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: Is the Crown Made of Gold?
Consider weighing a completely submerged object of mass m
and you want to determine its density.
The crown weighs 25N in air and 23.3N in water.
Hence ρ = (25/1.7)gm/cm3 = 14.7gm/cm3.
The density of Au is ρ = 19.3gm/cm3 and Ag is ρ = 10.5gm/cm3.
Crown is approximately 48%Au and 52%Ag.
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.
Thermal Resistance
Thermal resistance
is introduced as:
Δx
R=
kA
Heat-flow rate
equation is now:
ΔT
H =−
R
When heat flows through two
materials, it is analogous to two
resistors connected in a series.
To see this we first write down the heat flow
through each slab:
What about this temperature T2?
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
Kinetic theory of the ideal gas.
From the Kinetic Theory of gases
From the ideal gas law
The average kinetic energy is
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
λ = v/ f
k = 2π / λ
k = 2π f / v = ω / v ω = kv
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:
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
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?
Sound Waves in Review
Sound waves are longitudinal waves
propagating in gases, liquids, or solids.
The velocity of sound is given by
The intensity of sound depends on
the square of the pressure fluctuations
or the displacement velocity of the air.
Audible sound covers many orders of
magnitude. By convention the measure
of intensity is called decibel level:
Doppler effect for moving source
and moving observer:
Upper sign approaching
Lower sign receding.
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
Spherical Mirrors Form Images
l
Object located at l
Object height is h
Image located at l’
Image height is h’
The shaded triangles are similar triangles
Mirror Equation
Spherical Mirrors Form Images
Object located at l
Object height is h
Image located at l’
Image height is h’
As it turns out the image may also be virtual. This occurs when object is closer
to the mirror than the focal length. The image is then upright and virtual.
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:
What is going on here?
We had a plane light
wave, so why do the fronts
get curved after passing
through the slits?
Diffraction – the wave front
gets curved if the wave
passes through a narrow
slit (or a small hole).
Diffraction from a single slit
The intensity on a screen as a function of θ is
Diffraction Limit From a Single Slit
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.
If d is the slit width then
Electron microscopes versus optical microscopes
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.
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
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
Interference in Thin Films
Constructive Interference:
Destructive Interference:
Is the wavelength
inside the film and
Interference in Thin Films
Constructive Interference:
Destructive Interference:
(a) If the film were illuminated with 650nm light, how
many bright bands in the film with n=4/3 will appear?
There are 4 bright bands, why??
(b) What part of the film will be dark? Since the
smallest wavelength for visible light is λ=400nm,