Review for midterm 1
Ch. 17, 18, 19 and 20;
Three problems; 65 minutes; closed book; one
page formula sheet allowed. If you have a
question about problems, let me know.
Attendance will be taken from zoom list.
.
Use correct units –
• Pressure in Pascals;
• Volume in meter cube,
• Temp. in Kelvins
• Mass in Kilograms
• Height or Length in meters
• Q (heat) in Joules
If something is given in a different unit, convert it into these units
first before proceeding with solving the problem.
Ch.17: Review
• Thermal expansion Δ𝐿 = 𝐿 0 𝛼Δ 𝑇
• Calorimetry: Be careful to keep all terms that involve heat loss and heat gained;
sometimes using common sense is helpful. Remember this diagram
•
•
• calorimetry Eqn. Q1 +Q2 +….=0
1 gaining Q
2 losing Q
Q=mc(T2 –T1 ); Q=mL
• Make sure phase transition is included in the equation, if there is one.
Some useful hints:
• If phase transition not complete in the final equilibrium state,
the final T of the whole system is the phase transition
temperature e.g. ice not completely melted means Tf =00 C;
water only partially evaporated, Tf =1000 C. To check if phase
transition is complete, in the Q equation use Q=ymL where y
is the fraction (always <1) that has undergone phase
transition. If y>1, use a different expression for Q i.e.
mL+mc (Tf –Tmelt ) (c is the specific heat …)
since that means the phase transition is completed and water
or steam is getting heated up beyond melting or boiling point.
• To repeat: Include all contributions to Q’s correctly to get any
credit.
Example: m Kg of Ice at -5C added to 100 Kg of water at
70C. What is the final state:
First equation: mcI (Tf –(-5))+100cW (Tf -70)=0
Suppose you get Tf >0C above eq is wrong and add
the latent heat term since part or all of the ice melted.
How do you know part or all? Take fraction y of mass m
of ice and write: mcI (0 –(-5))+ymL+100cW (0 -70)=0
Suppose you get y <1; your work is finished and Tf =0C;
Suppose you get y>1 that means the above is not the
right equation and write and solve for Tf.
mcI (0 –(-5))+mL+mc W (Tf-))+ 100cW (Tf -70)=0
• Heat conduction: remember that H (heat flow rate)
is same when heat flows from one layer to another
(e.g. conducting layers in series).
• Two parallel conductors between same initial and
final TH and TC imply Htotal = H1 +H2
• Heat radiation: H=Ae𝜎 T4
Review
Ch.18: Ideal gas law and kinetic theory: appears in
many forms: (T in Kelvin, P in pascals, V in cubic meters)
PV=nRT;
PV=NkT;
N=no. of moles; N=number of atoms or molecules
When using formulae like:
the very beginning.
use T in K from
• If a problem has gauge pressure, add atmospheric
pressure to it (recall problem in mastering physics).
• Correct units give correct answers and you will lose
points if you don’t use correct units.
Ch. 18: Review
Kinetic theory
per atom, Kave =3/2 kB T
;
Total thermal energy U=nCV T
translation
Translation+two rotations
Ch. 19
First law of thermodynamics: Q=W+ΔU
Thermodynamic processes denoted by a PV diagram:
Each vertex has its own Pi , Vi , Ti satisfying ideal gas law.
Always ΔU=nCV (Tf –Ti); does not depend on the path;
-Q and W however depend on type of process and path.
-W
is given by the area under the line in the PV diagram.
isochoric: W=0;
Isobaric: W=P(Vf –Vi); Q=nCP (Tf –Ti)
Isothermal: ΔU=0; Q=W= nRTln(Vf /Vi)
Adiabatic: Q=0; W=-ΔU;
CP = CV + R > CV  more heat
required in an isobaric process to
increase T by same amount compared
to isochoric process
atomic
SVeummary of Ideal-Gas ProcessesVery
Slide 19-50
Second law of thermodynamics: (stated in 4 different ways)
(i) Heat flows from hot bath to cold bath
(ii) Entropy
increases
of an isolated system always
(iii) Heat engine cannot be 100% efficient or e < 1.
(iv) Refrigerator cannot have infinite coefficient of
performance
More on entropy calculation: When the temperature
change is large, use
Other Entropy
cases: change for free expansion of gas
T dS = dQ=U + dW=U +pdV
Gradual free isothermal expansion of gas U=0
Free exapansion of gas  entropy increases.
Entropy change in free expansion!
S –cSoSme hange in general thermodynamic
process
Entropy change for
ice melting:
Engines and refrigerators
(i)
Engine takes in heat QH ; does work W and
discards the remaining heat QC ;
(ii) Refrigerator takes in heat QC from the refrigerator,
(QC >0), work W is done on the fluid to heat it up
(W < 0) and system then discards heat QH to
environment (QH. <0)
W=|QH |-QC
RefrRefigerators
Refrigerators
vs
Engines
Engines
to calculate
efficiency for a typical engine
How toHow
do engine
problems:
problem
Draw the P-V diagram for the engine
(0) Calculate T at each vertex using ideal PV=nRT.
(i) To calculate heat Q across a line in PV diagram, using
Q=nCV (Tf –Ti ) for isochoric line;
Isobaric Q=nCP (Tf –Ti ); across an isothermal line, use
Q=W since change in U is zero; across adiabatic line: Q=0.
(ii) Keep track of the signs of all the Q’s correctly.
(iii) Add the positive Q’s which gives QH added to system;
add negative Q’s which give QC; heat going out of engine;
find efficiency using e=1-|QC |/QH
(iv) Carnot engine e=1-TC /TH
(a)e=1-300/500= 0.4
(b) Power output = W/sec =1000Watt
To get the Rate of heat input recall that W=eQH ;
 QH /sec = (W/sec)/e=1000/.4=2500Watt
(c) Heat output |QC| = (QH-W)/sec =1500 Watts
For an engine,
heat and
coming
into the system
is QH andQheat
going out is QC;
Engines
refrigerators:
calculating
H and QC
For a refrigerator, Heat coming in to the system (hence positive) is denoted
by QC and going out (hence negative) is QH since that is the heat being
discarded to the environment.
For a refrigerator, Heat coming in is QC since the system is
absorbing heat from the refrigerator and heat given out is QH
since that is the heat being discarded to the environment
Qca =QH
|QH |=QC + |W|
Qab +Qbc =QC
Problem on refrigerator: 20.51
H-diatomic: CV =5/2R; CP =7/2R;
Ta=Pa Va /nR=300K; Tb = 300K; Tc = Ta Vc /Va = 1000K
QC =Qab +Qbc ; QH=Qca=nCP (Ta –TC)=-17000J; Qbc =nCV
x700=12000J; Qab =nRTa ln (Vb /Va)=2528J
K=(Qab +Qbc ) /Qca -(Qab +Qbc ) =6.2
Review for midterm 2
Ch. 15 Key concepts and formulae:
Transverse waves: speed:
•
•
•
Also
for sinusoidal waves.
When waves travel from one medium to
another, speed changes and as a result
wavelength changes and not frequency
Wave function for traveling wave
Particle velocity & acceleration in sinusoidal
wave: different from wave speed
Green vy and yellow ay
.
Superposition principle for Waves
What happens when waves meet
at one place?
When two waves are at the same
point, their displacements (or wave
functions) add up.
Net result is a phenomenon called
interference. (unique to waves)
Phase difference and constructive-destructive interference
If k(x1 –x2 )=π, 3π,… y1 +y2=0
Destructive (x1 –x2 )=
=0, 2π,…. y1 +y2=MaxConstructive
(x1 –x2 )=0,
In the case of constructive interference,
amplitude doubles and therefore intensity
increases 4 times. In case of destructive
Interference, Intensity=0.
• Standing waves are created by the interference of two
waves when one wave travels in a confined region, gets
reflected by the end points creating the second wave, one
travelling forward and one backward.
Standing Waves on a String
 Shown are the first four possible
standing waves on a string of
fixed length L.
 Each mode, numbered by the
integer n, has a unique
wavelength and frequency.
 Modes have nodes and antinodes
 Wavelengths decrease and
frequencies increase as n
increases
 Wave function for standing
wave:
Wave function for the nth harmonic:
 Speed of all harmonics is same.
 Maximum value of the particle speed in
the medium:
Ch. 16: sound waves
Wave function:
Decibels:
Standing sound waves
Standing sound waves: One end
closed, one open
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Both ends open
•.
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Beats:
•.
Doppler effect
• When a source of sound or the listener of sound or both are moving,
the frequency the listener hears is different from the actual frequency
emitted by the source. That is called Doppler effect.
• Next slide gives a master formula for determining the listener
frequency.
Master formula for Doppler shift of sound
vL,S can be positive or
negatiev
Positive x direction
• Join a line from listener to source and call that positive
x direction. Use that to fix the signs of vL and vS .
• Sometimes you may have to break up problem into
two parts (like the bicycle problem in quiz 5)
Ch. 21 and 22
• Electric fields and electric forces between electrically
charged systems
• How to sketch electric fields for a system
• Important to realize that if there are two
charges, magnitude of force on both
charges is same but directions opposite
• The acceleration a of each charge,
which is given by a=F/m may be
different depending on the mass m of
the individual charged object.
Coulomb’s law for multiple charges: superposition
Coulomb’s Law
 Forces are same on each charge when two
different charges are involved (Newton’s 3rd law).
 Forces on a single charge due to different
charges add vectorially !
 This is the superposition law for electric forces.
 Need to know how to calculate x, y
components of vectors
Concept of electric field: if force on a
charge q is F, the E-field at the location
of the charge q is E=F/q and in same
direction as F.
Need to know how to give a sketch of
the magnitudes of E-field around
charge distributions.
Some examples of sketches we discussed in class last week:
Some examples of sketches we discussed in class last week:
E-sketch (r-profile)
E-sketch (x-profile)
Ex is E field along
the x-axis; Note
how left arrow
means Ex is
negative and right
arrow means, it is
positive. 2 +ve
charges
2 –ve charges
Effect of electric field on a charge q
Gauss’s Law
For any closed surface enclosing total
charge Qin, the net electric flux through
the surface satisfies the relation:
This result for electric flux is known as
Gauss’s Law.
The closed surface is called Gaussian
surface. The shape of surface irrelevant!
Slide 27-68
Specifically, you should know how to
calculate charge distribution in charged
systems, charge inside conducting
shell, sphere inside shell, cylindrical
shell, insulating sphere, insulating
cylinder etc.
Specifically, you should know how to
calculate charge distribution in charged
systems, charge inside conducting shell,
sphere inside shell, cylindrical shell,
insulating sphere, insulating cylinder etc.
E field inside an insulating sphere or infinite cylinder
with uniform charge distribution:
sphere
Cylinder
Review for midterm 3
Ch.23. 23
Potential energy due to force is given
by:
If work done by force is negative, U
increases and vice versa.
The Potential Energy of Two Point Charges
 Take two like charges q1 and q2.
 The electric field of q1 pushes q2 as
it moves from xi to xf.
 The work done is positive and U
decreases:
 By matching both sides, we identify that
Slide 28-37
Potential energy for system of multiple charges
If there are three charges, what is their PE? This is the work
done in assembling the system of charges (or the total
potential in the system)
Case of three charges:
where
Important application of the concept of PE: conservation of
energy helps to determine speeds without using F=ma
equation e.g. how much speed to give -3q charge to move
it to infinity?
Ki + Ui = Kf + Uf
U(4 charges)+1/2 mv2 =U(3 charges)
PotPential energy of a charge between two plates
+
-
For a negative
charge potential
energy U
increases from + to
– plate and reverse
for +ve charge
From Potential energy U to potential V
V=U/q
Potential energy of q0 in
the presence of q1,2,3
Potential at location of q0 due
to q1,2,3
Potential for a cylindrical conductor:
Need to know how to do simple integrations
e.g.
VU for system: spherical conductor + point
V forcharge
a sphere with charge
Inside conductor, E=0 Its
integral V is a constant. Outside,
E~1/r2; so V~1/r
Q
Q
q
r
At the surface of conductor
V=U/q= potential
Equipotential surfaces and field lines
Surface of a conductor is an equipotential
surface. It obeys the relation V=ER. Less
radius means stronger E field.
Ch. 24: Beginning formulae in this chapter
(If separation between plates goes down, C goes up
and vice versa.)
Capacitors in series and parallel:
HHow to solve capacitor problemso solve
multiple capacitor problems
(i) Find the effective capacitance of the whole capacitor system;
Ceq
(ii) Given the applied voltage across the whole system, V, find the
charge on the whole system with effective capacitance i.e.
Q=Ceq V.
(iii) If the system of capacitors consists of capacitors in series, this
is the charge on each capacitor and you can calculate voltage
drop Vi across individual capacitors by Qi /Ci = Vi
(iv) If there is a subsystem where capacitors are in parallel, the
charge divides i.e. Q=Q1 +Q2 +...
Use this equation plus the equations for same
V across the whole system to find Qi
The Energy Stored in a Capacitor
 In terms of the capacitor’s potential difference, the
potential energy stored in a capacitor is:
Slide 29-81
Energy density inside a capacitor with dielectric
If there is a dielectric inside a capacitor
with dielectric constant
Then the effective capacitance C is given
by
(where C0 is capacitance
without dielectric). Use in every formula C
instead of C0 e.g
Capacitor filled with two slabs of dielectric with dielectric constants 2 and 3 each taking half
the distance d between the original plates d. What is the final total energy stored if
connected to a battery with voltage V?
Capacitance C without dielectrics: C0 =
A/d
After insertion of slabs: Ceq : This is a combination of two capacitors with capacitances in
series:
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Ch. 25+26: Resistance and Ohm’s law
• Current density: J=I/A; J=nqvd
•The potential across a conductor is given by Ohm’s law
•
V = IR.
• Formula for conductivity
Review for mid 3 contd
• Resistances in parallel and series
• Power delivered in a resistance
• Circuits with resistances and capacitances after current flows for a
long time:
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Resistances in series and parallel from Ch. 26
Series:Same current flows
all and
resistances
RResesistances
inthrough
series
parallel
Resistors in series and parallel
Parallel: Same voltage across each resistance
but the current divides:
How to deal with complicated circuits with resistances?W
Second example:
Power formula
• The rate of energy transfer to a circuit
element is power and is denoted by P:
• Using Ohm’s law, we can write power
across a resistor as P=I2 R=V2 /R (unit
Watts)
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Wattage on an appliance: what does it mean?
• Conceptual problems
• Which of the two light bulbs is brighter? 60W or 100W?
• What is the current I in the circuit
and the potential drop Vac
Find current in the circuit. (16-8 )V=I (1.6+5+1.4+9 ) I=.47 A
Find Vac = 8+0.47 (5+1.4)=11 V
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