS)
Assume this heat engine has the
ideal Carnot efficiency:
T1 –T2 / T1
= 600-300/ 600 = 50%
ΔS = -100/600 + 50/300
= 0
Zero change in entropy
This is an ideal engine, it works
in theory but friction would not
allow this in practice
ΔQ is negative
because heat is lost
600K
100J
Q hot
50J
ΔW
50J
300K
Q cold
Greater than Carnot efficiency





What if efficency = 60%?
600K
ΔS = -100/600 + 40/300
= - 0.033JK-1
100J
Entropy decreases!
The positive
bit got
Impossible
smaller
That’s why we need
40J
Carnot’s theorem
300K
Q hot
60J
ΔW
Q cold
A real engine





If efficiency is less than
50%, say 40%
ΔS = -100/600 + 60/300
= - 0.167 + 0.02
= + 0.033 JK-1
Change in entropy is
positive so this machine
does not break the
second law
600K
100J
Q hot
40J
ΔW
60J
300K
Q cold