1 / 28

PHYSICS 231 Lecture 30: review

PHYSICS 231 Lecture 30: review. Remco Zegers Walk-in hour: Monday 9:15-10:15 am Helproom. chapter 9. Young’s modulus. Solids:. Shear modulus. Bulk modulus Also fluids. General:. P=F/A (N/m 2 =Pa) F pressure-difference = PA =M/V (kg/m 3 ).

kristopher-tannis
Télécharger la présentation

PHYSICS 231 Lecture 30: review

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. PHYSICS 231Lecture 30: review Remco Zegers Walk-in hour: Monday 9:15-10:15 am Helproom PHY 231

  2. chapter 9 Young’s modulus Solids: Shear modulus Bulk modulus Also fluids General: P=F/A (N/m2=Pa) Fpressure-difference=PA =M/V (kg/m3) Pascal’s principle: a change in pressure applied to a fluid that is enclosed is transmitted to the whole fluid and all the walls of the container that hold the fluid. PHY 231

  3. P0 h B w Pressure at depth h P= P0+ fluidgh h: distance between liquid surface and the point where you measure P h P Buoyant force for submerged object 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 Buoyant force for floating object 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 231

  4. Bernoulli’s equation P1+½v12+gy1= P2+½v22+gy2 P+½v2+gy=constant The sum of the pressure (P), the kinetic energy per unit volume (½v2) 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 of continuity. PHY 231

  5. Poiseuille’s Law How fast does a fluid flow through a tube? R4(P1-P2) (unit: m3/s) Rate of flow Q= v/t= 8L PHY 231

  6. Bulk modulus A rubber ball, with bulk modulus B, volume V at 1 atm. increases its volume by 1 cm3 when put in a vacuum chamber (P=0). If a ball of the same material but 5 times larger in volume at 1 atm, is put under a pressure of 3 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 231

  7. Same material, same Young’s modulus! Young’s modulus • Consider 2 steel rods, A and B. B has 3 times the area • and 2 times the length of A, so Young’s modulus for B • will be what factor times Young’s modulus for A? • 3.0 • 0.5 • 1.5 • 1.0 PHY 231

  8. Buoyant forces When submerged in water an object weighs 1.6N. At the same time, the water level in the water container (with A=0.01 m2) rises 0.01 m. What is the specific gravity (sg) of the object? (water=1.0x103 kg/m3) Use the fact that the Buoyant force on 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 A=0.01 m2 PHY 231

  9. Keep it coming. A plastic bag contains a glucose solution. The part of the bag that is not filled is under vacuum. If the pressure in a blood vein is 1.33x104 Pa, how high must one hang the bag to make sure the solution (specific gravity 1.02) enters the body? (w=1.0x103kg/m3) P=P0+gh 1.33x104=0+1.02*1.0x103*9.8*h h=1.33 m PHY 231

  10. Titanic: After the Titanic sunk, 10 people manage 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 231

  11. A=5cm2 Bernoulli =1.0x103 kg/m3 2m A=2cm2, P=1 atm 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/s B) Use Bernoulli. Ptop+½vtop2+ghtop= Pbot+½vbot2+ghbot Ptop+0.5*(1E+03)*0.42+(1E+3)*9.8*2=(1E+05)+0.5*(1E+03)12 Ptop=80820 Pa. PHY 231

  12. Chapter 10 Temperature scales Conversions Tcelsius=Tkelvin-273.5 Tfahrenheit=9/5*Tcelcius+32 We will use Tkelvin. If Tkelvin=0, the atoms/molecules have no kinetic energy and every substance is a solid; it is called the Absolute zero-point. Celsius Fahrenheit Kelvin PHY 231

  13. Thermal expansion L length L=LoT surface A=AoT =2 V=VoT =3  volume L0 : coefficient of linear expansion different for each material Some examples: =24E-06 1/K Aluminum =1.2E-04 1/K Alcohol T=T0+T T=T0 lead bell PHY 231

  14. 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: PHY 231

  15. Microscopic Macroscopic Temperature ~ average molecular kinetic energy Average molecular kinetic energy Total kinetic energy rms speed of a molecule M=Molar mass (kg/mol) PHY 231

  16. Ch. 10 Metal hoop A metal (thermal expansion coefficient =17x10-6 1/0C) hoop of radius 0.10 m is heated from 200C to 1000C. By how much does its radius change? 0.1m L =L0 T =17x10-6(2r0)80=8.5x10-4m rnew=(L0+L)/2=L0/2+L/2=r0+1.35x10-4m PHY 231

  17. 10: Moles Two moles of Nitrogen gas (N2) are enclosed in a cylinder with 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 231

  18. Gas law One way to heat a gas is to compress it. A gas at 1.00 atm at 250C is compressed to one tenth of its original volume and it reaches 40.0 atm pressure. What is its new temperature? use P1V1/T1=P2V2/T2 P1=1.00 atm P2=40.0 atm V2=V1/10 T1=273+25=298 K T2=P2V2T1 /(P1V1)=40.0*(0.1*V1)*298/(1.00*V1)= =1192 K=919oC PHY 231

  19. Calorimetry If we connect two objects with different temperature energy will transferred from the hotter to the cooler one 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 231

  20. Phase Change Q=cgasmT Q=csolidmT Solid (low T) GAS(high T) liquid  solid Gas  liquid Q=mLv liquid (medium T) Q=mLf Q=cliquidmT make sure you can calculate cases like water and gold shown in earlier lectures PHY 231

  21. Heat transfer via conduction Conduction occurs if there is a temperature difference between two parts of a conducting medium Rate of energy transfer P P=Q/t (unit Watt) P=kA(Th-Tc)/x=kAT/x k: thermal conductivity Unit:J/(msoC) metal k~300 J/(msoC) gases k~0.1 J/(msoC) nonmetals~1 J/(msoC) more than 1 layer: PHY 231

  22. Radiation Nearly all objects emit energy through radiation: P=AeT4 : Stefan’s law (J/s) =5.6696x10-8 W/m2K4 A: surface area e: object dependent constant emissivity (0-1) T: temperature (K) P: energy radiated per second. If an object would only emit radiation it would eventually have 0 K temperature. In reality, an object emits AND receives radiation. P=Ae(T4-T04)where T: temperature of object T0: temperature of surroundings. PHY 231

  23. 11: Heat transfer A hot block and a cold block are thermally connected. Three different methods to transfer heat are proposed as shown. Which one is the most efficient way (fastest) to transfer heat from hot to cold and what are the relative rates of transfer? A: cross section surface of black wire,L:its length Area: A Length L Use: P=kAT/L Case 1: P~A/L Case 2: P~0.1A/0.2L=0.5A/L Case 3: P~4A/5L=0.8A/L P1:P2:P3 = 1:0.5:0.8 First case is most efficient. 1 0.1A 0.2L 2 3 4A 5L PHY 231

  24. Qfluid=-Qsolid mfluidcfluid(Tfinal-Tfluid)=-msolidcsolid(Tfinal-Tsolid) Cfluid -msolid(Tfinal-Tsolid) -20(30-70) = = = 0.8 Csolid mfluid (Tfinal-Tfluid) 100(30-20) Csolid>Cfluid Thermal equilibrium • 20g of a solid at 700C is placed in 100g of a fluid at 20oC. • After waiting a while the temperature of the whole system • is 30oC and stays that way. The specific heat of the solid is: • Equal to that of the fluid • Less than that of the fluid • Larger than that of the fluid • Unknown; different phases cannot be compared • Unknown; different materials cannot be compared PHY 231

  25. radiation An object at 270C has its temperature increased to 370C. The power than radiated by this object increases by how many percent? P= AeT4 Ti=273+27=300 K Tf=273+37=310 K P~T4 Pi~3004 Pf=3104 Pf/Pi=1.14 increase by 14% PHY 231

  26. First Law of thermodynamics U=Uf-Ui=Q+W U=change in internal energy Q=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=-PV (area under P-V diagram This law is a general rule for conservation of energy PHY 231

  27. Types of processes A: Isovolumetric V=0 B: Adiabatic Q=0 C: Isothermal T=0 D: Isobaric P=0 PV/T=constant PHY 231

  28. 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 was 1 m3 P=1 atm, how much work has been done on the gas? P1V1/T1=P2V2/T2 P1=P2 T1=270 K T2=300 K V2=V1T2/T1=1*400/300=1.33 m3 Isobaric, so: W=-PV=-1x105*0.33=-3.3x104 J 1 atm=1x105 Pa. The work done on the gas is negative, so the gas has done work (positive). PHY 231

More Related