Things to write on the board at the startAug 07, 2017 · Problem 2 (p.8) Q: A diatomic ideal gas...

Things to write on the board at the start

Objects are “hot” when their atomic constituents are moving energetically.

Objects are “cold” when their atomic constituents

are moving less energetically.

Equipartition Thm. kTdE particle )2/(=

Where d is the number of quadratic energy d.o.f.

kTKE

ntranslatioparticle )2/3(= for a particle of any kind.

From 7A: First two are key:

2)2/1( mvKE ntranslatio = 2)2/1( ωIKErotation =

2

2

1

∆=dt

xdmKEvibration

( )2)2/1( xkPEvibration ∆=

Will add d = 3,5,7 for translation, rotational, vibrational (3D) in the middle

column after explaining to them.

Ideal Diatomic Gas: Vibration = 2 q.e.d.o.f.

I need to correct something that I said on Monday. I said that for a diatomic gas, we have 6 quadratic energy degrees of freedom. The more accurate statement is that there is, at least naively speaking, a MAXIMUM of 6 q.e.d.o.f. available, but whether or not the gas acquires all 6 does not have a general answer. But, for a broad class of diatomic gases, which we might call ideal diatomic gases, there is a generic answer that is good over a broad range of P,V and T. That answer is 2 q.e.d.o.f. for vibration. This is related to the fact that what opens up the other e.d.o.f. of vibration are relativistic effects (when the vibrational frequencies become very high). This means ultra-high temperature. But, just as a you may know or will learn, when we take relativistic effects into account, the energy even for translational motion is no longer (1/2)mv2 but

2

2

)/(1 cv

mcE

−=

and this is not even quadratic. So, we would have to throw out the equipartition theorem anyway. The point is, for our purposes, and for most reasonable situations, a diatomic gas only picks up 2 q.e.d.o.f. from vibration. (Add to chart on board)

Summary of Q.D.O.F. for Diatomic

So, my upper division thermodynamics textbook has this graph which indicates the average internal energy per particle for diatomic hydrogen gas. And it roughly looks like this (DRAW IT ON BOARD) . So, below 100 K, only translational modes are present. But, that’s pretty cold (-173 C). At “normal” temperatures, rotational and translation modes are both present. At very high temperatures around 1000 K, we start getting vibrational modes. So, I believe that, for the purposes of this class, if you’re given a monoatomic gas, you have (3/2)kT, if you’re given a diatomic gas and not given any temperature etc. then you assume a relatively “normal” temperature and say (5/2)kT. Then, if you are actually told to include the vibrational modes for a diatomic gas then it’s (7/2)kT. But, I will ask this question during our group meeting on Friday and let you know.

Policy on questions in class

The next thing I want to briefly talk about are questions during section. So last time we got a bit bogged down and fixated on technical details. I have the awful tendency to question everything I learn and qualify every statement with when we expect them to hold and when not. Then, it’s usually very interesting and exciting to study the case when it does not hold. However, if we follow this pattern here, we might never actually understand the subject material of this class. Thus, I will try my best not to deviate so much from the main material and leave more intricate details to future courses, which I implore you to take. Because of this, I may at times answer your question with a glib remark like, “we would need quantum mechanics to answer this.” I’m not just being mean or dismissive. I just don’t want to take section time discussing such a tangential topic. However, I would love to talk about this more tangential stuff outside of section because that’s what I do all day anyway. Send me an email to set up meeting times, or come to office hours and I’d love to discuss that then.

Problem 2 (p.8)

Q: A diatomic ideal gas has N gas molecules, initial pressure p1 and initial volume V1. The gas then undergoes a series of three transformations:

(1) a Bunsen burner causes the gas to expand, at constant pressure, to volume 7V1 (2) the volume is held constant while an ice bath lowers the pressure to p1/4 (3) the water bath allows the gas to be compressed along a straight line in the pV

plane, until the pressure and volume return to their initial values

a) Sketch this on a p-V diagram b) Find the temperature at the three corners

In terms of p1, V, and N. NkVpT /111 = 12 7TT = 4/7 13 TT =

c) Find the change in internal energy During each stage (in p1 and V1)

1115 VpEi =∆ 8/105 11VpEii −=∆ 8/15 11VpEii −=∆ d) Add up the delta E’s. Why 0?

012012015105)8*15( =−=−− . Since 0=∆T .

V

P

p1

p1/4

V1 7V1

T1 T2

T3

(i)

(ii )

(iii)

Energy in an Ideal Gas

Discussion Problem 1 (p.9)

This is Conceptual Example 17-4 (p.460) of the textbook.

Q: Imagine a disk with a hole cut out of it. If you increased the temperature, does the hold get bigger or smaller?

Since it’s a question in the text and I expect that you’ve read the text, I’ll just discuss the answer.

A: It gets bigger. The usual argument is that you can do the experiment in two ways. First, heat the ring. Or second, draw the outline of the hole on the disk, heat the disk, then cut out the hole. For the latter, it’s obvious that the hole is larger than before.

Another way to think about this is that the constituents of the metal want to spread away from each other when heating up. If the hole is shrinking, then the metal constituents near the hole edge are actually getting closer and closer!

Thermal Expansion

Discussion Problem 2 (p.9)

This is Conceptual Example 17-6 (p.461) of the textbook.

Q: Metal has a larger coefficient of linear expansion than glass. If you wish to remove a metal lid from a glass jar, do you run it under cold or hot water?

Since it’s a question in the text and I expect that you’ve read the text, I’ll just discuss the answer.

A: Hot water. Firstly, the water may contact the metal before the glass and so the metal simply expands sooner. But even if not, the larger coefficient of linear expansion for the metal means that it will expand more than the glass, which is roughly the same size.

Thermal Expansion

Problem 1 (Easy Interpretation) (p.9) This Conceptual Example 17-5 (p.461) of the textbook. Q: You have a metal ring with 6

1 102 −×=α )/1( oC . You wish to put it around a pipe made of metal with 6

2 103 −×=α )/1( oC . If at 25oC, the inner radius of the ring is 10.0 cm and the outer radius of the pipe is 10.001 cm, what is the temperature that will allow you to slip the ring around the pipe?

Interpretation: Only the ring is to be heated to get bigger. The pipe remains at 25oC. Remark: Since we have not been told anything else, I will suppose that the ring can slip around the pipe when its inner radius is equal to the outer radius of the pipe (realistically it should be bigger).

A: Let CT i o25)( = the initial temp. and )( fT be the final temp. that we are looking for. Let 0.10)( =i

ringr cm and

001.10=piper cm be the initial inner radius of the ring and outer radius of the pipe. The change in ringr is ( ))()()(

1ifi

ringring TTrr −=∆ α . Note that the coefficient of linear expansion measures the FRAC-TIONAL increase

in length. Thus, ( ) ( )( ))()(1

)(1

)()()( 11 ifiring

iringring

iring

fring TTrTrrrr −+=∆+=∆+= αα is the final radius if the ring. Setting this

equal to piper and solving for )( fT gives

CCCcm

cmCC

r

rTT

iring

pipeif ooo

o

o 7510105.025110

001.10

102251

1 466)(

1

)()( =××+=

−×

+=

−+= −

−α

Remark: you might want to do this problem by cooling the rod instead.

Thermal Expansion

Problem 1 (Hard Interpretation) (p.9) Interpretation: the pipe and ring are heated or cooled together. That’s why we’re given 2α . Q1: Must they be cooled or heated? A1: They must be cooled because if heated the outer radius of the pipe will keep getting bigger and at a faster rate than the inner radius of the ring.

A: Let CT i o25)( = be the initial temp. and )( fT be the final temp. that we are looking for. Since we know that it must be cooled we know 0)()( <−=∆ if TTT .

Let 0.10)( =iringr cm and 001.10)( =i

piper cm be the initial inner radius of the ring and initial outer radius of

the pipe. The change in ringr is Trr iringring ∆=∆ )(

1α and for the pipe is Trr ipipepipe ∆=∆ )(

2α . Note that the coefficient of linear expansion measures the FRACTIONAL increase in length.

Thus, the final radii )( fringr and )( f

piper are given by ( ) ( )( ))()(1

)(1

)()()( 11 ifiring

iringring

iring

fring TTrTrrrr −+=∆+=∆+= αα and

( ) ( )( ))()(2

)(2

)()()( 11 ifipipe

ipipepipe

ipipe

fpipe TTrTrrrr −+=∆+=∆+= αα . Note that since 0<∆T , the radii decrease. The pipe starts

out bigger, the pipe and ring both shrink, but the pipe shrinks at a faster rate since 12 αα > . At some final temp., we will have )()( f

pipef

ring rr = . Solving for )( fT gives

CCCcmCcmC

cmcmC

rr

rrTT

iring

ipipe

iring

ipipeif ooo

oo

o 7510025)10)()(102()001.10)()(103(

10001.1025

1616)(1

)(2

)()()()( −=−=

×−×−−=

−−

−= −−−−αα

Remark: Let’s skip Problem 2. In this case, the radius of a sphere increases and you’re asked what happens to the volume.

Thermal Expansion

Kinetic Theory: Minilecture On pp.477-478 of the textbook is a classical “derivation” of the equipartition theorem for translational degrees of freedom in 3D by considering the change in momentum due to a particle bouncing against the container walls. I’m not going to discuss this derivation, which Professor Freedman might do in class. Instead, we will discuss what’s called the Maxwell distribution of speeds. I tend to favor taking this as more elementary and “deriving” the translational equipartion theorem from it as we will do in Problem 1. However, I should make one important remark. The v2 that appears in the equipartition is actually the square of the root-mean-square speed. Before we talk about the Maxwell distribution of speeds, let me just make the distinction between average speed and root-mean-square speed or rms speed. Average speed is just the usual mean: Nvvv N /)( 1 ++= L . The rms is Nvvvv Nrms /)( 22

12 ++== L . In general, for 2≥N ,

rmsvv ≤ . In fact, 222 σ+= vvrms where σ is the standard deviation. Try two random numbers and see. If you are unconvinced, check out example 18-4 (p.480) of the text.

Kinetic Theory

Maxwell Speed Distribution: Minilecture The Maxwell speed distribution is a function of v, the speed, which, when integrated from v1 to v2 gives us the number of particles in our sample whose speed lies between v1 and v2. As in your text, I won’t derive it for you because that would take us along a very long tangent: roughly the first 250 pages of my undergrad thermo textbook. The distr. Is

( ) kTmvev

kT

mNvf

/2

12

2/3 2

24

−

=π

π

Where N is the number of molecules, T is the temperature, m is the mass of each particle. Let me say a word or two about the expression. The exponential factor is called the Boltzmann factor. In general, the probability for a particle to occupy a particular state of energy Es is proportional to the Boltzmann factor kTEe /2− . Since the kinetic energy of the molecule in the “state with velocity v

r ” is 2)2/1( mv , the exponential measure the probability that a particle will have velocity v

r . Remark: the particle states here are distinguished by velocity, not speed. The number of velocity vectors that correspond to a single speed v grows like 2v (like the number of spatial vectors of a given length d scales as d2, like the surface area of a sphere of radius d). This explains the v2 factor. The remaining factors can be determined by ( ) Ndvvf =∫

∞

0 (# with speed between 0 and infinity is N).

Kinetic Theory

Discussion 1/2 (p.10) – In groups 1. Why do puddles evaporate, even if the temperature is much colder than the boiling

point of water? Why do sealed jars never evaporate? Even at the lower temperature, there are still some molecules with speeds high enough to break the surface tension and leave the liquid water. When the jar is sealed, eventually, just as many gas molecules reenter the liquid as fast liquid molecules leave the liquid. 2. Plot a typical Maxwell Distribution for some value of N and T. What would it look like

if you increased the temperature, keeping N constant. What if we increase N with T constant.?

Increase T at constant N � the most probable speed increases. The amplitude decreases since the total integral must be the same (N) Increase N at constant T � just increase the amplitude throughout.

Kinetic Theory

Problem 1 (p.11) – In groups 1a) # particles with speed between v and v + dv is ( )dvvf . 1b) Finite: Ni particles with speed vi with i = 1, 2 … Then the average of v is

( ) NvNvi ii /∑= where N is the total number of particles. Generalize to ( ) Ndvvvfv /

0

= ∫

∞ . For a

function of v we simply have ( ) ( ) ( ) Ndvvfvgvg /0

= ∫

∞ .

1c) By definition, ( ) Ndvvf =∫

∞

0 and so the average of 1 is indeed 1.

1d) The average of the kinetic energy is

( ) kTm

kT

kT

mmdvev

kT

mmNdvvfvmKE vkTm

2

32

8

3

22

24

2/

2/52/3

0

)2/(42/3

0

221

2

=

=

=

= ∫∫

∞ −∞ ππ

ππ

π

1e) Yes, 1(d) agrees with equipartition.

Kinetic Theory

Problem 2 (p.11) – In groups 2a) smv /8.1210/)1715)14(2)12(4)10(2( =++++= and smvrms /97.1210/)289225)196(2)144(4)100(2( =++++= 2b) kTmvrms )2/5()2/1( 2 = and so kmkmvT rms /64.335/2 == 2c) mmmvNE rms 841)97.12()2/1(10)2/1( 22

int === 2d) There are so few particles.

Kinetic Theory

Discussion 1-3 (p.12) – In groups 1) Is it possible for H2O to be in liquid form at 0 oC?

Yes. You must take out more energy to actually make ice.

2) Glass of water at 0 oC. Is it possible to draw heat out without lowering its temperature? Where does the heat energy come from? Yes. The heat comes from making bonds between water molecules (to form ice).

3) Why can you get a more sever burn from steam at 100 oC than water at same temp.? Because the steam has more energy that of the water plus latent.

SKIP 4 until end; time permitting.

Calorimetry

Calorimetry: Minilecture Write the question on the board: When you add heat to a gas, does the temperature go up? And take a vote. Then, draw an isochoric and an isothermal process on a pV diagram: Both of these have heat added to the gas. But, the isochoric Process has increase in T whereas there is no change for the Isothermal process. It is also actually possible to add heat and DECREASE the temperature! But we won’t discuss this now. Write next to the question: IT DEPENDS! For a gas, there is no fundamental relationship between heat and Temperature, or heat and internal energy. What’s missing? WORK! Now write the first law: WQE −=∆ int . This W can be almost anything and so the answer the question is that it depends. But, what if you had a material that couldn’t expand much? Work is negligible! This is the case for most solids and liquids. So, most of the heat goes directly into raising the internal energy or temperature, reflected in the formula TmcQ ∆= . Explain the quantities.

Calorimetry

Problem 1 (p.12) – In groups

Calorimetry


Thermal Expansion Kinetic Theory Coeff. of Thermal Linear

Expansion - ! is the fractional length change.

Coeff. of Thermal Vol. Expansion - ! is fractional vol. change.

TΔα

TT Δ≈Δ αβ 3

Maxwell Distribution

! The integral from v1 to v2 is the # of particles whose speed is ! .

kTmvevkTmNvf /2

2/32

21

24)( −

⎟⎠

⎞⎜⎝

⎛=π

π

21 vvv ≤≤

Three key speeds ! ! ! Means of functions

!

mkTmkTvp /41.1/2 ≈=

mkTmkTv /60.1/8 ≈= π

mkTmkTvrms /73.1/3 ≈=

∫∞

=0

)()(1)( dvvfvgN

vg

Problem 1 (p.11) – Clarification of (b) and (d) 1a) # particles with speed between v and v + dv is ! .

1b) Finite: Ni particles with speed vi with i = 1, 2 … Then the average of v is ! where N is the total number of particles. Discretize the v axis by pieces of length dv. Let Ni be the number of particles with speed between v + idv and v + (i+1)dv. From part (a), this is roughly ! and ! . The average speed is

! , which in the continuum limit is ! .

For a function of v we simply have ! .

1d) The average of the kinetic energy is

!

Used: ! .

1e) Yes, 1(d) agrees with equipartition.

( )dvvf

( ) NvNvi ii /∑=

dvvfdvidvvfN ii )()( =+= dvvfvvN iiii )(=

( ) ( ) NdvvfvNvNvi iii ii /)(/ ∑∑ == Ndvvvfv /)(

0∫∞

=

( ) ( ) ( ) Ndvvfvgvg /0

⎟⎠⎞⎜

⎝⎛= ∫

∞

( ) kTmkT

kTmmdvev

kTmmNdvvfvmKE vkTm

232

83

22

24

2/

2/52/3

0

)2/(42/3

0

221 2

=⎟⎠

⎞⎜⎝

⎛⎟⎠

⎞⎜⎝

⎛=⎟⎠

⎞⎜⎝

⎛=⎟⎠⎞⎜

⎝⎛= ∫∫

∞ −∞ π

ππ

ππ

2/5

0

4 8/32

adxex ax π=∫∞ −

Kinetic Theory

Problem 2 (p.11) – In groups

2a) ! and !

2b) ! and so !

2c) !

2d) There are so few particles.

smv /8.1210/)1715)14(2)12(4)10(2( =++++= smvrms /97.1210/)289225)196(2)144(4)100(2( =++++=

kTmvrms )2/5()2/1( 2 = kmkmvT rms /64.335/2 ==

mmmvNE rms 841)97.12()2/1(10)2/1( 22int ===

Kinetic Theory

Discussion 1-3 (p.12) – In groups 1) Is it possible for H2O to be in liquid form at 0 oC?

Yes. You must take out more energy to actually make ice.

2) Glass of water at 0 oC. Is it possible to draw heat out without lowering its temperature? Where does the heat energy come from? Yes. The heat comes from making bonds between water molecules (to form ice).

3) Why can you get a more sever burn from steam at 100 oC than water at same temp.? Because the steam has more energy that of the water plus latent.

SKIP 4 until end; time permitting.

Calorimetry

Calorimetry: Minilecture Write the question on the board: When you add heat to a gas, does the temperature go up? And take a vote. Then, draw an isochoric and an isothermal process on a pV diagram: Both of these have heat added to the gas. But, the isochoric Process has increase in T whereas there is no change for the Isothermal process. It is also actually possible to add heat and DECREASE the temperature! But we won’t discuss this now.

Write next to the question: IT DEPENDS! For a gas, there is no fundamental relationship between heat and Temperature, or heat and internal energy. What’s missing? WORK! Now write the first law: ! . This W can be almost anything and so the answer the question is that it depends. We won’t yet discuss exactly what heat and work are.

WQE −=Δ int

Calorimetry

But, what if you had a material that couldn’t expand much? Work is negligible! This is the case for most solids and liquids. So, most of the heat goes directly into raising the internal energy or temperature, reflected in the formula ! . Explain the quantities:

Calorimetry: Minilecture (cont.) Here m is the mass of the substance, c is a constant called the specific heat and ! is the change in temperature, which can be negative for cooling!

There is a broad class of problems wherein a certain number of materials at various temperatures are made to come into contact with each other and to settle at an equilib. Temperature. This is called calorimetry. The steps to solve them are (text p.504)

(1) Is the system isolated, or sufficiently nearly so? (Isolated means no interaction with the rest of the universe – no matter or energy transfer).

(2) ! for each subsystem i. Note, by conservation of energy, some of these Q’s will be negative!

(3) If no phase changes occur, then just use ! where ! . (4) If phase changes occur, additional terms of the form ! must appear. Be

careful of signs: ! (condensation or freezing) and ! (vaporization or melting). Determine in which phase the final state will be before #(2).

TmcQ Δ=

TΔ

0=∑i itoaddedQ

TmcQ Δ= if TTT −=Δ

|| mLQ ±=

0<Q 0>Q

Calorimetry

(5) Remember that the final temperatures of all the substances should be the same.

Note: there are differences between this and that on p.504 of the text!

Problem 2 (p.12) – Me on board Since we really want to get to heat transfer and the First Law on Wednesday, we’ll just do Problem 2. Part of the reason is that calorimetry questions are most difficult when phase transitions come in. However, Problem 1 is good practice because it has more than two combining substances. You can try to do Problem 1 on your own.

(1) We assume that the system is reasonably well isolated. (4) The ice melts and thus has ! added to it at ! . The steam condenses and so has ! added to it (i.e. ! taken away) at ! . (3) The ice is heated from ! to ! . Thus ! is added to the ice to get it to the melting point (note: since ! , ! ). Then, it is melted to water. The mass does not change and is still ! even though it’s now water. This water is heated to ! and thus there is a heat input of ! . The steam does not first cool since it’s already at ! . It is liquefied and then the water is cooled to ! with a heat output of ! .

ficemelt LmQ = C�0

vsteamcondense LmQ −= vsteamLm C�100

CT iice�10−= C�0

iiceiceiceice TcmQ −=

0<iiceT 0>iceQ

icem CT f �50=f

watericewater TcmQ =)1(

C�100

CT f �50= )100()2( −= fwatersteamwater TcmQ

Calorimetry

(2) Then, ! . The only unknown is

! . Solving gives ! .

Heat Flow by Conduction (Minilecture) You should all be intimately familiar with the topic of heat flow. You are constantly maintaining an average temperature commensurate to the rate of heat flow out of your body. Shivering when it’s cold or sweating when it’s hot are just two methods. You might also be familiar with the sensation of getting burned, for example by accidentally touching a metal baking tray that has been in the oven. Presumably the air in the oven is just as hot and yet you don’t normally get burned by it. Some of these very questions will hopefully be fleshed out by the discussion problems. More quantitatively, the conduction of heat through a material is described via the heat conduction equation and by a “constant” peculiar to that material called the thermal conductivity. The picture to keep in mind here is a rod whose ends sit at different temps.:

I like the remember the differential form: ! (text p.516). Here the variable x is a position variable across the rod from hot to cold and Q is heat and is positive flowing from hot to cold, k is the thermal conductivity and A is the cross-sectional area.

0)100( =++−−+− fwatericefice

iiceiceice

fwatersteamvsteam TcmLmTcmTcmLm

steammg

TcLTcLTcm

m fwaterv

iiceicef

fwaterice

steam 34)100()(=

−−

−+=

dxdT

kAdtdQ

−=

Heat Flow by Conduction

We will always assume steady state. This does not mean that the temperature is the same throughout the rod. It simply means that the temperature at a given point in the rod does not change in time. This means that the heat flow rate into that point and out of that point are the same! We will return to this point again and again.

DP 1 (p.13) Q: You are standing in your bathroom with bare feet, one foot on the tile floor, and the other on a rug. You notice that the tile feels colder than the rug. Are they not at the same temperature? Explain.

A: The skin does not really sense temperature, per se. Instead, it is more like a sensor of heat flow rate. A high rate of heat flow outwards through the skin feels “cold” and a high rate of heat flow inwards feels “hot”. This is why a hot shower feels “hotter” initially than it does after a while of getting equilibrated. When your skin is closer to the temp. of the water, the heat flow rate in through the skin decreases and hence the shower no longer feels so “hot”. Therefore, the tile feels colder because, since it has a higher thermal conductivity (~0.84 J/smCo for concrete/brick?) than does rug (0.024 J/smCo for polyurethane?).


DP 2 (p.13) Q: You are able to reach into a hot oven without getting burned, but you will be burned in your hand brushes the metal rack or a baking dish inside. Explain.

A: Same reason. The thermal conductivity of air (0.023 J/smCo) is much lower than that of glass (0.84 J/smCo) or metal (200 J/smCo for Aluminum). Thus, even though the air and the metal or glass have the same temperature, less heat is transferred through the skin via the hot air than via the hot glass or metal. Even a brief exposure to the hot metal or glass can transfer a large amount of heat and raise the local temperature of the tissue to the point of causing damage.


DP 3 (p.14) A wooden rod has length L and cross-sectional area A. One end of the rod is kept at 200 oC by an oven. The other end of the rod is kept at 0 oC by a refrigerator. Because of the temperature difference across the rod, heat flows through the rod. As usual, we assume that everything has settled into a steady state, so that the temperature distribution in the rod is not changing with time. We suppose that the rate of heat flow is 50 J/s.

Q: Will the temperature at the rod midpoint be greater than 100 oC, less than or equal? Neglect thermal expansion subtleties. Take a vote.

A: Equal to 100 oC. Divide the rod into two halves. Since we are in the steady state, the heat flows in the two half-rods must be the same. Since the two half-rods obviously have


the exact same thermal conduction properties (A, l, k), it follows that the temperature difference between the ends and the midpoints must be the same. Hence, the center temperature must be the average of the two endpoints.

We can also see this from the differential equation ! (explain this, especially the sign, briefly). In our case, H is constant (50 J/s). Thus, ! , a constant and this is solved by a linear function.

DP 4 (p.15) Q: Now suppose that the original wooden rod is replaced by another wooden rod with the same length L, but with cross-sectional area 2A.

(a) At what rate will heat flow through this new rod? (Numerical answer) (b) Will the midpoint temperature be greater than 100 oC, less than or equal? Vote.

A: (a) Heat flow rate is now H = 100 J/s because ! and the only thing that has changed is A, which was doubled. Hence H is doubled. All other things equal, increasing the area to 2A is just like putting up a second rod of area A. Each has heat flow rate 50 J/s and so the two together give a total flow rate of 100 J/s.

dxdTkAH

dtdQ

−==

CdxdT −=/

dxdTkAH

dtdQ

−==


(b) Still at 100 oC by the same argument as before.

DP 5 (p.15) Q: Suppose that the wooden rod is replaced by a metal rod with cross-sectional area A and length L.

(a) Will the midpoint temperature be greater than 100 oC, less than or equal? Vote. (b) Will the heat flow rate be greater than 50 J/s, less than or equal? Vote.

A: (a) Still 100 oC by the same reason. (b) The heat flow rate will be greater by a factor ! . woodmetal kk /


DP 6 (p.15-16) I switched (b) and (c) around.

Q: Suppose that the metal rod is replaced by a composite rod, with cross-sectional area A and length L. The “hot” half is metal, the “cold” half is wood.

(a) How does the heat flow rate into the junction compare toe the flow rate out? (b) How will the heat flow rate through the wood part compare to the rate of heat

flow through the metal part? (c) Will the heat flow rate be greater than 50 J/s, less than or equal? (d) Will the midpoint temperature be greater than 100 oC, less than or equal? Vote.


A: (a) Steady state, the junction cannot be changing temperature. Thus, the heat flow in must be equal to the heat flow out. (b) The heat flow rate must be equal in the wood and metal at steady state. (c) Greater than 50 J/s but less than ( ! )(50 J/s). (d) The midpoint temperature must be higher than 100 oC in order for the heat flow rate in the metal and wood to be equal.

Problem 1 (p.16) Q: For the composite rod of DP 6 (p.15-16), each section has length 0.75 m and cross-sectional area 4 cm2. Suppose the thermal conductivities of wood and metal are J/smCo and J/smCo.

(a) Find the midpoint temperature. (b) Find the heat flow rate across the composite rod.

A: The heat flow through the metal is ! and through the wood is

! .

woodmetal kk /

1=wk

14=wk

lTT

AkHdtdQ hotmidmmm

−−==)/(

lTT

AkHdtdQ midcoldwww

−−==)/(


(a) Setting these two equal (at steady state) and solving for ! yields

!

(b) Plugging back in to either Hm (or Hw) and simplifying yields

! J/s or W

Problem 2 (p.16) Q: A container of water has been outdoors in cold weather until a 5cm-thick slab of ice has formed on its surface. The air above the ice is at -10oC. Calculate the rate of formation of ice (in cm/hr) on the bottom surface of the slab. Assume the walls of the container are thermally insulating. Data: kice = 1.7 W/m K 0.92 g/cm3 kJ/kg

A: Let the water at the lower ice surface be at ! and let ! . Let the area of the ice slab be A (this will drop out). Let the present thickness be ! m. All the heat taken

midT

CkkTkTk

Twm

coldwhotmmid

�187=+

+=

1.011)(1

=⎟⎟⎠

⎞⎜⎜⎝

⎛+

+=

−

wm

coldhot

kklTTA

H

=iceρ 333=fL

CTin�0= CTout

�10−=

05.0=l


out of the water turns it into ice since it’s already at 0oC. Thus, and so where m is the mass of water that is frozen and dm/dt is the rate of freezing in kg/sec. We

know that ! and so ! . The rate of freezing in volume/sec is

just ! . Finally, the rate of freezing in length/sec, that is, the rate

at which the ice thickness is increasing is ! cm/hr.

Heat Flow by Radiation - Minilecture Empirically, the rate of heat flow radiated out of an object at temp. T is given by ! where ! , the emissivity, is a number between 0 and 1 and measures how well a substance absorbs and emits, W/m2K4 is the Stefan-Boltzmann constant, and A is the surface area of the emitting object. The rate at which a substance absorbs radiation of a given power flux S is ! . Here S has units of power/area. For the radiation from the sun reaching the earth, S is

fmLQ = fLdtdm

dtdQ

=

lTT

AkdtdQ inout

ice

−−= )( inout

f

ice TTlLAk

dtdm

−−=

)(1inout

fice

ice

ice

freeze TTlLAk

dtdm

dt

dV−−==

ρρ

4.0)(1=−−== inout

fice

icefreeze TTlL

kdt

dV

Adtdl

ρ

4/ ATdtdQPrad εσ== ε81067.5 −×=σ

ASPin ε=

Heat Flow by Radiation

given in your workbooks as 1350 W/m2. We will actually compute this number given known quantities in Problem 1.

DP 1 (p.17) Q: For a spherical blackbody, describe the distribution of emitted radiation? A: It is spherically symmetric. That is, it is constant: just total emitted radiation divided by the surface area of a unit sphere.

DP 2 (p.17)


Q: Consider a sheet of metal. How does the power emitted on one side compare to the power emitted on the other? What about if you paint one side black and the other side white?

A: Assuming that the faces are at the same temperature, the power emitted on either side should be equal. Black things tend to have a higher ! (emissivity) than white things and thus the black side will emit more radiation than the white side (good absorbers are also good emitters).

Issue: You might think that the black side cools faster than the white side and so we never have thermal equilibrium. This is okay because the black side also absorbs more radiation than the white side. This difference powers the temperature gradient.

DP 3 (p.17) Q: Why is it better to wear a white shirt than a black shirt on a hot day out in the sun?

A: White reflects more of the radiation away. Black absorbs a lot of the radiation and maintains a higher temperature.

ε


DP 4 (p.17) Q: Given a system in steady state, how does the power absorbed compare to the power emitted?

A: They are equal. Otherwise, heat will build up or drain out thus changing the temperature.

DP 5 (p.18) Q: For the power absorbed by the earth from the sun, what area should we use? Why?

A: Use the cross-sectional area of the earth ( ! ) because this is the area exposed to the direct radiation.

2rπ


DP 6 (p.18) Q: Given two bodies giving off the same total power, how do their temperatures compare if one body has four times the emissivity of the other? How about if one body has twice the radius of the other?

A: If ! then ! . If ! , then ! and so ! .

DP 7 (p.18) Q: A body at temperature T1 is immersed in a heat bath at temperature T2. What is the net rate of heat loss due to radiation in this case?

21 4εε = 2/21 TT = 21 2rr = 21 4AA = 2/21 TT =


A: Power emitted is ! where ! is the emissivity of the body. The power flux due to the environment is ! where ! is the emissivity of the environment (answers seem to say ! ). and so the power absorbed by the body is ! . Thus, the net rate of heat loss due to radiation is ! .

Issue: in the text Eq. 19-18 (p.518), there is no ! factor. Why?

Problem 1 (p.18) Q: Consider a simplified version of the Earth-Sun system in which both bodies are perfect blackbodies at uniform temperatures and in a steady state.

411 ATPout σε= 1ε

422/ TAPS env

rad σε== 2ε

12 =ε 42211 ATSPin σεεσε ==

)( 422

411 TTAPPP inout εσε −=−=

2ε


(a) Given the radius of the earth’s orbit, r0, the radius of the sun, rs, and the temperature of the sun, Ts, find the solar constant.

(b) Using ! K, ! km, ! km, find S. (c) Why is your answer for S different from the one quote at the beginning of this

section? (d) Given the earth radius ! km, calculate the average earth temperature.

A: (a) ! . By the time this reaches the earth, it will have spread evenly across a sphere of radius ! . Thus, ! . (b) Thus, plug it in. We get S = 1360 W/m2 (c) ? I do get roughly 1350? (d) ! and ! . In steady state, these are equal. Thus, ! K!

Problem 2 (p.18) Q: An object with surface area A is placed in an oven which is maintaining temp. Toven. The object has a specific heat c, a mass m, and initial temp. T0. Ignore any heat transfer

5770=sT6

0 106.149 ×=r 51096.6 ×=sr

31038.6 ×=Er

424 sssunrad TrP πσ=

0r42

020 )/(4/ ss

sunrad TrrrPS σπ ==

SrP Eearthabs

2σπ= 424 eEearthrad TrP πσ= 278)4/( 4/1 == σSTE


by conduction or convection, and assume the oven system is isolated from the environ. Also assume that the time for the objects to reach a stead state is much faster than any other time scales in the problem (assume steady state). At time t the object has temperature T(t).

(a) At time t, what is the net rate of heat gain by the object? (b) Find the temperature of the object at a time t.

A: (a) From DP 7 (p.18), ! .

(b) ! and so ! . Let ! . Solve the homogeneous equation without the Toven. This gives

!

))(( 44 tTTAdtdQ

oven −= εσ

TmcQ Δ=))((1 44 tTT

mcA

dtdQ

mcdtdT

oven −==εσ

mcAB /εσ=

tBTT

BttT oven4

3/1

20313)( +⎟

⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛+=

−


Heat Flow by Radiation Power Radiated

(Stefan-Boltzmann Law):

4ATPdt

dQrad

rad εσ==

Power Absorbed: S = rad. power being abs. per unit area of absorber.

ASPdt

dQabs

abs ε==

Administratives: Keith Licardo and Jennifer Hsu were dropped because they were absent on the first day. If they still want to be in the course, they have to add it again by this Friday the 6th.

DP 7 (Review) (p.18) Q: A body at temperature T1 is immersed in a heat bath at temperature T2. What is the net rate of heat loss due to radiation in this case? A: Power emitted is 4

11 ATPout σε= where 1ε is the emissivity of the body. The power flux due to the environment is 4

22/ TAPS envrad σε== where 2ε is the emissivity of the environment

(answers seem to say 12 =ε ). and so the power absorbed by the body is 42211 ATSPin σεεσε == .

Thus, the net rate of heat loss due to radiation is )( 422

411 TTAPPP inout εσε −=−= .

Issue: in the text Eq. 19-18 (p.518), there is no 2ε factor. Why? Resolution: the other GSI’s, I think, now believe I’m correct. The head GSI said that she will post something on bspace indicating this. But this is only true if you think of the environment as a smaller system within the whole universe. But if for the environment we take the whole universe, then, really we must set 1=ε because there is nowhere else for the radiation to go. All the radiation that is re-emitted by the object must be completely absorbed by the whole universe.


Problem 1 (p.18) Q: Consider a simplified version of the Earth-Sun system in which both bodies are perfect blackbodies at uniform temperatures and in a steady state.

(a) Given the radius of the earth’s orbit, r0, the radius of the sun, rs, and the temperature of the sun, Ts, find the solar constant.

(b) Using 5770=sT K, 60 106.149 ×=r km, 51096.6 ×=sr km, find S.

(c) Why is your answer for S different from the one quote at the beginning of this section?

(d) Given the earth radius 31038.6 ×=Er km, calculate the average earth temperature. A: (a) 424 ss

sunrad TrP πσ= . By the time this reaches the earth, it will have spread evenly across a

sphere of radius 0r . Thus, 420

20 )/(4/ ss

sunrad TrrrPS σπ == .

(b) Thus, plug it in. We get S = 1360 W/m2 (c) ? I do get roughly 1350? (d) SrP E

earthabs

2σπ= and 424 eEearth

rad TrP πσ= . In steady state, these are equal. Thus, 278)4/( 4/1 == σSTE K!


Problem 2 (p.18) Q: An object with surface area A is placed in an oven which is maintaining temp. Toven. The object has a specific heat c, a mass m, and initial temp. T0. Ignore any heat transfer by conduction or convection, and assume the oven system is isolated from the environ. Also assume that the time for the objects to reach a stead state is much faster than any other time scales in the problem (assume steady state). At time t the object has temperature T(t).

(a) At time t, what is the net rate of heat gain by the object? (b) Find the temperature of the object at a time t.

A: (a) From DP 7 (p.18), ))(( 44 tTTA

dt

dQoven −= εσ .

(b) TmcQ ∆= and so ))((

1 44 tTTmc

A

dt

dQ

mcdt

dToven −== εσ . Let mcAB /εσ= . Solve the homogeneous

equation without the Toven. This gives

tBTT

BttT oven4

3/1

203

13)( +

+=

−


First Law - Minilecture Remember that on Monday, we briefly introduced the First Law to describe the more complicated relationship between heat and temperature, or, equivalently, heat and internal energy for a gas. The new and important term that obstructed a direct relation between heat and temperature or internal energy was WORK and the fact that a gas is capable of doing a lot of work on the environment or experiencing work done on it by expanding and contracting by a significant amount. This is quite a different behavior than that exhibited by liquids and solids. However, in principle, the first law applies to liquids and solids as well since, as we will see, the first law is just a statement of conservation of energy. Now, you could look at a gas and tell me how much internal energy it has. *Draw a box of gas with particles whizzing around. You could just look at one particle, figure out its energy, and do the same for all the particles, then add it up. But, even if that were practically possible, it’s certainly unwieldy. So, we opt for the next best thing: a more statistical average viewpoint. Physically, we might stick in a thermometer and just measure the temperature. Of course, equipartition does the rest: NkTdE )2/(int = . Here N and T are STATE VARIABLES, which are just the variables that describe the system intrinsically.

First Law of Thermo

First Law – Minilecture (cont.) Since Eint is a function of state variables it is called a STATE FUNCTION. If the gas is not ideal, we might not have such a simple relationship between Eint and N and T. But, nevertheless, we expect that if we know P and V, as well, we should also be able to figure out what Eint is. So, in general, Eint is a state function of all of these state variables. Now, let me ask you a question. “How can we increase the internal energy of a gas? Or thus, how can we make the particles speed up?” Maybe, “Add heat to it.” Maybe, “Do work on the gas by compressing it.” This commonsense is stated in the first law:

byadded WQE −=∆ int

So, to increase the internal energy of the gas, you have to add heat to it or do work on it by compressing it or both, or some delicate balance of adding heat and letting it do a little work by expanding. Note on signs: You may often see the first law without all the subscripts. You should always be sure of the sign conventions because different people and texts use different ones. Here, by Q we will mean heat ADDED to the gas. If positive heat is taken AWAY from the gas then it will just be negative. The work is defined to be that done BY the gas. If positive work is done ON the gas then it will just be negative work done BY the gas.

First Law of Thermo

Qadded Wby Eint

First Law – Minilecture (cont. 2) While Eint is a state function, work and heat are not! You cannot look at a gas and tell me how much work or heat it has. That doesn’t even make sense. Heat and work only make sense in the context of energy TRANSFERS and are thus properties of transformations, not of the gases themselves. So now, let’s describe some of these transformation processes that involve heat input or output and work. Firstly, it should be clear by the arrows diagram that the First Law is just a statement of energy conservation. But you’re probably very comfortable with conservation laws where only two expressions show up: for example, for an isolated system consisting of two subsystems, the energy gained by one is the energy lost by the other – just TWO terms. The First Law has three terms!!! So, the easiest way to begin understanding it is just considering situations where one or more vanish. To this end, we consider isothermal, adiabatic, and isovolumetric processes, which respectively get rid of intE∆ , Q and W. Firstly, here’s a summary of what might be a standard strategy for analyzing these transformations. DISCUSS BIG POST-IT.

First Law of Thermo

Isothermal Process – Minilecture For an isothermal process, 0=∆T and thus 0int =∆E (why?). Here we find the first instance of the quasistatic assumption: we assume that the process is sufficiently slow so that sufficient enters or leaves the gas to maintain constant T at each time step. One way to accomplish this process is to have a piston compressed or expanded slowly in a very large heat bath (DRAW THIS ). The first law then says byin WQ = . We understand this physically as just energy conservation: for example, if the gas is expanding, then it is doing work, which means that the particles must be slowing down and so the temperature should drop. However, if the temperature is to remain the same (as in an isothermal process), then the heat intake must just be enough to compensate.

First Law of Thermo

Adiabatic Process – Minilecture For an adiabatic process, Q = 0. Physically, this is accomplished by compression or expansion in an extremely well insulated environment (thermos) so as to get rid of heat loss or gain, or the process can be very fast (often cannot even be drawn on a PV diagram). The first law then says byWE −=∆ int . Wby > 0 � expansion Wby < 0 � compression

Isovolumetric Process – Minilecture For this process, Wby = 0. Thus, inQE =∆ int .

Isobaric Process – Minilecture For this process, VPWby ∆= and so VPQE in ∆−=∆ int .

First Law of Thermo

DP 1 (p.19) – in groups Q: When an ideal gas undergoes adiabatic expansion, the temperature _________?

Would things differ if the gas underwent adiabatic compression? A: goes down. Have them draw this on the board. This is because Qin = 0 while Wby > 0 and so 0int <−=∆ byWE . Thus, 0<∆T . Physically, no heat is going in, but since the gas is doing work, it must be losing energy. If it was compressed, then Wby < 0 and so 0int >−=∆ byWE . Thus, 0>∆T . T goes up.

DP 2 (p.19) – in groups Q: When an ideal gas undergoes isothermal compression, the internal energy _______? Would things differ if the gas underwent isothermal expansion? A: stays the same. Since isothermal, 0=∆T and so 0int =∆E . Same if expansion.

First Law of Thermo

DP 3 (p.20) – in groups Q: When an ideal gas undergoes isothermal expansion, ____________? Compression? A: heat flows into the gas. 0>= byin WQ since expansion.

We understand this physically as just energy conservation: for example, if the gas is expanding, then it is doing work, which means that the particles must be slowing down and so the temperature should drop. However, if the temperature is to remain the same (as in an isothermal process), then the heat intake must just be enough to compensate.

DP 4 (p.20) – in groups Q: If heat is added to a gas while the gas is held at constant volume, T ________? What about constant pressure? A: must increase. No work, so 0int >=∆ inQE . Physically, the added heat can either go into increasing the internal energy or is expended as work via expansion. But, there is no expansion, hence, the heat must go entirely into raising the temperature of the gas. Now, the gas may expand and so VPQE in ∆−=∆ int , which is less than just inQE =∆ int .

First Law of Thermo

DP 5 (p.20) – in groups Q: When you let air out of a tire, the air feels cool. Explain. A: Since air is poor thermal conductor, this process is approximately adiabatic. Hence, no heat is lost or gained by the released air. However, it goes from higher to lower pressure (typical car tire pressures are around 2-3 atm), thus it expands when it exits doing work on the surrounding air. By DP 1 (p.19), this air must cool down. This explains some common phenomena. Have you ever opened a can of soda and seen what sort of looks like fog emerge? Again, the gas was at a higher pressure before the can was opened, it expands adiabatically and thus cools. If the cooling is sufficient, some of the gas condenses into “fog.” This also explains why mountain air tends to be cool despite the fact that hot air rises. Air currents carry air up the mountainside to a lower pressure. The air expands approximately adiabatically and thus cools.

First Law of Thermo

Problem 1 (p.21) – in groups First : with NkT

dE

2int = , the equation defining an adiabat is .constPV =γ where d

d 2+=γ .

Q: An ideal gas of N diatomic molecules undergoes three consecutive transformations: (1) isobaric expansion from V0 to 5V0; (2) adiabatic expansion; (3) isothermal comp.

(a) In terms of P0, V0 and N, find the temperatures Ti at the three corners. (b) In terms of V0 and γ , find the volume V3 at corner 3. (c) In terms of P0 and γ , find the pressure P3 at corner 3. (d) For each of the transformations, find the change in internal energy. (e) For each transformation, find the work (by gas) in terms of P0 and V0. (f) For each transformation, find the amount of heat flow into the gas.

A: (a) Nk

VPTT 00

31 == and 100

2 5)5(

TNk

VPT ==

(b) Adiabat gives γγ3300 )5( VPVP = and isotherm gives 3300 VPVP = . Dividing these two and solving for V3

gives 00)1/(

3 2805 VVV ≈= −γγ . (c) Plugging back to isotherm gives 280/5 00

)1/(3 PPP ≈= −− γγ .

(d) 001221 10)(2

5VPTTNkE =−=∆ → and similarly, 002332 10)(

2

5VPTTNkE −=−=∆ → and 013 =∆ →E .

(e) 0000021 4)5( VPVVPW =−=→ and 003232 10 VPEW =∆−= →→ and 000030113 6.55ln2

7)/ln( VPVPVVNkTW −≈−=−=→ .

(f) 00212121 14 VPWEQ =+∆= →→→ and 032 =→Q and 001313 6.5 VPWQ −== →→ .

First Law of Thermo

Problem 2 (p.21) – in groups Q: One mole of monoatomic gas undergoes: (1) isothermal expansion from V1 to 3V1; (2) isochoric cooling from P2 to P3; (3) adiabatic compression back to initial state.

(a) Find P2 and P3 in terms of P1, V1 and γ . (b) Find T1, T2 and T3 in terms of P1, V1 and γ . (c) What is the direction of heat flow at each step? (d) Find the amount of heat flow into the gas during steps 1 and 2 in terms of P1, V1 and γ . (e) Is the net work done by the gas during the cycle pos. or neg.? Is the net heat pos. or neg.? (f) Do you think this cycle represents a heat engine or a refrigerator? Why?

A: (a) )3( 1211 VPVP = � 3/12 PP = . Now, γγ )3( 1311 VPVP = and so 113 16.03 PPP ≈= −γ . (b) RVPTT /1121 == (since n = 1). 11

1133 48.03/)3( TTRVPT ≈== −γ .

(c) Step 1: in. Step 2: out. Step 3: 0. (d) 11111112121 1.13ln)/3ln( VPVPVVRTWQ ≈=== →→ and 11

1112

3232

32121 78.0)13()( VPVPTTREQ −≈−=−=∆= −

→→γ .

(e) 032.0 11 >≈= VPQW netnet since 0=∆ netE since 0=∆ netT . Talk about integrals here. (f) Heat engine since heat in and work out. Refrigerator would be work in heat out.

First Law of Thermo

Problem 3 (p.22) – in groups First: Area of ellipse is abπ where a and b are the semimajor and semiminor axes. Q: A monatomic ideal gas undergoes a cyclic transformation (see diagram in workbook).

(a) When the gas goes form A to B, what is the change in internal energy? (b) When the gas goes form A to B, how much work is done by it on its environment? (c) How much heat flows into the gas during the transformation A � B? (d) Answer the same questions for the return transformation B � A. (e) What is the net work done by the gas on its environment over the cycle?

A: (a) NkVPTA /)2( 00= and AB TNkVPT 5/)5)(2( 00 == . Thus, 002

3 12)4( VPTNkE ABA ==∆ → . (b) Area under upper curve: 00002

100 )8()2()4)(2( VPVPVPW BA ππ +=+=→ .

(c) 00)20( VPWEQ BABABA π+=+∆= →→→ . (d) 0012 VPEE BAAB −=∆−=∆ →→ . Area under lower curve: 00002

100 )8())2()4)(2(( VPVPVPW AB ππ +−=−−=→ . Then,

00)20( VPWEQ ABABAB π+−=+∆= →→→ . (e) 002 VPWWW ABBAnet π=+= →→ .

Challenge Problem 4 (p.23) – in groups Sol. In workbook.

First Law of Thermo


Calorimetry Conduction Radiation When negligible or no

expansion/compression: TmcQ ∆= (solid/liquid)

Or mLQ = (phase trans.) TCQ V∆= (gas-isochoric)

dxdTkA

dtdQ

−=

4ATP

dtdQ

radrad εσ==

ASPdt

dQabs

abs ε==

First Law: byin WQE −=∆ int Isothermal: 0=∆T ! 0int =∆E and PV = const. (ideal gas) Adiabatic: 0=Q ! .constPV =γ where dd /)2( +=γ and d = # q.e.d.o.f. Isovolumetric: 0=∆V ! 0=W and ./ constTP = Isobaric: 0=∆P ! VPW ∆= and ./ constTV = General work: d ∫=→

B

A

V

VBA PdVW

Section 6 – Mon., Feb. 9, 2009

Things to write on the board at the start z

zrzrqF

outin

annulus ˆ)/(1

1)/(1

12 22

0

+−

+=

εσ!

rin = 0 (Disk): zzR

qFdisk ˆ)/(1

112 2

0

+−=

εσ! with R = disk rad.

rout = ∞ also (∞ plane): zqFplane ˆ2 0εσ

=!

y

yLyqQF linefinite

ˆ)2/(1

14 22

0 +=

πε

! rr

qQFpo ˆ4 2

0int πε=

! (Coulomb)

yy

qF lineiniteˆ1

2 0inf πε

λ=

!

z

zLzLzqQF loopsquare

ˆ)2/(21

1)2/(1

14 222

0 ++=

πε

!

Section 11 – Wed., Mar. 4, 2009

Finite Square Surface Charge Consider a square centered at origin of side length 2x And a slightly larger one of side length 2(x + dx). The Difference in area is xdxxdxxdxxdA 84)(484 222 =−++= where We drop all terms of higher than linear order in dx. This Little square loop has charge xdxdQ σ8= . This plays the Role of Q in the square loop case and x plays the role of L/2. So, we have z

zxzxd

zxzxqz

zxzxzxdxqFd ˆ

)/(21)/(

)/(1/2ˆ

)/(211

)/(11

48

220

2220 ++

=++

⋅=

πεσ

πεσ! . Integrating from

0/ == zxu to zLu 2/= gives, using the integral 222

21arctan1

121

uuu

udu+=

++∫ and the facts that

4/)1arctan( π= and 2/)arctan( π=∞ , we have

zzLqFsquare ˆ12

21arctan42

2

0

−

−=

πεσ! and so zqFplane ˆ

2 0εσ

=!

Setting up Integrals

Remark on Setting up Integrals Please keep safe the formulas that we have derived here, especially for the forces felt by a point charge some distance away from an infinite line charge or an infinite plane of charge. We will come back to them without rederiving them each time. I wanted you to know where these formulas came from and to see that you have the capacity to derive them by setting up the appropriate integrals by breaking up the objects into little differential bits, using Coulombs law on each bit, and integrating over all of the bits. There is often no substitution for this procedure and it in principle works in any situation, although you might have to compute some integrals numerically using a computer. Another reason why I wanted us to go through the painful process of actually computing some of these integrals is because, in particularly symmetric situations, we can use other techniques to compute these forces that require MUCH less work. Thus, you will learn to appreciate these tricks all the more if you have an idea of how much work it would take without them.


Minilecture Having spent so much time computing forces felt by point charges will hopefully save us a bit of time getting used to the idea of an electric field. One way to define the electric field produced by some charged body is as follows: imagine placing a very small positive charge q at some point in space around the charged body. This small positive charge will feel a force due to that charged body, which you could compute the same way we computed it for the disk or the line, etc. Then, the field is defined to be that force divided by the charge q of the small positive charge feeling the force. Since we’re dividing q out, the field is something intrinsic to the charged body producing the field and is independent of the small charge q that you used to probe the electric field. For example, from our previous work,

zzrzr

Eoutin

annulus ˆ)/(1

1)/(1

12 22

0

+−

+=

εσ! z

zREdisk ˆ

)/(111

2 20

+−=

εσ! zEplane ˆ

2 0εσ

=!

yyLy

QE linefiniteˆ

)2/(11

4 220 +

=πε

! yy

E lineiniteˆ1

2 0inf πε

λ=

! rr

QEpo ˆ4 2

0int πε=

!

zzLzLz

QE loopsquareˆ

)2/(211

)2/(11

4 2220 ++

=πε

! zzLEsquare ˆ12

21arctan42

2

0

−

−=

πεσ!

Electric Fields

Minilecture The virtue in the electric field is that it is a property only of the object creating the field. We can divide by q (charge of the probe) after we compute the force or we could just as well divide right at the very beginning in which case, if you think about it, we never really computed any force to begin with. Then, if you are given a point charge around a charged body and you are asked what is the force felt by the point charge, you just need to multiply the electric field produced by the charged body at the position of the point charge by the charge of the point charge: EqF

!!=

We draw electric field vectors as follows: consider a point around the charged body. The electric field at that point has magnitude and direction (it is a vector). Simply draw that vector as an arrow starting at the position in space pointing in the direction of the field and having length commensurate to the magnitude of the field. You can figure out the direction by imagining placing a small positive charge there and then determining which direction it would move.

For aesthetic reasons, we usually pick some symmetrically distributed bunch of points at which we draw the field vectors. We obviously cannot draw all vectors! Let’s do an example.

Electric Fields

DP 1 (p.49) – At the board Q: Sketch the electric field created by +1C and –2C point charges. A:

Electric Fields

DP 2 (p.49) – At the board Q: Sketch the electric field created a positive line charge. How would your picture look if the line were negatively charged? A:

Electric Fields

DP 3 (p.50) – At the board

Q: Sketch the electric field created a positive plane charge. How would your picture look if the line were negatively charged? A:

Electric Fields

DP 4 (p.50) – At the board Q: Sketch the electric field created two parallel charged planes, one positive and one negative. Remember that there are two field vectors, one due to each charged body and the net field vector is the sum of the two vectors (superposition principle). A:

Electric Fields

DP 5 (p.50) – At the board Q: Repeat q.4 if the sheets are both positive. A:

Electric Fields

Problem 2 (p.51) – Groups Q: Three point charges (+Q, +Q and –2Q) are equidistant, d, from one another.

(a) Sketch the net electric field vector at the center of the triangle. (b) Calculate the electric field at this point. The distance from each charge to the

center is 3/d . A: (a) (b) 20

34

121 d

QEE πε==!! and 20

64

13 d

QE πε=! . The dir. are ( )0,3,12

11 =d! , ( )0,3,12

12 −=d! and ( )0,1,03 −=d

! . Thus, ( ) ( ) ( ) ( ) yE

dQ

dQ

dQ

dQ ˆ0,1,00,3,10,3,1 20202020 4

23364

1213

41

213

41 ⋅=−⋅+−⋅+⋅= −

πεπεπεπε

! . Answer is ( ) ydQE ˆ

4323

20πε−−

=! .

Electric Fields

Problem 4 (p.52) – Groups Q: A point mass m with positive charge q is suspended from the ceiling by a thread. Nearby, a charged object causes the point charge q to deflect from the vertical. Find the angle of deflection in each case when the nearby charge is

(a) another point charge Q located a horizontal distance D away. (b) A large positively charged sheet, with σ and perp. Distance D away. (c) A long positively charged line, with λ horizontal D away.

A: (a) The gravitational force is ymgFg ˆ−= and the electric force is x

DqQFe ˆ

41

20πε

= . The

sum of these forces must by along the string or else the tension cannot possibly cancel them. Thus,

mgDqQ2

041tanπε

θ = .

(b) This time, xqFe ˆ2 0εσ

= and so mg

q02

tanεσ

θ = .

(c) This time, x

DqFe ˆ

2 0πελ

= and so Dmgq02

tanπε

λθ = .

Electric Fields

Problem 1 (p.21) – in groups First: with NkTdE

2int = , the equation defining an adiabat is .constPV =γ where d

d 2+=γ .

Q: An ideal gas of N diatomic molecules undergoes three consecutive transformations: (1) isobaric expansion from V0 to 5V0; (2) adiabatic expansion; (3) isothermal comp.

(a) In terms of P0, V0 and N, find the temperatures Ti at the three corners. (b) In terms of V0 and γ , find the volume V3 at corner 3. (c) In terms of P0 and γ , find the pressure P3 at corner 3. (d) For each of the transformations, find the change in internal energy. (e) For each transformation, find the work (by gas) in terms of P0 and V0. (f) For each transformation, find the amount of heat flow into the gas.

A: (a) NkVPTT 00

31 == and 100

2 5)5( TNkVPT ==

(b) Adiabat gives γγ3300 )5( VPVP = and isotherm gives 3300 VPVP = . Dividing these two and solving for V3

gives 00)1/(

3 2805 VVV ≈= −γγ . (c) Plugging back to isotherm gives 280/5 00

)1/(3 PPP ≈= −− γγ .

(d) 001221 10)(25 VPTTNkE =−=∆ → and similarly, 002332 10)(

25 VPTTNkE −=−=∆ → and 013 =∆ →E .

(e) 0000021 4)5( VPVVPW =−=→ and 003232 10 VPEW =∆−= →→ and 000030113 6.55ln27)/ln( VPVPVVNkTW −≈−=−=→ .

(f) 00212121 14 VPWEQ =+∆= →→→ and 032 =→Q and 001313 6.5 VPWQ −== →→ .

First Law of Thermo

Problem 2 (p.21) – in groups Q: One mole of monoatomic gas undergoes: (1) isothermal expansion from V1 to 3V1; (2) isochoric cooling from P2 to P3; (3) adiabatic compression back to initial state.

(a) Find P2 and P3 in terms of P1, V1 and γ . (b) Find T1, T2 and T3 in terms of P1, V1 and γ . (c) What is the direction of heat flow at each step? (d) Find the amount of heat flow into the gas during steps 1 and 2 in terms of P1, V1 and γ . (e) Is the net work done by the gas during the cycle pos. or neg.? Is the net heat pos. or neg.? (f) Do you think this cycle represents a heat engine or a refrigerator? Why?

A: (a) )3( 1211 VPVP = ! 3/12 PP = . Now, γγ )3( 1311 VPVP = and so 113 16.03 PPP ≈= −γ . (b) RVPTT /1121 == (since n = 1). 11

1133 48.03/)3( TTRVPT ≈== −γ .

(c) Step 1: in. Step 2: out. Step 3: 0. (d) 11111112121 1.13ln)/3ln( VPVPVVRTWQ ≈=== →→ and 11

1112

3232

33232 78.0)13()( VPVPTTREQ −≈−=−=∆= −

→→γ .

(e) 032.0 11 >≈= VPQW netnet since 0=∆ netE since 0=∆ netT . Talk about integrals here. (f) Heat engine since heat in and work out. Refrigerator would be work in heat out.

First Law of Thermo

Problem 3 (p.22) – in groups First: Area of ellipse is abπ where a and b are the semimajor and semiminor axes. Q: A monatomic ideal gas undergoes a cyclic transformation (see diagram in workbook).

(a) When the gas goes form A to B, what is the change in internal energy? (b) When the gas goes form A to B, how much work is done by it on its environment? (c) How much heat flows into the gas during the transformation A ! B? (d) Answer the same questions for the return transformation B ! A. (e) What is the net work done by the gas on its environment over the cycle?

A: (a) NkVPTA /)2( 00= and AB TNkVPT 5/)5)(2( 00 == . Thus, 002

3 12)4( VPTNkE ABA ==∆ → . (b) Area under upper curve: 00002

100 )8()2()4)(2( VPVPVPW BA ππ +=+=→ .

(c) 00)20( VPWEQ BABABA π+=+∆= →→→ . (d) 0012 VPEE BAAB −=∆−=∆ →→ . Area under lower curve: 00002

100 )8())2()4)(2(( VPVPVPW AB ππ +−=−−=→ . Then,

00)20( VPWEQ ABABAB π+−=+∆= →→→ . (e) 002 VPWWW ABBAnet π=+= →→ .

Challenge Problem 4 (p.23) – in groups Sol. In workbook. Skip this and move on to heat engines, etc.

First Law of Thermo

Questions to ask them about Lecture (1) Have you talked about heat engines, refrigerators and heat pumps? (2) If so, have you talked about efficiency and coefficient of performance? (3) Have you talked about the Carnot cycle? (4) If so, do you know what the efficiency of a Carnot cycle is?

Engines and Efficiency

Engines – Minilecture (if no to all above) In all generality, a heat engine is just a mechanism by which we can extract work out of the natural tendency for heat energy to transfer from something hot to something cold. So, let’s look at the schematic heat engine diagram on p.25 of the workbook. Beside the “hot reservoir” you might write “burning fuel”. Beside the “working substance” you might write “steam”. Beside the work arrow, you might write “piston turns drive shaft” and beside “cold reservoir” you might write “outside air.”

The refrigerator is essentially the same, but with all the arrows reversed. This time, we have to supply work in order to move heat from cold to hot, which

is, of course, unnatural.


Hot Reservoir

Cold Reservoir

Working substance

QH

QC

Wnet

Burning fuel

Steam

Outside air

Piston turns drive shaft

Hot Reservoir

Cold Reservoir

Working substance

QH

QC

Wnet

Kitchen air

Freon

Ref. air

Electric motor

Carnot Cycle & Efficiency – Minilecture A useful heat engine or refrigerator must be able to work along a cycle. That is, it must return to the initial state periodically so that we can keep repeating the process and actually drive something like an automobile. The quintessential example is the Carnot cycle, which, for an ideal gas, looks like Figure 20-7 (p.533): (1) isothermal expansion (2) adiabatic expansion (3) isothermal compression (4) adiabatic compression Efficiency is defined to be “what you want” / “what you have to put in”. So, in the case of a heat engine, Hnet QWe /= . Don’t say this since it’s DP 1: By conservation of energy, CHnet QQW −= , so

H

CQQe −= 1 . For

a Carnot engine, we find H

CTTe −= 1 .


Misconceptions – Minilecture Now, let’s disabuse ourselves of some common misconceptions: (1) “The efficiency of any engine is

Hottest

ColdestTTe −= 1 .”

This is only true for the Carnot cycle. (2) “All reversible engines have the efficiency of the Carnot engine.”

I’m not entirely sure why it says this on p.535 of the text. But, the Otto cycle is an example of a reversible cycle whose efficiency is not the same as that of a Carnot cycle working between the Otto cycle’s extreme temperatures! And we will encounter others in the worksheet.


DP 1 (p.25) - Groups Q: (a) What is the efficiency of a heat engine in terms of Wnet, QH, and/or QL? (b) Use the First Law and the cyclic running of the engine to show that )/(1 HL QQe −= . (c) What is Kref in terms of Wnet, QH, and/or QL? (d) What is Kheat pump in terms of Wnet, QH, and/or QL? A: (a) Hnet QWe /= (b) 0int =∆E across the cycle and so CHnet QQW −= . Thus, )/(1/)( HCHCH QQQQQe −=−= . (c) netCref WQK /= (d) netHhp WQK /= Note: the sign convention here is that QH is positive added to the working substance whereas QC is positive taken out of the working substance. Therefore, when we plug into the first law, the sign of QC is actually negative since the first law uses positive Q as being added to the working substance.


Skip DP 2 move on to DP 3 (p.26) - Groups Q: Does the working substance of a Carnot engine have to be an ideal gas? If a Carnot engine uses a different substance, then can we still find the efficiency using the formula )/(1 HC TTe −= ? A: No, the working substance does not have to be an ideal gas. For example, it could be a van der Waals gas. All that matters is that it follow the steps of a Carnot cycle reversibly. However, the cycle would not look the same on a PV diagram as it does for an ideal gas. For example, some isotherms are plotted for the van der Waals gas in Figure 18-11 (p.487). The adiabats would look quite different as well as we approach the yellow shaded region. Admittedly, the formula )/(1 HC TTe −= was derived from )/(1 HC QQe −= explicitly using the ideal gas equation of state: PV = nRT. Deriving this from a general equation of state is much more difficult. In fact, it took almost five hours of my weekend time for me to devise a general proof that satisfied me.


Skip DP 4 move on to Problem 1 (p.28) - Groups Q: A sample of monoatomic ideal gas undergoes (1) isobaric compression from VA to VB; (2) isochoric heating; (3) isothermal expansion.

(a) During each step, does heat flow into or out of the gas? (b) Calculate intE∆ at each step in terms of P0, VA and VB. What about W and Q? (c) Does this cycle represent a heat engine or refrigerator/heat pump? (d) Find the efficiency.

A: Ideal gas: NkVPT A /01 = and NkVPT B /02 = .

0)()( 023

1223 <−−=−=∆ → BABA VVPTTNkE

0)()( 023

2123 >−=−=∆ → BACB VVPTTNkE and 0=∆ →ACE

0)(0 <−−=→ BABA VVPW , 0=→CBW and 0lnln 01 >==→ B

A

B

AVV

AVV

AC VPNkTW 0)(02

5 <−−=→ BABA VVPQ , 0)(023 >−=→ BACB VVPQ , and 0ln0 >=→ B

AVV

AAC VPQ )1)1(ln(0 +−=

B

A

B

AVV

VV

Bnet VPW and ))(ln( 23

23

0 −+=+= →→ B

A

B

AVV

VV

BACCBin VPQQQ ))(ln/()1)1(ln(/ 2

323 −++−==

B

A

B

A

B

A

B

AVV

VV

VV

VV

innetnet QWW This is a heat engine: heat in work out.


P0

VB VA

T1 T2

T1

Problem 2 (p.29) - Groups Q: A heat engine uses an ideal gas of N monoatomic particles. The cycle runs as: (1) isobaric expansion; (2) adiabatic expansion; (3) isobaric compression; (4) isochoric heating. (See drawing). (a) Find the efficiency. (b) What are the highest and lowest temperatures? (c) what is the equivalent Carnot efficiency? (d) Are the results consistent? A: NkVPTT /0040 =≡ , 01 2TT = , 02 6TT = . For adiabat, .)2/(2 constTP d =+− , so 3

5/200

5/20 )6()2( TPTP −− = and so

05/3

3 32 TT = . Thus, 001223

21 6)( VPTTNkE =−=∆ → & 005/2

2323

32 )12(9)( VPTTNkE −=−=∆ −→ & 00

5/323

43 )321( VPE −=∆ → & 002

314 VPE =∆ → . Now, 0021 4 VPW =→ & 3232 →→ ∆−= EW & 00

5/343 )321( VPW −=→ & 014 =→W . Finally, we get

0021 10 VPQ =→ & 032 =→Q & 005/3

25

43 )321( VPQ −=→ & 1414 →→ ∆= EQ . So, 0000

5/35/2 63.2))321()12(94( VPVPWnet ≈−+−−= − and 000023

2114 5.11)10( VPVPQQQin =+=+= →→ and so 23.05.11/63.2/ === innet QWe .

The highest temp. is NkVPTTH /66 000 == and the coldest is NkVPTTC /000 == . The Carnot efficiency would be 83.06/5)6/1(1)/(1 ==−=−= HCC TTe , which is indeed greater than 0.23.


Challenge Problem T-5 (p.30) - Groups (a) AinAA QeW ,= & AinAAout QeQ ,, )1( −= . (b) AinABAoutBBinBB QeeQeQeW ,,, )1( −=== & AinABBout QeeQ ,, )1)(1( −−= . (c) AinBABABAnet QeeeeWWW ,)( −+=+= (d) )1)(1(1)(/ , BABABAAinnetnet eeeeeeQWe −−−=−+== (e) If 1, <BA ee , then, 11,1 <−− BA ee and so 1)1)(1( <−− BA ee and so 1)1)(1(1 <−−−= BAnet eee . (f) d

H

C

M

C

H

M

M

C

H

MTT

TT

TT

TT

TT

nete −=−=−−−−−= 1)(1))1(1))(1(1(1 = Carnot efficiency.



The refrigerator is essentially the same, but with all the arrows reversed. This time, we have to supply work in order to move heat from cold to hot, which

is, of course, unnatural.

Carnot Cycle Don’t say this since it’s DP 1: By conservation of energy, CHnet QQW −= , so

H

C

QQe −= 1 . For

a Carnot engine, we find H

C

TTe −= 1 .

Section 7 – Wed., Feb. 18, 2009

Hot Reservoir

Cold Reservoir

Working substance

QH

QC

Wnet

Burning fuel

Steam

Outside air

Piston turns drive shaft

Hot Reservoir

Cold Reservoir

Working substance

QH

QC

Wnet

Kitchen air

Freon

Ref. air

Electric motor

DP 1 (p.25) - Groups Q: (a) What is the efficiency of a heat engine in terms of Wnet, QH, and/or QL? (b) Use the First Law and the cyclic running of the engine to show that )/(1 HL QQe −= . (c) What is Kref in terms of Wnet, QH, and/or QL? (d) What is Kheat pump in terms of Wnet, QH, and/or QL? A: (a) Hnet QWe /= (b) 0int =∆E across the cycle and so CHnet QQW −= . Thus, )/(1/)( HCHCH QQQQQe −=−= . (c) netCref WQK /= (d) netHhp WQK /= Note: the sign convention here is that QH is positive added to the working substance whereas QC is positive taken out of the working substance. Therefore, when we plug into the first law, the sign of QC is actually negative since the first law uses positive Q as being added to the working substance.


Skip DP 2 move on to DP 3 (p.26) - Groups Q: Does the working substance of a Carnot engine have to be an ideal gas? If a Carnot engine uses a different substance, then can we still find the efficiency using the formula )/(1 HC TTe −= ? A: No, the working substance does not have to be an ideal gas. For example, it could be a van der Waals gas. All that matters is that it follow the steps of a Carnot cycle reversibly. However, the cycle would not look the same on a PV diagram as it does for an ideal gas. For example, some isotherms are plotted for the van der Waals gas in Figure 18-11 (p.487). The adiabats would look quite different as well as we approach the yellow shaded region. Admittedly, the formula )/(1 HC TTe −= was derived from )/(1 HC QQe −= explicitly using the ideal gas equation of state: PV = nRT. Deriving this from a general equation of state is much more difficult. In fact, it took almost five hours of my weekend time for me to devise a general proof that satisfied me.


Skip DP 4 move on to Problem 1 (p.28) - Groups Q: A sample of monoatomic ideal gas undergoes (1) isobaric compression from VA to VB; (2) isochoric heating; (3) isothermal expansion.

(a) During each step, does heat flow into or out of the gas? (b) Calculate intE∆ at each step in terms of P0, VA and VB. What about W and Q? (c) Does this cycle represent a heat engine or refrigerator/heat pump? (d) Find the efficiency.

A: Ideal gas: NkVPT A /01 = and NkVPT B /02 = .

0)()( 023

1223 <−−=−=∆ → BABA VVPTTNkE

0)()( 023

2123 >−=−=∆ → BACB VVPTTNkE and 0=∆ →ACE

0)(0 <−−=→ BABA VVPW , 0=→CBW and 0lnln 01 >==→ B

A

B

A

VV

AVV

AC VPNkTW 0)(02

5 <−−=→ BABA VVPQ , 0)(023 >−=→ BACB VVPQ , and 0ln0 >=→ B

A

VV

AAC VPQ )1)1(ln(0 +−=

B

A

B

A

VV

VV

Bnet VPW and ))(ln( 23

23

0 −+=+= →→ B

A

B

A

VV

VV

BACCBin VPQQQ ))(ln/()1)1(ln(/ 2

323 −++−==

B

A

B

A

B

A

B

A

VV

VV

VV

VV

innetnet QWW This is a heat engine: heat in work out.


P0

VB VA

T1 T2

T1

Problem 2 (p.29) - Groups Q: A heat engine uses an ideal gas of N monoatomic particles. The cycle runs as: (1) isobaric expansion; (2) adiabatic expansion; (3) isobaric compression; (4) isochoric heating. (See drawing). (a) Find the efficiency. (b) What are the highest and lowest temperatures? (c) what is the equivalent Carnot efficiency? (d) Are the results consistent? A: NkVPTT /0040 =≡ , 01 2TT = , 02 6TT = . For adiabat, .)2/(2 constTP d =+− , so 3

5/200

5/20 )6()2( TPTP −− = and so

05/3

3 32 TT = . Thus, 001223

21 6)( VPTTNkE =−=∆ → & 005/2

2323

32 )12(9)( VPTTNkE −=−=∆ −→ & 00

5/323

43 )321( VPE −=∆ → & 002

314 VPE =∆ → . Now, 0021 4 VPW =→ & 3232 →→ ∆−= EW & 00

5/343 )321( VPW −=→ & 014 =→W . Finally, we get

0021 10 VPQ =→ & 032 =→Q & 005/3

25

43 )321( VPQ −=→ & 1414 →→ ∆= EQ . So, 0000

5/35/2 63.2))321()12(94( VPVPWnet ≈−+−−= − and 000023

2114 5.11)10( VPVPQQQin =+=+= →→ and so 23.05.11/63.2/ === innet QWe .

The highest temp. is NkVPTTH /66 000 == and the coldest is NkVPTTC /000 == . The Carnot efficiency would be 83.06/5)6/1(1)/(1 ==−=−= HCC TTe , which is indeed greater than 0.23.


Challenge Problem T-5 (p.30) - Groups (a) AinAA QeW ,= & AinAAout QeQ ,, )1( −= . (b) AinABAoutBBinBB QeeQeQeW ,,, )1( −=== & AinABBout QeeQ ,, )1)(1( −−= . (c) AinBABABAnet QeeeeWWW ,)( −+=+= (d) )1)(1(1)(/ , BABABAAinnetnet eeeeeeQWe −−−=−+== (e) If 1, <BA ee , then, 11,1 <−− BA ee and so 1)1)(1( <−− BA ee and so 1)1)(1(1 <−−−= BAnet eee . (f) d

H

C

M

C

H

M

M

C

H

M

TT

TT

TT

TT

TT

nete −=−=−−−−−= 1)(1))1(1))(1(1(1 = Carnot efficiency.


Entropy – Minilecture I So far we have only used one state function – the internal energy. Recall that this meant that you could look at a system and tell me how much internal energy it has, as opposed to work or heat. In the case of an ideal gas, we just had NkTdE )2/(int = and so intE∆ depended only on the starting and ending points of your path in the PV diagram and not on how you got there. Another important state function is entropy, S. This is roughly a measure of the disorder of a system. Intuitively, ice, which has a crystalline structure, is much more ordered than liquid water. So, we say that the water has more entropy than does the ice. What do we do to the ice to form liquid water thus increasing entropy? Well, we melt it by adding heat to it. Indeed, Adding heat to a system always increases its entropy This is written as TQddS /= or TdSQ = (provisionally, we will replace the equality with an inequality in the second minelecture) where the slash just indicates that Q is not a state function. The important and useful formula for ideal gases is (we won’t derive this and it’s not even in your textbook):

+=∆

i

f

i

fideal V

V

T

TdNkS lnln

2

Entropy and Second Law

DP 1 (p.31) 1a) What is the relation between inS∆ and outS∆ ? Why is it not possible to develop a cyclic engine that converts heat entirely into work? b) In a Carnot engine, Qin enters the system only along the isotherm TH and Qout leaves the system along the isotherm TL. Write inout QQ / in terms of TH and TL. c) Explain why the Carnot engine gives the highest efficiency. A: (a) outin SS ∆=∆ since cyclic and S is a state function. (b) Hinin TQS /=∆ and Loutout TQS /=∆ . Thus, outin SS ∆=∆ implies HLinout TTQQ // = (c) Since TdSQ = , in order to minimize Qout we must minimize the temperature at which that heat is ejected. Similarly, to minimize inS∆ , we want to maximize the temperature at which Qin enters the system. This is achieved by the Carnot cycle.


Entropy – Minilecture II Now let’s talk about the issue of reversibility. Earlier, we said that adding heat to a system increases its entropy. However, entropy can increase even without the addition of heat. For example, initially, I have warm water and a few ice cubes inside a thermos. Some time later, what will I have in the thermos? Liquid water . But, no heat entered or left the whole system since it was in a thermos. Thus, we really should write TQddS /≥ or TdSQ ≤ . When the equality holds, we say that the process is reversible. Since there is more entropy increase than is just due to heat for an irreversible process, these processes increase the overall entropy of the universe. Spontaneous processes tend to increase the entropy of the universe.


DP 2 (p.32) Q: A cyclic heat engine uses an ideal gas as its working substance. Which are true: (a) gasS∆ over an entire cycle is 0. (b) gasS∆ over an entire cycle is 0 only if the engine operates reversibly. If irreversible then 0>∆ gasS . (c) Over an entire cycle, 0=∆+∆=∆ tenvironmengasuniverse SSS . (d) Over an entire cycle, 0=∆+∆=∆ tenvironmengasuniverse SSS only if the engine is reversible. If irreversible then 0>∆ universeS . A: True: (a) & (d). False: (b) & (c). Main ideas:

(1) Entropy is a state function – so (a) must be true. (2) Irreversible processes add to the entropy of the universe. The universe does

not in general return to its initial state unless the process is reversible.


DP 3 (p.32) Q: A box with total volume V0 is divided in half by a partition. On the LHS, there is a sample of ideal gas with initial pressure P0 and temp. T0. The RHS is empty. The partition is suddenly removed and the gas expands freely to fill the entire box. Then, the gas reaches thermal equilibrium.

(a) Intuitively, what do you think happens to the entropy of the gas when it expands freely? Does it increase, decrease or stay the same?

(b) What is the change in entropy? A: (a) It should increase. Spontaneous processes tend to increase the overall entropy. Also, the final state is less orderly than the initial state with half empty. (b) 2ln

2ln0lnln

2Nk

V

VNk

V

V

T

TdNkS

i

i

i

f

i

fideal =

+=

+=∆ . The other is wrong because all we know is

that ∫≥∆ TQdS / .


Problem 1 (p.34)

(a) Heat is flowing in since 0int >∆E and W > 0. (b) latmlatmT ⋅== 8)8)(1(1 and 12 432)32)(1( TlatmlatmT =⋅== . Therefore,

6.572ln104

ln4

ln2

3)2(

1

1

1

1 ==

+==∆ R

V

V

T

TRnRS J/K

(c) 0=∆ →XAS since there is no heat and it is reversible. (d) 6.572ln1032ln2// =====∆ →→→ RRTWTQS BXBXBX J/K (not: S is a state variable). (e) 6.57=∆ →→ BXAS J/K (f) S is a state variable and so S∆ only depends on endpoints of path. (g) 6.57=∆ →→ BYAS J/K


Problem 2 (p.35)

(a) Temperature does not change during free expansion because there is nothing that actually slows down the particles as they fill in the extra space.

(b) AABB VpVp = and so 2/0ppB = . (c) Net work is negative! The gas did no work to expand. Then, we do work to

compress it! (d) 2lnNkS BA =∆ → , 2ln

2

7

2

1ln

2

1ln

2

5NkNkS CB −=

−=∆ → , 2ln2

52ln

2

5NkNkS AC =

=∆ →

(e) 0=∆∑ S since cyclic. (f) 0=∆ →

envBAS , 2ln

2

7NkS env

CB =∆ → (since reversible), 2ln2

5NkS env

AC −=∆ → (since reversible)

(g) 2ln2ln0 NkNkSSS envgasuniv =+=∆+∆=∆ . (h) Yes. The gas has no entropy change over a cycle. But the universe entropy

increases.


Problem 3 (p.36) (a) 51068.3ln ×−=−

−=+

−=∆ ∫

C

Hw

H

vT

T

wH

vsteam T

Tmc

T

mL

T

dTmc

T

mLS

C

H

J/K.

(b) Since the process is reversible, the working substance must increase in entropy by the same amount and lose all of this entropy to the cold reservoir which is at the fixed temp. 0 oC. Thus, 810004092.1 ×=∆−= steamCout STQ J. The total heat intake was 810337500.1)( ×=−+= CHwvin TTmcmLQ J. Thus, the total work is the difference 710334082.3 ×=−= outin QQW J = 33 MJ.



zzrzr

qF

outin

annulus ˆ)/(1

1

)/(1

1

2 220

+−

+=

εσ�

rin = 0 (Disk): zzR

qFdisk ˆ

)/(1

11

2 20

+−=

εσ� with R = disk rad.

rout = ∞ also (∞ plane): zq

Fplane ˆ2 0ε

σ=�

y

yLy

qQF

linefiniteˆ

)2/(1

1

4 220 +

=πε

� rr

qQFpo ˆ

4 20

int πε=

� (Coulomb)

yy

qF

lineiniteˆ

1

2 0inf πε

λ=�

z

zLzLz

qQF

loopsquareˆ

)2/(21

1

)2/(1

1

4 2220 ++

=πε

�

Section 11 – Wed., Mar. 4, 2009

Finite Square Surface Charge Consider a square centered at origin of side length 2x And a slightly larger one of side length 2(x + dx). The Difference in area is xdxxdxxdxxdA 84)(484 222 =−++= where We drop all terms of higher than linear order in dx. This Little square loop has charge xdxdQ σ8= . This plays the Role of Q in the square loop case and x plays the role of L/2. So, we have z

zx

zxd

zx

zxqz

zxzxz

xdxqFd ˆ

)/(21

)/(

)/(1

/2ˆ

)/(21

1

)/(1

1

4

822

0222

0 ++=

++⋅=

πεσ

πεσ� . Integrating from

0/ == zxu to zLu 2/= gives, using the integral 222

21arctan1

1

21u

uu

udu +=++∫ and the facts that

4/)1arctan( π= and 2/)arctan( π=∞ , we have

zz

LqFsquare ˆ1

221arctan

4

2

2

0

−

−=πε

σ� and so zq

Fplane ˆ2 0ε

σ=�


Remark on Setting up Integrals Please keep safe the formulas that we have derived here, especially for the forces felt by a point charge some distance away from an infinite line charge or an infinite plane of charge. We will come back to them without rederiving them each time. I wanted you to know where these formulas came from and to see that you have the capacity to derive them by setting up the appropriate integrals by breaking up the objects into little differential bits, using Coulombs law on each bit, and integrating over all of the bits. There is often no substitution for this procedure and it in principle works in any situation, although you might have to compute some integrals numerically using a computer. Another reason why I wanted us to go through the painful process of actually computing some of these integrals is because, in particularly symmetric situations, we can use other techniques to compute these forces that require MUCH less work. Thus, you will learn to appreciate these tricks all the more if you have an idea of how much work it would take without them.


Minilecture Having spent so much time computing forces felt by point charges will hopefully save us a bit of time getting used to the idea of an electric field. One way to define the electric field produced by some charged body is as follows: imagine placing a very small positive charge q at some point in space around the charged body. This small positive charge will feel a force due to that charged body, which you could compute the same way we computed it for the disk or the line, etc. Then, the field is defined to be that force divided by the charge q of the small positive charge feeling the force. Since we’re dividing q out, the field is something intrinsic to the charged body producing the field and is independent of the small charge q that you used to probe the electric field. For example, from our previous work,

zzrzr

Eoutin

annulus ˆ)/(1

1

)/(1

1

2 220

+−

+=

εσ� z

zREdisk ˆ

)/(1

11

2 20

+−=

εσ� zE plane ˆ

2 0εσ=

�

yyLy

QE

linefiniteˆ

)2/(1

1

4 220 +

=πε

� yy

Elineinite

ˆ1

2 0inf πε

λ=� r

r

QE po ˆ

4 20

int πε=

�

zzLzLz

QE

loopsquareˆ

)2/(21

1

)2/(1

1

4 2220 ++

=πε

� zz

LEsquare ˆ1

221arctan

4

2

2

0

−

−=πε

σ�

Electric Fields

Minilecture The virtue in the electric field is that it is a property only of the object creating the field. We can divide by q (charge of the probe) after we compute the force or we could just as well divide right at the very beginning in which case, if you think about it, we never really computed any force to begin with. Then, if you are given a point charge around a charged body and you are asked what is the force felt by the point charge, you just need to multiply the electric field produced by the charged body at the position of the point charge by the charge of the point charge: EqF

��

= We draw electric field vectors as follows: consider a point around the charged

body. The electric field at that point has magnitude and direction (it is a vector). Simply draw that vector as an arrow starting at the position in space pointing in the direction of the field and having length commensurate to the magnitude of the field. You can figure out the direction by imagining placing a small positive charge there and then determining which direction it would move.

For aesthetic reasons, we usually pick some symmetrically distributed bunch of points at which we draw the field vectors. We obviously cannot draw all vectors! Let’s do an example.

Electric Fields

DP 1 (p.49) – At the board Q: Sketch the electric field created by +1C and –2C point charges. A:

Electric Fields

DP 2 (p.49) – At the board Q: Sketch the electric field created a positive line charge. How would your picture look if the line were negatively charged? A:

Electric Fields

DP 3 (p.50) – At the board Q: Sketch the electric field created a positive plane charge. How would your picture look if the line were negatively charged? A:

Electric Fields

DP 4 (p.50) – At the board Q: Sketch the electric field created two parallel charged planes, one positive and one negative. Remember that there are two field vectors, one due to each charged body and the net field vector is the sum of the two vectors (superposition principle). A:

Electric Fields

DP 5 (p.50) – At the board Q: Repeat q.4 if the sheets are both positive. A:

Electric Fields




0

34

121 d

QEE πε==�� and 2

0

64

13 d

QE πε=� . The dir. are ( )0,3,12

11 =d� , ( )0,3,12

12 −=d� and ( )0,1,03 −=d

� . Thus,

( ) ( ) ( ) ( ) yEd

Q

d

Q

d

Q

d

Q ˆ0,1,00,3,10,3,1 20

20

20

20 4

23364

1213

41

213

41 ⋅=−⋅+−⋅+⋅= −

πεπεπεπε

� . Answer is ( )y

d

QE ˆ

4

3232

0πε−−=

� .

Electric Fields




D

qQFe ˆ

4

12

0πε= . The


mgD

qQ2

04

1tan

πεθ = .

(b) This time, xq

Fe ˆ2 0ε

σ= and so mg

q

02tan

εσθ = .

(c) This time, x

D

qFe ˆ

2 0πελ= and so

Dmg

q

02tan

πελθ = .

Electric Fields

Things to write on the board at the start We computed the forces and electric fields pertaining to certain charged bodies.

EqF��

=




0

34

121 d

QEE πε==�� and 2

0

64

13 d

QE πε=� . The dir. are ( )0,3,12

11 =d� , ( )0,3,12

12 −=d� and ( )0,1,03 −=d

� . Thus,

( ) ( ) ( ) ( ) yEd

Q

d

Q

d

Q

d

Q ˆ0,1,00,3,10,3,1 20

20

20

20 4

23364

1213

41

213

41 ⋅=−⋅+−⋅+⋅= −

πεπεπεπε

� . Answer is ( )y

d

QE ˆ

4

3232

0πε−−=

� .

Section 12 – Mon., Mar. 9, 2009




D

qQFe ˆ

4

12

0πε= . The


mgD

qQ2

04

1tan

πεθ = .

(b) This time, xq

Fe ˆ2 0ε

σ= and so mg

q

02tan

εσθ = .

(c) This time, x

D

qFe ˆ

2 0πελ= and so

Dmg

q

02tan

πελθ = .

Electric Fields

Gauss’ Law - Minilecture Suppose you were sitting by a river and for some reason a question pops in your mind: I wonder how much water is flowing through the river in some given amount of time, say one second. What would you do to begin tackling this question? To do a more careful experiment, you would probably want a flow rate sensor. But, what do you suppose a flow rate sensor essentially does? Well, it essentially passes water through a tube of a certain cross-sectional area. The water pushes against a turbine the rate of whose rotation reflects the rate of passage of water through the tube. If we abstract the situation, we essentially have vectors in space indicating the direction of flow of water as well as the magnitude of flow of water and we are considering how much water flows through an imaginary loop in one second. We note that the way that we orient the loop with respect to the water flow is crucial. For example, if we place the loop so that its face is perpendicular to the flow, then no water flows the loop at all! This more abstract picture is valuable because it generalizes to arbitrary vector fields that might have nothing to do with flow rates – for example, we need to consider the Electric Field. In this case, there isn’t really anything flowing, but we can still talk roughly about “how much electric field” passes through a loop in some amount of time. This is called electric FLUX.

Gauss’ Law

Gauss’ Law - Minilecture Let me write down the general formula for electric flux: ∫∫∫∫ =⋅=Φ

surfacesurface

surface EdAAdE θcos��

. Here

the left hand side is the electric flux through some imaginary surface, which we have expressed as a two-dimensional surface integral of the dot product of the electric field with the differential surface area vector. This last object is the vector whose magnitude is the area of the bit of surface and whose direction is the normal to the bit of surface. Gauss’ Law tells us that “The net electric flux outward through at imaginary closed surface reflects the amount of electric charge contained within the surface.” Or, in math,

0/ εencsurface Q=Φ where kπε 4/10 = . I should emphasize that Gauss’ law is true for any surface you could choose, but it’s only useful for calculating electric fields when the surface and the field are simple enough. For example, if the electric field is constant throughout the surface and or if the

θcos term is constant throughout the surface. Otherwise, you can’t really use Gauss’ law for computational purposes. I should also say that the procedure we used so far of breaking objects up into differential pieces and integrating always works (unless the integral cannot be performed), but Gauss’ law can make some computations much simpler.

Gauss’ Law

DP 2 (p.55) - Groups A: (b)ii) 0/ εQ=Φ + iii) The field lines are mostly going out of the surface. (c) ii) 0=Φ ellipsoid iii) Every flux line that goes out comes back in! (d) ii) 0=Φ weird iii) Same.

DP 3 (p.57) – Groups A: 02/ εQplane =Φ

DP 4 (p.58) – Groups A: (a) 0

2 /εσπRcube =Φ (b) No. The field is not constant throughout the cube nor is the orientation of the field relative to the cube!

DP 5 (p.58) - Groups Q: Use symmetry to find the most general form of the electric field when

(a) Spherically symmetric – charge distribution only depends on r (b) Cylindrically symmetric – infinitely long and depends on radial axial distance (c) Planar symmetry – infinite in two directions and only depends on third dir.

A: (a) rE ˆ∝

� (b) rE ˆ∝� (c) zE ˆ∝

�

Gauss’ Law

Problem 1 (p.59) – Groups Note: (f) & (g) should read (b) – (e) not (c) – (e) A: (b) Yes. The field is constant throughout the surface since the field only depends on radial axial distance and all points on the surface are at the same radial axial distance r. (c) angle between E field and surface is always 0 (d) Yes, angle is always the same throughout the surface (e) RLEEA π2==Φ since E is constant throughout the surface, it can be taken out of the integral (f) No, E field not the same at all points on the lids; 900; Yes, angle is constant; 0=Φ lid . (g) same as (f). (h) RLEEA π2==Φ (i) LQenc λ= (j) RE 02/ πελ= Note: The answer you get for the electric field corresponds to the answer we computed from before for using the more difficult procedure of setting up integrals. So, why did we do it using the harder procedure first? Part (k) gives a hint. (k) If the line were of finite length, then the field would no longer be constant on the surface we chose (magnitude and direction).

Board Organization for Problem 1 Calculating Φ directly

∫∫∫∫ =⋅=Φsurfacesurface

surface EdAAdE θcos��

RLEbottomlidlabel π2=Φ+Φ+Φ=Φ

Calculating encQ

LQenc λ=

0/ εencQ=Φ

0/2 ελπ LRLE =

Gauss’ Law

Problem 2 (p.60) – Groups

Board Organization for Problem 2 for r>R Calculating Φ directly

rLEπ2=Φ Calculating encQ

LRQenc2ρπ=

0/ εencQ=Φ 0

2 /2 ερππ LRrLE = (b) From above, the field is rrRE ˆ)2/( 0

2 ερ=� .

Board Organization for Problem 2 for r<R

Calculating Φ directly rLEπ2=Φ

Calculating encQ LrQenc

2ρπ= 0/ εencQ=Φ

02 /2 ερππ LrrLE =

(c) From above, the field is rrE ˆ)2/( 0ερ=

� . (d) Yes, the electric field is continuous at the surface.

Gauss’ Law

Problem 3 (p.60) – Groups

Board Organization for Problem 3 for r<RA

Calculating Φ directly Er 24π=Φ

Calculating encQ 0=encQ

0/ εencQ=Φ 04 2 =Erπ

(a) From above, the field is 0.

Board Organization for Problem 3 for r>RB


Calculating encQ 3/)(4 33 ρπ ABenc RRQ −=

0/ εencQ=Φ 0

332 3/)(44 ερππ AB RREr −= (b) From above, the field is ))(3/( 332

0 AB RRrE −= ερ .

Board Organization for Problem 3 for RA < r < RB


Calculating encQ 3/)(4 33 ρπ Aenc RrQ −=

0/ εencQ=Φ 0

332 3/)(44 ερππ ARrEr −= (c) From above, the field is ))(3/( 332

0 ARrrE −= ερ . (d) Only BRr > is affected and 2332

0 )/())(3/( rRRRrE BAB σερ +−= .

Gauss’ Law

Problem 6 (p.61) – Groups (a) Below the plane, the field vector points directly away from the plane (positive) or towards the plane (negative). Above is the same.

Board Organization for Problem 6

Calculating Φ directly EA2=Φ

Calculating encQ AQenc σ=

0/ εencQ=Φ 0/2 εσAEA =

(a) From above, the field is 02/ εσ=E , which agrees with out previous result.

Gauss’ Law

Things to write on the board at the start Gauss’ Law: 0/cos εθ enc

surfacesurface

surface QEdAAdE ==⋅=Φ ∫∫∫∫��

.

Type of Symmetry Generic Gaussian Surface Spherical (point, sphere, solid ball) Sphere with same center Cylindrical (line, cyl. Shell, solid cyl.) Cylinder with same axis Planar (plane) Perp. Cyl. Or pillbox Conductors: E field vanishes in the material of a conductor. Work on Problem 3 first then this if there is time.

Problem 1 (p.64) – Me (a) Like four infinite planes: xxQQQQE A

QA

ˆˆ))3()5((0

1

0

422112

11 εε

−=−−−−−=�

(b) 01 =E� and so 41 =Q C.

(c) xxQQQQE AQ

Aˆˆ))5()3((

0

2

0

411222

12 εε

−=+−+−+−=� . Now, 02 =E

� and so 42 =Q C. (d) 15 1 =− Q C and 13 2 −=− Q C.

Section 14 – Mon., Mar. 16, 2009

Electric Potential - Minilecture Let’s think for a minute about the electric field. What is it, and what is it good for? Electric Field: What is it? The electric field is a vector field: a vector attached to every point in space. It’s created by a charged object. What is the electric field good for? Once you know the E-field created by an object, you can find the force on any point charge that happens to be around using EqFq

��= .

Electric Potential

Electric Potential - Minilecture Now let’s think about a new kind of field that is related very closely to the electric field. It is called electric potential. The word potential is potentially confusing. This field is intimately related to potential energy, but it is NOT the same thing. Electric Potential: What is it? V=7 V=20 V=20 V=7 V=3 V=4 V=10 The electric potential is a number field: a number attached to every point in space. It’s created by a charged object. What is the electric potential good for? Once you know the E-potential created by an object, you can find the potential energy of any point charge that happens to be around using qVU q = .

Electric Potential

Electric Potential - Summary The Electric Field, E What is it? What is it good for? A vector at Gives the force on any Every point point charge near the In space object using EqFq

��=

The Electric Potential, V What is it? What is it good for? A number at Gives the potential Every point energy of any point In space charge near the object using qVU q =

For Point Charge

204/ˆ rrQE πε=

� rQV 04/ πε=

We observe the relation for point charges: rdrdVE ˆ)/(−=

� . This is an example of a far more general fact VE −∇=

� where ∇ is the gradient operator. The integral way of writing this is ∫ ⋅−=− 2

1

)()( 12

r

rdErVrV ℓ�� . It is clear from VE −∇=

� that adding a constant to V

does not change anything. Thus, we often have to fix a value of V at some point to fix that constant. Usually, we insist that V goes to 0 at infinity.

Electric Potential

DP 2 (p.67) - Groups Q: A point mass m=0.05 kg with charge Q= +3 C is sitting in an external potential V i = 17 V. How much work will it take to move it to Vf = 22 V? A: W= Q(Vf – Vi) = 3 C (22 V – 17 V) = 15 J

DP 3 (p.67) – Groups A: (a) JVVCW BA 8))6(2)(1( =−−=→ (b) JWW BACA 8== →→ since B and C are equipotential. (c) 0 J. (d) Strongest near central peak since field lines are closest there. (e) Field lines are always perpendicular to equipotentials.

I’d like to move on to problems, but Discussion questions 1 and 4 are good exam studying questions.

Electric Potential

Problem 1 (p.69) – Groups Q: The source of the sun’s energy is a process called nuclear fusion. In this process, protons collide and create larger particles. As a result, huge amounts of energy are liberated. The problem, however, is that for fusion to occur, the protons have to “collide” – that is, they have to pass near to each other – say, within a proton diameter, 1510− m.

(a) Suppose that one proton is fixed in place. If another proton starts out from very far away, then how fast muts it be going initially, if it is to approach to within 1510− m of the fixed proton?

(b) Calculate how high the temperature would have to be inside the sun, in order for the speed you foind in part (a) to be a typical rms speed for particles in an ideal gas.

(c) The actual temperature in the solar interior is about 107 K. Is this consisten with the fusion mechanism?

A: (a) Initially, the energy is all kinetic. At minimum impact distance, the incoming proton will have been stopped, so all the energy is electric potential energy. Thus, we set them equal: 2

21

41 2

0mvr

e =πε . This gives 70

2 1066.12/ ×== mrev πε m/s. (b) 10

02 1011.12//3 ×=⇒⇒⇒= TmremkT πε K.

(c) No!

Electric Potential

Problem 5a (p.70) – Me I’ll do this problem, which we’ll use for 6. Q: For a charged disk with radius R and uniform surface charge density σ , the electric potential at the center is given by 02/ εσRV = (we have taken the potential to be 0 at infinity.) Find the electric potential energy of a gold nucleus (+79 e), if it happens to be located at the center of a charge annulus of inner radius Rin and outer radius Rout. A: The central potential due to σ disk of radius Rout is 02/ εσ outRV =+ and of σ− disk of radius Rin is 02/ εσ inRV −=− . Net is 02/)( εσ inout RRVVV −=+= −+ . The electric potential energy of a gold nucleus here would be 02/)(79 εσ inout RReqVU −== .

Electric Potential

Problem 6 (p.70) – Groups Q: Charged annulus of inner radius R1 and outer radius R2 and uniform surface charge density σ . Take potential to be zero at infinity.

(a) Find potential along the axis of a thin ring of radius r and width dr. (b) Integrate over the radius to find the potential along the axis of the annulus. (c) At the very center of the annulus, what is the potential? Compare with q.5.

A: (a) Let the point be at a height z (can be negative) above the plane of the annulus.

2204

1),,(

rz

rdrdrzdV

+= φσ

πεφ . So,

2202

),(rz

rdrrzdV

+=

εσ .

(b)

+−+=+=

+= ∫

21

222

2

0

22

022

0 222)(

2

1

2

1

RzRzrzrz

rdrzV

R

R

R

R εσ

εσ

εσ

(c) ( )1202

)0( RRV −=εσ agrees with 5.

Electric Potential

Problem 7 (p.71) – Groups Q: A very long line of charge with uniform positive charge per unit length λ , is held fixed in place. Located a distance D from the line, and also fixed in place, is a point mass m with positive charge Q.

(a) Use Gauss’ Law to calculate the strength of the electric field created by the line charge at any distance r from the line charge.

(b) If the positive point charge Q is suddenly released, it will accelerate away from the line charge. Will the acceleration be constant?

(c) How fast will the point charge be going by the time it has reached a distance 5D from the line charge?

A: (a) Use cylinder of radius r and length L. Then, 0/2 ελπ LrLE =× so rE 02/ πελ= . (b) Acceleration is not constant since E isn’t and so F isn’t. (c) From rVE ∂−∂= / we have )/ln()2/( 00 rrV πελ−= where 0r is some constant distance. Thus,

0000005 2/5ln)5/ln()2/()/ln()2/()/5ln()2/( πελπελπελπελ −==+−=∆ → DDrDrDV DD . So, 05 2/5ln πελQU DD −=∆ → . From DDDD UK 55 →→ ∆−=∆ , we have mQv 0/5ln πελ= .

Electric Potential

Problem 4 (p.70) – Groups Q: Thin conducting shell of radius R1 is centered inside a thick conducting shell with inner radius R2 and outer radius R3. The inner shell has positive charge 2Q0 on it, while the outer shall has net charge –Q0.

(a) What is charge on inner surface of thick shell? Outer surface? (b) Taking 0)( =∞V , find )0(),(),(),( 123 VRVRVRV . (c) Connect spheres by thin conducting wire. After a long time passes, which of the follows is true?

a. All charge resides on inner thin shell. b. All charge resides on outer thick shell. c. Charge on both shells, but there is more positive charge on inner shell. d. Same as c. but more positive charge on outer shell.

A: (a) Inner surface cancels inner thin shell. So, Qinner = –2Q0. So, Qouter = +Q0. (b) Outside, potential is of charge Q0 so 3003 4/)( RQRV πε= . Since E is zero in thick shell, the potential must be constant here. Thus, )()( 32 RVRV = . Without outer shell, potential outside inner shell would be rQ 00 4/ πε and so at radius R2 would have been 200 4/ RQ πε . Now it’s forced to be 300 4/ RQ πε . Thus, between inner shell and thick shell it must be

))(4/( 13

12

100

−−− +− RRrQ πε . Thus, ))(4/()( 13

12

11001

−−− +−= RRRQRV πε . Again, E is constantly 0 in inner shell and so )()0( 1RVV = . (c) b. Since now inner surface of outer shell and inner shell are equipotential.

Electric Potential

Things to write on the board at the start Capacitors: Capacitance: ability of a system of conductors to hold charge. Geometry-dependent. Capacitance-Potential Relation: Q = CV Electrostatic energy of capacitor: U = (1/2)CV2 = Q2/2C Parallel-plate capacitor: C = ε0A/d. Should be able to compute capacitance of two concentric spheres with –Q on inner sphere and +Q on outer sphere. Or two concentric cylinders. DC Circuits: Ohm’s Law: V = IR Power: P = IV Kirchoff’s Current Law: 0=∑node

I where I is positive if going IN and negative if going OUT of the node. Kirchoff’s Voltage Law: 0=∑loop

V around a closed loop.

Section 17 – Mon., Apr. 6, 2009

Magnetostatics - Overview There are two main skills that we want to learn in the magnetostatics section, which are worksheets M1 to M4. These are:

(1) Compute the dynamics of a particle moving in a given external B-field. (2) Visualize and compute the B-field produced by a fixed current distribution.

We will tackle skill 1 mostly in M2 and skill 2 in M3 and M4. Keep this in mind as you go through the worksheets.

Magnetostatics Overview

DP 1 – Vote Q: Suppose you have a compass that points North when you are standing in Paris. If you take that compass to Capetown (South Africa), will it point South? A: The key here is to visualize the magnetic field of the Earth. (Draw it on board). Now, a compass needle will point along the direction of the magnetic field, which always points away from the magnetic south pole and towards the magnetic north pole. An interesting point here is that if the compass needle is just sitting on a point fulcrum placed at the needle’s center of mass, so that it is perfectly balanced, then, the needle will dip at an angle to follow the Earth’s magnetic field. This is called the angle of declination. Knowing the angle of declination is important for a number of key experiments especially the experiment that first measured the charge to mass ratio of an electron. This is one of the labs that your counterparts in my undergrad university had to do and it was a lab that I performed in my high school in Manila. Of course, no one had bothered to include Manila in a list of magnetic field declination angles around the world. So, we had to estimate this in a makeshift fashion using a simple compass and an overhead projector.

M1 – Introduction to Magnetism

DP 2 – Me Q: A compass often comes in handy when you’re hiking. But, in order to find geog. North using a compass, you have to correct its reading. Why is this? Is there a place in the United States you could be hiking where you would not have to correct the compass reading? A: There isn’t much to discuss here. The geographic north pole is defined to be the point in the northern hemisphere (the one containing Europe, for instance) where, if you stood there at night and watched the stars move, you would see them revolve around the zenith. A priori, this is not related to the position of the north pole of the Earth’s magnetic field. In fact, every billion or so years (or something long like that), the polarity of the earth’s magnetic field switches. That the two types of poles are not that far apart (11o apart) indicates that there is some sort of relation between the two and that we really shouldn’t expect the magnetic north pole to be located near the equator, for example. Indeed, the magnetic field is produced largely by currents deep in the earth caused by circulating ionized molten rock or iron, which tends to revolve roughly in synch with the revolution of the outer crust.


DP 3 – Me asking them Q: It was considered a major discovery in 1820 when Oersted noticed that compasses deflect near current-carrying wires. What would you say was the main implication of this discovery? A: We know that compasses react to magnetic fields. Oersted showed that they also reacted to current in a wire. This suggests that currents, or moving charges, produce magnetic fields. This is the basis of the theory we will be developing for the next few weeks. Another important consequence of Oersted’s observation is that it helped unify the study of electricity and magnetism. Prior to his discovery, those who studied electricity studied charges and currents while those who studied magnetism looked at lodestones and compasses. After this discovery it was clear that these two fields were somehow fundamentally related. The ultimate unification of the two fields would have to wait until Maxwell and his equations around the 1860s, I think it was. We probably won’t get to Maxwell’s equations, but if you have time it’s a great thing to try to learn of Wikipedia or something.


DP 4 – Me asking them Q: The diagram (in worksheet) shows the readings of a compass placed in various locations near a current-carrying wire.

a) Based on this diagram, try to sketch some magnetic field lines. b) Which way is the current flowing in the wire?

A: a) circles in counter-clockwise direction. b) to the left. The important thing here is one of the right-hand rules. If you wrap your right-hand fingers along the direction of the magnetic field around a wire, then point your thumb out in the natural direction (like okay sign), then your thumb points in the direction of the current. This also works in reverse if you want to know the direction of the magnetic field given the direction of the current.


Lorentz Force – Minilecture (5 mins) Here are what I might consider to be the five main points of this section:

(1) Get re-acquainted with vector cross-products and right-hand rules (2) Practice using F = qvxB. (3) Solve kinematic problems involving magnetic fields. (4) Understand the torque formula τ = µxB for a loop. (5) Understand why magnetic forces do no work.

Here is a chart organizing some of the important concepts:

Effect of an external magnetic field B on various objects Object Effect Notes Diagram

Moving point charge q F = qvxB Current-carrying wire F = iLxB L dir. is current dir. Loop of current

τ = µxB

|µ|= iA (A = loop area) Direction by RHR.

Note that the last two can be derived from the first by setting up integrals.

µ

B

M2 – The Lorentz force law

DP 1 & 2 – Groups (10 mins.) Q1: The diagram (WS), shows a proton moving through an external magnetic field at a particular instant in time.

a) At the instant, would the magnetic field be exerting any force on the proton? If so, in which direction? Sketch the force vector on the diagram.

b) Would your answers change if the particle were an electron? A1: a) Yes. Into the page. The important thing here is the right hand rule for vector cross products. First, point right hand fingers along v, then curl the fingers towards the direction of B, then the extended thumb points in the direction of the cross-product. b) Yes. The force is opposite for an electron. Q2: A positively charged particle is moving with the indicated velocity through a magnetic field (not shown). The force exerted on the moving charge is shown.

a) What is the direction of the external magnetic field at the location of the particle? b) Answer part (a) again, this time assuming the particle is negatively charged.

A: a) Out of the page + any component parallel to v. b) Just the opposite.


DP 5 – Together (5 mins.) Q: When a particle moves under the influence of a magnetic field, the speed of the particle remains constant. How does this come about? A: Physical explanation: the magnetic force is always pointing perpendicular to the path of the particle. Therefore, it only turns the particle, it doesn’t push or pull it and hence it can do no work. By the work-kinetic-energy relation, this means that magnetic forces cannot change the speed of a particle. Math explanation: if a particle moves a differential displacement (that’s a vector) ds in a differential bit of time dt, then, if it has velocity v, we have ds = vdt. The bit of work done is dW = F·ds = q(vxB)·vdt. But, vxB is perpendicular to v (and B) and thus, dotting this into v gives 0. Again, this shows mathematically that magnetic forces do no work.


Problem 1 – Groups (2 mins. per part) a) Assuming Carbon ions start at rest, 0

221 )12( eVvmp = and so pmeVv 12/2 0=

b) The force is radial; the particles circle upward. pmeVeBevBF 12/2 000 ==� .

c) 02

0 /12 evBRvmF ==� and so 2

00012 /24/12 eBVmeBvmR pp == .

d) 20014 /28 eBVmR p=

e) Note that the circles are not concentric. So, the separation is the separation of diameters (NOT radii): 2

00 /8)1214( eBVmR p−=∆ . f) Let 2

min 10−=∆R m and 10 10−=B T. Then, min

200 /8)1214( ReBVmp ∆≥− , which gives an

inequality for V0 which is 1560 ≥V V. g) The diameter is 5.12/482 2

0012 == eBVmR p cm if we plug in V0 = 156 V.

Problem 3 – Groups (10 mins.) a) F = i0dB to the left. b) F = ma. If you jump to v = v0 + at, ask why this is okay (because a is constant).

Then, v = (i0dB/m)t since v0 = 0.


Problem 2 & 4 – Separate Groups (15 mins.) A2: a) F1 = i0aB0 into page. b) F3 = i0aB0 out of page and F2 = F4 = 0. c) Rotate about vertical central axis in clockwise direction. d) τ1 = (b/2) i0aB0 = i0(ab/2)B0 = i0(A/2)B0 where A = ab is the area of loop. This

points downward due to RHR for τ = rxF. e) τ3 = τ1 downward whereas τ2 = τ4 = 0. f) τ = i0AB0 downward. g) τ = µxB and µ = i0A into page. By RHR, τ = i0AB0 downward (same as in (f)). A4: a) Fhor = –i0dB sinθ and Fvert = i0dB cosθ. b) N = mg – i0dB cosθ and so fmax = µ(mg – i0dB cosθ). Set this equal to i0dB sinθ, the horizontal component of the magnetic force. Then,

)/sin(cos0 µθθ +=

di

mgB .

c) 0)/sin(cos

)/cossin(2

0

=++−−=

µθθµθθ

θ di

mg

d

dB gives µθ /1tan = so that 2/12 )1(sin −+= µθ and 2/12 )1(cos −+= µµθ . This

gives 2

0min

1 µµ+

⋅=di

mgB .

d) If B were vertical, then F would point to the left. But, Bmin is not vertical because B modifies N reducing f.


Things to write on the board at the start Here is a chart organizing some of the important concepts:

Effect of an external magnetic field B on various objects Object Effect Notes Diagram

Moving point charge q F = qvxB Current-carrying wire F = iLxB L dir. is current dir. Loop of current

τ = µxB

|µ|= iA (A = loop area) Direction by RHR.

Two main skills that we want to learn in the magnetostatics section:

(1) Compute the dynamics of a particle moving in a given external B-field. (2) Visualize and compute the B-field produced by a fixed current distribution.

We will tackle skill 1 mostly in M2 and skill 2 in M3 and M4. Keep this in mind as you go through the worksheets.

µ

B

Section 18 – Wed., Apr. 8, 2009

DP 1 – Groups (10 mins.) Q: Sketch the magnetic fields of (a) long straight wire; (b) circular current loop. A: a) b) side view:

M3 – Magnetic Fields

I

DP 2 – Groups (10 mins.) Q: Suppose we have two long, straight current-carrying wires running parallel to one another. If the two currents flow in the same direction, will the wires attract each other or repel each other? A: Denote the two wires as points with currents flowing into the page. Then, the magnetic field of L (left) on R (right) points down. Hence, a positive charge moving into the page at R feels a force to the left (i.e. towards wire L). Thus, the force is attractive. Similarly, the force on an inward moving positive charge at L due to R is to the right (i.e. towards wire R) Thus, the force is attractive. NOTE: At this point, make a note that current directions describe how a positive charge would move. Negative charges such as electrons move opposite to the current!


Problem 3 – Groups (15 mins. / 10 mins. board) Q: Two straight wires are 4d apart an carry equal currents i0 in opposite directions. a) sketch the two contributions to B at point P d above midpoint. b) Add these vectors graphically to obtain the net B field at P. c) Calculate the magnitude and direction of the magnetic field at P. (Note: They need Bline = µ0i / 2πr). A: a) sketch is easy; b) the net field points up; c) didddiB πµπµ 5/2)5/2)(52/(2 0000 == upward.

Problem 2 – Groups (15 mins. / 10 mins. board) Q: Two concentric circular wire loops have radii 4a and 6a. The larger loop carries current i0 in the ccwise direction as viewed from above. a) If we want B to vanish at the center, in which direction should the current in the smaller loop flow? b) How strong must that current be? (Hint: at the center of a circular loop of radius R, the B field is µ0i / 2R). A: a) clockwise; b) µ0iin / 8a – µ0io / 12a = (µ0 / 12a)((3/2)iin – io) = 0 and so we have iin = (2/3)i0.


Problem 1 – Groups (15 mins. / 5 mins. board) Q: A long wire is bent into the hairpin shape.

a) What is the direction of the magnetic field at the center of the circular end? b) What is the magnitude of the magnetic field at that point?

A: a) By superposition, 2Bp = Bloop + Bline above + Bline below. All of these give fields that point into the page. b) Bp = (1/2)(µ0i /2d + µ0i/2πd + µ0i/2πd) = (µ0i /4d)(1 + 2/π)


Things to write on the board at the startAug 07, 2017 · Problem 2 (p.8) Q: A diatomic ideal gas...

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Transcript of Things to write on the board at the startAug 07, 2017 · Problem 2 (p.8) Q: A diatomic ideal gas...