Today’s Lecture
Phase Changes
First Law of Thermodynamics
Summary of Kinetic Theory
Physical meaning of the absolute
temperature is a measure of the
average kinetic energy of a molecule.
From this we can express the
pressure of an ideal gas as
Maxwell Boltzmann Distribution
Mean free path, λ
Phase Changes
We start with 1gm of ice at -30°C, supply energy at a constant rate
and monitor its temperature.
Straight line with constant slopes – warming up of ice, water and
steam. The slopes are inversely proportional to their specific heats.
Horizontal lines – no temperature change as you add energy.
Phase transitions – melting and vaporization.
It takes some amount of energy to melt and vaporize substances.
Those energies per unit mass of the materials are called
heat of fusion Lf and heat of vaporization, Lv.
Q = Lm
Q
L=
m
Since no temperature change occurs during the phase transitions, those
energies are sometimes called latent heats of fusion/vaporization.
Latent Heats
Q = Lm
Q
L=
m
Positive energy is required to melt/vaporize. The same amount of
energy, however, is released when the substance solidifies/condenses.
The process is perfectly reversible.
Hot Compress
Paraffin compress – I hope you never had one applied to you!
Typical melting temperature is from 49 to 71oC (120 to 160oF).
So, if you melt it, wrap it in a cloth, and apply it to yourself (or your
kid) it is going to stay at its melting temperature, around 140ºF until
its heat of fusion is transferred to your body…
Q = Lf m
L f ≈ 140 kJ/kg
Compress
versus
Hot Water Bottle
So, if you melt it, wrap it in a cloth, and apply it to yourself (or your
kid) it is going to stay at its melting temperature, around 60ºC (140ºF)
till its heat of fusion is transferred to your body…
Q = Lf m
For water cooling down:
Q = cmΔT
L f ≈ 140 kJ/kg
c = 4.2 kJ/kg ⋅ °C
To give (reject) the same amount of heat 1kg of water has to cool
down by ~35ºC, from 60ºC to 25ºC (140 to 80ºF).
Nuclear Meltdown
A nuclear reactor in meltdown still generates heat, 120MW. The
core is 2.5x105kg of uranium. How much energy is required to melt
the core after it reaches it melting point?
How long will it take?
This is less than 3 minutes!
Nuclear reactors contain failsafe emergency cooling systems
Ice Meltdown
How much energy is required to melt 10kg of ice initially at -10oC?
The specific heat of ice (at this temperature) is 2050J/kg-K.
The latent of fusion for water is 334kJ/kg.
If this meltdown takes place inside of a 500W microwave, how long
does it take?
Phase Diagram for Water
Summary – Thermal Behavior of Matter
Ideal gas law
Can be derived from
kinetic theory!
Kinetic theory
Mean free path, λ
Phase Changes –
Latent heats of fusion
and vaporization
Thermal expansion,
linear, cross
sectional area, and
volume.
Qf = mLf and Qv = mLv
First Law of Thermodynamics
Kinetic energy:
mv 2
K=
2
Potential energy (V):
Conservation of energy in mechanics mechanical energy of a system is conserved
External Forces and Work
What if there is work done on the
system or if the system does work?
If the system is not totally isolated. Then
external forces can do work on it or it
can do work on external objects.
work of the external forces
Wsys is the work done by the system.
In this case work against friction.
If Wsys > 0 the energy decreases.
First Law of Thermodynamics
In thermodynamics we introduce a new type of energy – internal
energy of the system, U, which is a sum of all the internal
energies of the system (usually microscopic). Then
ΔU = Wext − Wsys
We further unify Wext and Wsys by introducing the net work W. By
convention this is assumed to be the work done by the system.
Hence it lowers the internal of the system. However, if there is work
done on the system by external forces, then W is negative.
ΔU = −W
First Law of Thermodynamics
Example of converting work
into internal energy.
ΔU = −W
500g of water in an uninsulated container is shaken until its temperature
rises 3oC. The mechanical work to do this is 9kJ.
(a) How much heat is transferred to the water during the shaking?
(b) How much heat is lost to the environment?
First Law of Thermodynamics
ΔU = −W
How do we convert work into internal energy? This is easy!
First Law of Thermodynamics
ΔU = −W
External work done by the
hands (negative W) is
converted into positive internal
energy (positive ΔU)
Potential energy of the weights
is converted into work
(external work, negative W)
and to increase in the internal
energy of water (positive ΔU)
Are not we forgetting something, though?
In order to increase internal energy of a system, we do not necessarily
have to have some work done on it. There is a simple and common
alternative – transferring heat, Q, from some hot object (heater)
ΔU = Q − W
The heat Q is positive if the energy is transferred to the system.
First Law of Thermodynamics
ΔU = Q − W
change in the internal
energy of the system
net heat
transferred
to the system
work done
by the system
The change in the internal energy of a system depends only on the
net heat transferred to the system and the net work done by the
system, and is independent of the particular processes involved.
The equation is deceptively simple… One of the forms of the general
law of conservation of energy. BUT! Be careful about the sign
conventions. Positive Q is heat transferred to the system. Positive W
is work done by the system.
ΔU = Q − W
change in the internal
energy of the system
net heat
transferred
to the system
work done
by the system
The heat, Q, is considered positive, when it is transferred
TO the system. – The system gains energy.
The work, W, is considered positive if it is done BY the system.
– The system loses energy.
WHY? – The history heavily influenced by the heat engines.
ΔU = Q − W
change in the internal
energy of the system
net heat
transferred
to the system
work done
by the system
Implication: both heat, Q, and mechanical work, W, are means of
energy exchange between the system and the outside world.
The both are kinds of energy in transit. That’s why they are on
the same side in the equation.
Nevertheless, the heat, Q, specifically relates to the energy
transferred due to temperature difference alone. While, W,
incorporates all other sorts of energy transfer, most commonly
mechanical work.
Statement #1 First Law = Conservation of Energy
ΔU = Q − W
change in the internal
energy of the system
net heat
transferred
to the system
work done
by the system
Statement #2 – Internal energy, U, is a function of internal state of
the system, a thermodynamic state variable – a quantity, whose
value does not depend on how a system got to a particular state.
In principle, you can measure internal energy of the system, by
measuring the sum of energies of all molecules, the same way as you
can measure its temperature and pressure – both the variables, which
are well defined in thermodynamics equilibrium.
Usual ambiguity with setting the zero level of energy, though…
What is energy of Uranium at absolute zero temperature?
Rate of Change in Internal Energy
dU dQ dW
=
−
dt
dt
dt
Continuous processes – we differentiate with respect to time to define
rates of energy flow, measured in Watts.
Gasoline burning in an automobile engine releases energy at a rate of
160kJ per second. Heat is exhausted through the car’s radiator at a rate
of 51kJ per second and out of the exhaust at 50kJ per second.
An additional 23kJ per second goes to frictional heating within the
machinery of the car.
What fraction of the fuel energy is available for propelling the car?
%E = (160 – 124) /160 = 22.5%
ΔU = Q − W
change in the internal
energy of the system
net heat
transferred
to the system
work done
by the system
Example #1 – A 40W heat source heats a gas for 25sec. During this
time the gas expands and does 750J of work. What is the change in the
internal energy of the gas?
Example #2 – If in an automobile engine 17% of the total energy
released in burning gasoline ends up as work, what is the engine’s
output if the waste heat produced is 68kW?
Thermodynamic processes…
We are interested in
changes in internal energy,
the heat transferred to or from the
system and the work done by the
system - ideally the gas under piston
In principle, all we need to know
are the ideal gas law and the 1st
law of thermodynamics:
BUT…
There are many,
processes, conditions
(idealistic or realistic),
and applications…
PV = nRT
ΔU = Q − W
Thermodynamic Processes
Usually presented as P-V diagrams…
A diagram suggests that both
pressure and volume are well
defined in every point.
⇒ The gas is in a thermodynamic
equilibrium in every point, every
moment of time…
We call that a quasi-static
process.
Equilibrium with what?..
At least with itself!
That is, different parts of the gas are in equilibrium with each other.
Can be implemented if we change things slowly….
Quasi-Static Processes
Heating and boiling water on a stove. Quasi-static? Why?
The least we can say is that the water is in contact with burning
gas from below and cool air from above.
So, its temperature is not likely to be the same at the top and
bottom, which precludes thermodynamic equilibrium and quasistatic process.
Quasi-Static Processes
Is this one any better?
Practically, how slowly you
should go to be quasi-static?
Quasi-static Processes are Reversible
You can go back and forth along the same line of
well defined equilibrium states.
Can there be multiple paths to get from 1 to 2?
You betcha!
Work Done by a Gas
Work done by a gas:
Work done and heat
transferred – our major
concerns!
W = FΔx = PAΔx = PΔV
P – pressure of the gas;
Δx – displacement of the piston,
A – area of the piston
Work is positive when the gas expands!
Differential form:
dW = Fdx = PAdx = PdV
Integral form – valid for varying pressure:
W = ∫ dW = ∫ PdV
Work Done by a Gas
Work done and heat
transferred – our major
concerns!
W = ∫ dW = ∫ PdV
P – (varying) pressure of the gas;
dV – differential volume change
ΔW = P ΔV
Work equals area under curve on a P-V diagram