Today’s Lecture
Review for Final
Hydrostatic Equilibrium with
External Forces: Gravity
Force from above – PA
Force from below – (P+dP)A
Gravitational force (pulling down)
dFg = ρgA dh
balance
PA + ρgA dh = ( P + dP) A
ρgA dh = AdP
=>
dP
= ρg
dh
Note that h measures increasing depth.
Hydrostatic Equilibrium
with Gravity
dP
= ρg
dh
This is a differential equation.
How do we calculate the actual pressure?
We integrate the equation
P = P0 + ρgh
And end up getting an integration constant, P0.
What is this constant about?
Mostly pain in the neck!..
In order to find pressure, P, in a given point, we need to know the
pressure P0 in some reference point, h0.
Then we compare the depth between the two points and calculate P
as:
Buoyant Force
In equilibrium the upward pressure
force, Fp, is equal to the weight of
the fluid, Fg. If the fluid is replaced
by a solid object, the pressure
force remains the same. In general
the gravitational force changes.
The net force on the solid object
depends on whether it is more or
less dense than the fluid.
Archimede’s principle: The buoyant
force on an object is equal to the weight
of the fluid displaced by the object.
Fp = mw g = ρ wVg
The buoyant force is applied to the center of gravity of the
displaced fluid (center of the submerged volume of the body).
Remember it is only the submerged volume (not necessarily the
total volume).
Archimedes Principle:
The buoyant force on an object is equal to the
weight of the fluid displaced by the object.
The density of ice is ρice=.917gm/cm3 .
The density of sea water is ρsea=1.027gm/cm3.
What fraction of iceberg is submerged?
So almost 90% of the iceberg is submerged,
not so good for those on the Titanic!
Mass of fluid, Δm1, entering the tube
from the left over the time interval Δt
By mass conservation, over the time
interval Δt, the same mass, Δm2, is
exiting the tube from the right
Therefore ρ1A1v1 = ρ2A2v2 or ρAv = const.
everywhere along a flow tube.
This is the “Continuity Equation”.
If the fluid is incompressible and its
density, ρ, is constant we have
vA = const
Total Energy Balance
1
2
2
m(v2 − v1 ) + mg ( h2 − h1 ) = P1 A1Δx1 − P2 A2 Δx2
2
A1Δx1 = A2 Δx2
Incompressible fluids – constant density and volume
1 2
1 2
P1 + ρv1 + ρgh1 = P2 + ρv2 + ρgh2
2
2
1 2
P + ρv + ρgh = const
2
Bernoulli’s equation
Example of Bernoulli’s Equation
Venturi Flowmeter
Consider a Venturi flowmeter with
cross-sectional areas of A1 and A2
Solve for v1 in terms of the change in
pressure.
From Bernoulli’s equation and the Continuity equation,
Demo of Venturi tube.
Heat capacity is an extrinsic parameter (depends on quantity):
When you bring two objects together, heat capacity of the system of the
two objects becomes the sum of the two individual heat capacities.
It is convenient to introduce specific heat, c, an intrinsic parameter,
which is heat capacity of a material per unit mass .
Specific heat is measured in J/(K⋅kg).
ΔQ = cmΔT
ΔQ
C
c= =
m mΔT
Heat capacity of a water ball is large due to both high
specific heat and large mass of the water inside the ball.
Heat Conduction
ΔT
H = −kA
Δx
A lake with a flat bottom and steep
sides has a surface area of 1.5km2 and
is 8m deep. In the summer the surface
is 30oC while the bottom is 4oC.
What is the rate of heat conduction in the lake?
Thermal Resistance
ΔT
H = − kA
Δx
Analogous to the
Ohm’s law:
V
I=
R
The current, I = Δq/Δt, amount of charge per unit time, is
analogous to the heat-flow rate, H = ΔQ/Δt, “thermal
current”. The voltage, V, the factor driving the electric current,
is analogous to temperature difference, ΔT, “thermal voltage” .
Continuing the analogy: electric resistance, R, is analogous to
Thermal resistance is introduced as
The electrical conductivity, σ, is analogous to the thermal conductivity, k.
Black Body Radiation
All matter radiates energy via black-body radiation
“Stephan-Boltzmann Law”
Blackbody Spectra for Different
Temperatures
Wien’s law states that the peak in the intensity of blackbody radiation
is inversely proportional to the temperature, λmax = a / T.
Defining x = hc/λkBT, we found that xmax= 4.965. From this we can
solve for the wavelength that corresponds to the maximum in the
intensity distribution:
For our Sun the wavelength for the maximum intensity is ~502nm.
Hence the surface temperature of our Sun is found from:
In general the intensity distribution of the black body radiation
from a star yields the surface temperature of that star!
V = NkT / P
The ideal gas law
Doubling the temperature, number of
molecules, pressure?
P = NkT / V
Keeping the volume and the number of particles
constant, but doubling the temperature?
N is normally very big, while k is a very
small number… k=1.38X10-23J/K
N = nN A
NA = 6.022×1023 – Avogadro number, number
of molecules in 1 mole of a substance;
n is the number of moles in the gas.
PV = nN A kT = nRT
R = 8.314 J/mol⋅K universal gas constant
Ideal Gas - example
What is the volume occupied by 1.00 mol of an ideal gas
at standard temperature and pressure (STP), where T=0oC
and P=101.3kPa=1.013x105Pa (1atm)?
The ideal gas law
An Avogadro's number of gas molecules occupies 22.4L at STP!
Kinetic theory of the ideal gas.
From the Kinetic Theory of gases
From the ideal gas law
The average kinetic energy is
Physical meaning of the absolute
temperature is a measure of the average
kinetic energy of a molecule.
The rms speed of a
molecule – thermal speed:
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.
Work done and heat transferred –
Work done by the gas
our major concerns!
W = ∫ dW = ∫ PdV
P – (varying) pressure of the gas;
dV – differential volume change
ΔW = PΔ V
P
Always the Ideal Gas Law
V
Area under curve on a P-V diagram
Review of our Thermodynamic Processes…
ΔU = Q − W
PV = nRT
W = ∫ PdV
Adiabatic processes – no heat transfer.
Q=0
ΔU = −W
Positive work, W, is done by the expanding gas at expense of reduction
of its internal energy…
Since there is no heat supplied from the outside to replenish the gas
energy the temperature declines.
Work can be expressed in terms of -ΔΤ or -Δ(PV)
Adiabatic processes – no heat transfer.
Q=0
ΔU = −W
We now know the work done by an ideal gas during an adiabatic
expansion, but how do pressure and temperature behave?
From the ideal gas law:
From the 1st law:
As an exercise for the student
Specific Heats of an Ideal Gas
From kinetic theory we showed the average (translational) kinetic energy
per molecule is:
For n moles the
internal energy due
to KE is:
This means we
can find Cv :
For inert gases we have:
Cyclic processes
What is so special about them?
The system returns to the same point (state)
P
What can we say about change ΔU, Q and W?
Since the system returns to the same state and
V
U is a function of state only,
Therefore by the 1st law
we have
A
B
W =Q
ΔU = 0
ΔU = Q − W
That is the system gets heat Q from the
outside and does an equal amount of work W.
Or the other way around! Both Q and W may
be negative!
What simple thermodynamic process are cyclic processes similar to?
In what are they different?
Carnot Cycle
What is the net work and
efficiency of a Carnot cycle?
Wtotal = WAB + WCD
A
B
D
C
Since ThVBγ−1=TcVCγ-1
and ThVAγ-1=TcVDγ-1
Qc Tc
=
Qh Th
Efficiency
We have VB / VA = VC / VD and
W Qh − Qc Th − Tc
=
=
e=
Qh
Qh
Th
Entropy
Consider the Carnot
Cycle where we found
If we change the definition of Qc so that it is
the heat added vs heat rejected then
Any closed reversible cycle can be made
up of incrementally small isothermal and
adiabatic cycles. This leads to
This means that the integral
representing the change in entropy,
is path independent!
Summary of the First Law of Thermodynamics
First Law of Thermodynamics
Thermodynamic processes,
Quasi-static - work is given by
Isothermal – constant temperature
Isochoric – constant volume
Isobaric – constant pressure
Adiabatic – no heat transfer
Equipartition theorem –
1/(2kT) average molecular energy
for each degree of freedom
y
x
λ
- equation of a harmonic wave
ω = 2π / T = 2π f
ω = kv
k = 2π / λ
y( x, t ) = Ay cos(kx − ωt ) = Ay cos(kx − kvt) =
= Ay cos[k ( x − vt)]
A crest corresponds to a point, where
The general wave equation:
Velocity on a string:
Wave intensity is the wave
power per unit area
P
I=
A
Measured in
J
W
or 2
2
s⋅m
m
For a plane wave the
intensity remains constant.
For a spherical wave it
decreases with the distance,
r, from the source, like
P
I=
2
4πr
Does it mean that the power is lost as the wave propagates?
Speed of Sound in a Gas
v=
B
ρ
Can we calculate the bulk modulus of elasticity, B,
of a gas?
What process should we assume?
Sound waves are propagating quickly.
No time for heat exchange!
⇒ Adiabatic process.
For an adiabatic expansion:
Hence:
With the result:
PV γ = c = const
Speed of Sound in Air
γP
v=
ρ
Air is essentially 79% nitrogen (N2) and 21% oxygen (O2).
The masses of nitrogen and oxygen molecules are 28au
32au respectively.
At one atmosphere of pressure and room temperature, 20oC, one
mole of gas under these conditions occupies V=22.4x293/273L=24L,
and the density is:
For a diatomic molecule γ =1.4 and
In air the speed of sound must satisfy
Measuring sound intensity in
decibels, dB
I 0 = 10
−12
W/m
I = I 0 ⋅ 10
β / 10
⎛I ⎞
β = 10 log ⎜⎜ ⎟⎟
⎝ I0 ⎠
2 - reference value, corresponding to
threshold of sensitivity of a normal
ear at about optimal frequency
An example: TV is turned down from 75 dB to 72 dB, -3dB.
How does its sound intensity change?
General Expression for Doppler Shift
For a moving observer:
For a moving source:
Both of these expressions are included in the general relationship:
Here uo is the velocity of the observer and us is the velocity of the source.
It is usually best to use physical intuition to determine the signs. The
Doppler shift results in an increase in the frequency when the object and
the source are approaching each other with an increase in velocity.
Police Radar – sends frequency f.
The car moving toward the radar (moving observer) at a
speed u receives a frequency:
v+u
f '=
f
v
It reflects the received frequency f’, but it is a source moving
toward the radar, so that the radar receives a frequency
f'
1+ u / v
= f
f ''=
1− u / v
1− u / v
If u/v is small, which is typical for electromagnetic waves (v =
light speed).
f ' ' = f (1 + 2u / v)
The frequency of the received signal is higher by a fraction
2u
v
Frequency and Wavelength at Refraction.
• As light travels from one medium to
another, its frequency does not
change
– Both the wave speed and the
wavelength do change
– The wavefronts do not pile up, nor
are created or destroyed at the
boundary, so ƒ must stay the
same
1
2
1
2
1
2
f = f
v
λ
=
v
λ
λ1 v1 c / n1 n2
= =
=
λ2 v2 c / n2 n1
So we see that
Snell’s Law
•
In terms of indices of refraction the
law of refraction becomes
•
Sine of angle of incidence (refraction)
is inversely proportional to the index
of refraction of the medium
Snell’s law of refraction is written in a form
symmetric to the incident and refracted
beams:
http://www.physics.uoguelph.ca/applets/Intro_physics/refraction/LightRefract.html
Displaced Laser Beam
l
Find the displacement of a laser beam
that is approaching a 5cm thick layer
of ice with an angle of incidence of
θ1 = 30o when ice has an index of
refraction of n=1.3.
Defining the diagonal through the slab
as l we find:
From Snell’s law:
Hence the displacement is:
Critical Angle
• A particular angle of
incidence will result in an
angle of refraction of 90°
– This angle of incidence is
called the critical angle
• For angles of incidence greater than the critical angle, the
beam is entirely reflected at the boundary
– This ray obeys the Law of Reflection at the boundary
• Total internal reflection occurs only when light attempts to move
from a medium of higher index of refraction to a medium of
lower index of refraction – key to fiber optics
Summary for
Converging Lens
Image real, inverted, reduced
(magnified for l < 22f )
Virtual, upright, magnified
Lens Magnification – Diverging lens
1 1 1
+ =
l l' f
f  0, ℓ − anything, ℓ ′  0 and |ℓ ′ |  ℓ;
M  −ℓ ′ /ℓ, M  1
Virtual, upright, reduced
To Summarize for both types of lenses:
Imagine a camera with a single lens with a focal distance f = 35mm.
By how much and in what direction should the lens be moved to
move its focus from an object, which is far-far away to an object at a
distance of 1.5m? Far-far away means l = ∞, and l’= f = 35mm.
New distance l = 1.5m; l’= ?
A healthy human eye can
clearly see (focus) objects at
distances from infinity to about
25cm. How is that achieved?
By changing the focal
distance of the lens! We
always have l’≈ 22mm.
Far-far away means l = ∞, and f = l’ ≈ 22mm.
An object at l = 25cm means
The adaptive lens driven by the eye muscles changes the focal
distance of the eye by “only” 9%. But this is quite a lot!..
Can we see any interference without a laser?
Approximation used:
d << L
θ
r2 − r1 = d sin θ
Constructive (a bright strip)
Destructive (a dark strip)
r2 − r1 ≈ dy / L
for small
d sin θ = mλ
d sin θ = (m + 1 / 2)λ
y/L
Approximation used:
d << L
θ
Constructive (a bright strip)
r2 − r1 = d sin θ
d sin θ = mλ
Consider two slits .075mm apart located
1.5m from screen. If the 3rd order bright fringe is
3.8cm from the screen center what is λ?
Approximation used:
d << L
θ
Destructive (dark strip)
r2 − r1 = d sin θ
Consider two slits .075mm apart located
1.5m from screen. Given λ =633nm what is the distance
from screen center to the 3rd order dark fringe?
Diffraction Limit For Circular Apertures
Diffraction imposes a
fundamental limit on the ability
of optical systems to resolve
closely space objects.
Rayleigh Criterion
Two peaks are can barely be
distinguished if the central
maximum of one coincides
with the first minimum of
the other.
For a circular aperture
(a) Two point sources are far enough apart so that they are well resolved
(b) Two point sources are separated enough to barely satisfy Rayleigh Criterion
(c) Two point sources are so close that they are not resolved.
Resolution of the Eye
Assume light of wavelength of λ = 500nm
and a pupil diameter of d = 2mm.
(a) Estimate the limiting angle of resolution
for a human eye.
For a circular aperture
(b) Find the minimum distance of separation, d, for two objects that are
25cm from the observer.
At 25cm objects can not be resolved if they are closer than 80μm!
Multiple-Slit Interference and Diffraction Gratings
From the diffraction limit with only two
slits the intensity pattern is
πd sin θ
S = 4 S 0 cos (
)
λ
2
What if you require better resolution?
Consider multiple slits.
For multiple slits
The location of the bright fringes still satisfies
d sin θ = mλ
The location of the dark fringes now satisfy
Here N is the number of slits. This means there
are N-1 dark fringes between each bright fringe.
Resolving Power and Diffraction Gratings
For a grating with N slits the criteria for the mth order
maximum is
The adjacent minimum occurs at
Maximum for λ’ = minimum for λ
Multiple-Slit Interference and Diffraction Gratings
A binary star system has one massive star at rest and
another smaller star rotating about it. The hydrogen α line
for the rotating star is Doppler shifted from λ=656.272nm
to λ=656.215nm. What order spectrum is required for
astronomers to resolve these lines when using a grating
that is 2.5cm wide with 2000 slits/cm?
Since the order must be an integer, the resolution of the
two wavelengths is third order, m = 3.
Interference in Thin Films
Constructive Interference:
Destructive Interference:
Is the wavelength
inside the film and