PHY 2311

PHYSICS 231Lecture 30: review

Remco ZegersWalk-in hour: Monday 9:15-10:15 am

Helproom

PHY 2312

chapter 9

Solids: LA

FL

LL

AFY

0

0/

/ Young’s modulus

xA

Fh

hx

AFS

/

/Shear modulus

pressureP

VV

P

VV

AFB

00 //

/ Bulk modulusAlso fluids

P=F/A (N/m2=Pa) Fpressure-difference=PA=M/V (kg/m3)

General:

Pascal’s principle: a change in pressure applied to a fluid that is enclosed is transmitted to the wholefluid and all the walls of the container that hold the fluid.

PHY 2313

P = P0+ fluidghh: distance between liquid surface and the point where you measure P

P0

P

h

B = fluidVobjectg = Mfluidg = wfluid

The buoyant force equals the weight of the amount of water that can be put in the volume taken by the object.If object is not moving: B=wobject object= fluid

Pressure at depth h

Buoyant force for submerged object

Buoyant force for floating objecthB

w

The buoyant force equals the weight of the amount of water that can be put in the part of the volume of the object that is under water.objectVobject= waterVdisplaced h= objectVobject/(waterA)

PHY 2314

Bernoulli’s equation

P1+½v12+gy1=

P2+½v22+gy2

P+½v2+gy=constant

The sum of the pressure (P), the kinetic energy per unit volume (½v2) and the potential energy per unit volume (gy)is constant at all points along a path of flow.

Note that for an incompressible fluid:A1v1=A2v2

This is called the equation ofcontinuity.

PHY 2315

Poiseuille’s Law

How fast does a fluid flowthrough a tube?

Rate of flow Q= v/t=R4(P1-P2)

8L(unit: m3/s)

PHY 2316

Bulk modulus

A rubber ball, with bulk modulus B, volume V at 1 atm.increases its volume by 1 cm3 when put in a vacuumchamber (P=0). If a ball of the same material but 5times larger in volume at 1 atm, is put under a pressure of3 atm, how much will its volume shrink?

B=-P/(V/V)

First case (1 atm -> vacuum): B=(-1 atm)/(-1 cm3/V)

Second case (1 atm -> 3 atm): B=(2 atm)/(V/5V)

B is a constant, so must be equal in both cases: V=10 cm3

If you are not sure whether you need to convert to SI units,just do it: it is a bit of extra work, but at least you are sure it’s okay.

PHY 2317

Young’s modulus

Consider 2 steel rods, A and B. B has 3 times the areaand 2 times the length of A, so Young’s modulus for Bwill be what factor times Young’s modulus for A?

a) 3.0b) 0.5c) 1.5d) 1.0

Same material, same Young’s modulus!

PHY 2318

Buoyant forces

When submerged in water an object weighs 1.6N. At thesame time, the water level in the water container (withA=0.01 m2) rises 0.01 m. What is the specific gravity (sg) of the object? (water=1.0x103 kg/m3)

A=0.01 m2

Use the fact that the Buoyant forceon a submerged object equals the weight of the displaced water.W =Fg-B

=Mobjectg-Mwater,displacedg =objectVobjectg-waterVobjectg

=Vobjectg (object-water)1.6N =0.01*0.01*g(object-water)=1.0x10-4*9.8*water(sg-1)sg=2.63

PHY 2319

Keep it coming.A plastic bag contains a glucosesolution. The part of the bag thatis not filled is under vacuum. If thepressure in a blood vein is 1.33x104 Pa,how high must one hang the bag tomake sure the solution (specific gravity1.02) enters the body? (w=1.0x103kg/m3)

P=P0+gh1.33x104=0+1.02*1.0x103*9.8*hh=1.33 m

PHY 23110

Titanic: After the Titanic sunk, 10 peoplemanage to seek refuge on a 2x4m wooden raft. It is still 1.0 cm above water. A heavy debate follows when another person (60 kg) wants to board as well. Fortunately, a PHY231 student is among the 10. Can she convince the others that it is safe to pull the person on board without the whole raft sinking? (w=1.0x103 kg/m3)

With 10 people: Fg=B (Mraft+M10)g=Vdisplaced,before wg

With 11 people: Fg=(Mraft+M10+M1)g B=(Vdisplaced,before+Vextra) wg

stationary if Fg=B (Mraft+M10)g+M1g=(Vdisplaced,before+Vextra)wg M1g=Vextra wg so Vextra=(M1/w)

Vextra=60/1.0x103=0.06m3

Vextra=LxWxH=8H=0.06 so H=0.75cm so there is still 0.25cm to spare!

PHY 23111

BernoulliA=5cm2

A=2cm2, P=1 atm

2m

Water flows over a height of 2m through an oddly shaped pipe.A) If the fluid velocity is 1 m/s at the bottom, what is it at the top? B) What is the water pressure at the top?A) Use the equation of continuity: A1v1=A2v2

5*vtop=2*1 vtop=0.4 m/sB) Use Bernoulli. Ptop+½vtop

2+ghtop= Pbot+½vbot2+ghbot

Ptop+0.5*(1E+03)*0.42+(1E+3)*9.8*2=(1E+05)+0.5*(1E+03)12

Ptop=80820 Pa.

=1.0x103 kg/m3

PHY 23112

Temperature scales

ConversionsTcelsius=Tkelvin-273.5Tfahrenheit=9/5*Tcelcius+32

We will use Tkelvin.

If Tkelvin=0, the atoms/moleculeshave no kinetic energy and everysubstance is a solid; it is called theAbsolute zero-point.

Kelvin

Celsius Fahrenheit

Chapter 10

PHY 23113

Thermal expansion

L=LoT

L0

L

T=T0T=T0+T

A=AoT =2

V=VoT =3

length

surface

volume

Some examples:=24E-06 1/K Aluminum=1.2E-04 1/K Alcohol

: coefficient of linear expansion different for each material

lead bell

PHY 23114

Boyle & Charles & Gay-LussacIDEAL GAS LAW

PV/T = nR

n: number of particles in the gas (mol)R: universal gas constant 8.31 J/mol·K

If no molecules are extracted from or added to a system:

2

22

1

11 constant T

VP

T

VP

T

PV

PHY 23115

M

RT

m

Tkvv

nRTTNkE

Tkvm

vmk

T

TNkPV

vmNPV

brms

Bkin

B

B

B

33

2

3

2

32

3

2

1

)2

1(

3

2

2

1

3

2

2

2

2

2

Microscopic

Macroscopic

Temperature ~ average molecular kinetic energy

Average molecular kinetic energy

Total kinetic energy

rms speed of a moleculeM=Molar mass (kg/mol)

PHY 23116

Ch. 10 Metal hoop

A metal (thermal expansion coefficient=17x10-6 1/0C) hoop of radius 0.10 m is heated from200C to 1000C. By how much does its radius change?

0.1mL =L0 T=17x10-6(2r0)80=8.5x10-4m

rnew=(L0+L)/2=L0/2+L/2=r0+1.35x10-4m

PHY 23117

10: Moles

Two moles of Nitrogen gas (N2) are enclosed in a cylinderwith a moveable piston. A) If the temperature is 298 K and the pressure is 1.01x106 Pa, what is the volume (R=8.31 J/molK)?b) What is the average kinetic energy of the molecules? kB=1.38x10-23 J/K

A)PV=nRT V=nRT/P =2*8.31*298/1.01x106=4.9E-03 m3

B) Ekin,average=½mv2=3/2kBT=3/2*1.38x10-23*298=6.2x10-21 J

PHY 23118

Gas law

One way to heat a gas is to compress it. A gas at 1.00atm at 250C is compressed to one tenth of its original volume and it reaches 40.0 atm pressure. What is itsnew temperature?

use P1V1/T1=P2V2/T2

P1=1.00 atmP2=40.0 atmV2=V1/10T1=273+25=298 K

T2=P2V2T1 /(P1V1)=40.0*(0.1*V1)*298/(1.00*V1)= =1192 K=919oC

PHY 23119

Calorimetry

If we connect two objects with different temperatureenergy will transferred from the hotter to the coolerone until their temperatures are the same. If the system is isolated:

Qcold=-Qhot

mcoldccold(Tfinal-Tcold)=-mhotchot(Tfinal-Thot)

the final temperature is: Tfinal=

mcoldccoldTcold+mhotchotThot

mcoldccold+mhotchot

PHY 23120

Phase Change

GAS(high T)

liquid (medium T)

Solid (low T)Q=cgasmT

Q=cliquidmT

Q=csolidmT

Gas liquid

liquid solid

Q=mLf

Q=mLv

make sure you can calculate cases likewater and gold shownin earlier lectures

PHY 23121

Heat transfer via conduction

Conduction occurs if there is a temperature difference betweentwo parts of a conducting medium

Rate of energy transfer PP=Q/t (unit Watt)P=kA(Th-Tc)/x=kAT/x

k: thermal conductivityUnit:J/(msoC)

metal k~300 J/(msoC)gases k~0.1 J/(msoC)nonmetals~1 J/(msoC)

iii

ch

kL

TTA

t

QP

)/(

)(more than 1 layer:

PHY 23122

Radiation

Nearly all objects emit energy through radiation:P=AeT4 : Stefan’s law (J/s)

=5.6696x10-8 W/m2K4

A: surface areae: object dependent constant emissivity (0-1)T: temperature (K)

P: energy radiated per second.

If an object would only emit radiation it would eventuallyhave 0 K temperature. In reality, an object emits ANDreceives radiation.

P=Ae(T4-T04) where

T: temperature of objectT0: temperature of surroundings.

PHY 23123

11: Heat transfer

3

A hot block and a cold block are thermally connected.Three different methods to transfer heat are proposed

as shown. Which one is the mostefficient way (fastest) to transfer heat from hot to cold and what arethe relative rates of transfer?Area: A Length L

0.1A 0.2L

4A 5L

A: cross section surfaceof black wire,L:its length

Use: P=kAT/LCase 1: P~A/LCase 2: P~0.1A/0.2L=0.5A/LCase 3: P~4A/5L=0.8A/LP1:P2:P3 = 1:0.5:0.8First case is most efficient.

1

2

PHY 23124

Thermal equilibrium20g of a solid at 700C is placed in 100g of a fluid at 20oC.After waiting a while the temperature of the whole systemis 30oC and stays that way. The specific heat of the solid is:a) Equal to that of the fluidb) Less than that of the fluidc) Larger than that of the fluidd) Unknown; different phases cannot be comparede) Unknown; different materials cannot be compared

Qfluid=-Qsolid

mfluidcfluid(Tfinal-Tfluid)=-msolidcsolid(Tfinal-Tsolid)

Cfluid -msolid(Tfinal-Tsolid)

Csolid mfluid (Tfinal-Tfluid)= =

-20(30-70)

100(30-20)= 0.8

Csolid>Cfluid

PHY 23125

radiation

An object at 270C has its temperature increased to 370C. The power than radiated by this objectincreases by how many percent?

P= AeT4

Ti=273+27=300 KTf=273+37=310 K

P~T4 Pi~3004 Pf=3104

Pf/Pi=1.14 increase by 14%

PHY 23126

First Law of thermodynamics

U=Uf-Ui=Q+W

U=change in internal energyQ=energy transfer through heat (+ if heat is

transferred to the system)W=energy transfer through work (+ if work is

done on the system) if P: constant then W=-PV (area under P-V diagram

This law is a general rule for conservation of energy

PHY 23127

Types of processes

A: Isovolumetric V=0B: Adiabatic Q=0C: Isothermal T=0D: Isobaric P=0

PV/T=constant

PHY 23128

work on a gas.

A gas, kept at constant pressure all of the time,is heated from 300 to 400 K. If the original volume was1 m3 P=1 atm, how much work has been done on the gas?

P1V1/T1=P2V2/T2

P1=P2 T1=270 K T2=300 KV2=V1T2/T1=1*400/300=1.33 m3

Isobaric, so: W=-PV=-1x105*0.33=-3.3x104 J1 atm=1x105 Pa.The work done on the gas is negative, so the gas has donework (positive).