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

2. PHY 231 2 chapter 9 Solids: Young’s modulus Shear modulus Bulk modulus Also fluids P=F/A (N/m 2 =Pa) F pressure-difference =  PA  =M/V (kg/m 3 ) General: 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.

3. PHY 231 3 P=P 0 +  fluid gh h: distance between liquid surface and the point where you measure P P0P0 P h B =  fluid V object g = M fluid g = w fluid 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=w object  object =  fluid Pressure at depth h Buoyant force for submerged object Buoyant force for floating object h B 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.  object V object =  water V displaced h=  object V object /(  water A)

4. PHY 231 4 Bernoulli’s equation P 1 +½  v 1 2 +  gy 1 = P 2 +½  v 2 2 +  gy 2 P+½  v 2 +  gy=constant The sum of the pressure (P), the kinetic energy per unit volume (½  v 2 ) and the potential energy per unit volume (  gy) is constant at all points along a path of flow. Note that for an incompressible fluid: A 1 v 1 =A 2 v 2 This is called the equation of continuity.

5. PHY 231 5 Poiseuille’s Law How fast does a fluid flow through a tube? Rate of flow Q=  v/  t=  R 4 (P 1 -P 2 ) 8L8L (unit: m 3 /s)

6. PHY 231 6 Bulk modulus A rubber ball, with bulk modulus B, volume V at 1 atm. increases its volume by 1 cm 3 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 cm 3 /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 cm 3 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.

7. PHY 231 7 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? a)3.0 b)0.5 c)1.5 d)1.0 Same material, same Young’s modulus!

8. 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 m 2 ) rises 0.01 m. What is the specific gravity (sg) of the object? (  water =1.0x10 3 kg/m 3 ) A=0.01 m 2 Use the fact that the Buoyant force on a submerged object equals the weight of the displaced water. W=F g -B =M object g-M water,displaced g =  object V object g-  water V object g =V object g (  object -  water ) 1.6N=0.01*0.01*g(  object -  water )=1.0x10 -4 *9.8*  water (sg-1) sg=2.63

9. 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 1.33x10 4 Pa, how high must one hang the bag to make sure the solution (specific gravity 1.02) enters the body? (  w =1.0x10 3 kg/m 3 ) P=P 0 +  gh 1.33x10 4 =0+1.02*1.0x10 3 *9.8*h h=1.33 m

10. 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 =1.0x10 3 kg/m 3 ) With 10 people: F g =B (M raft +M 10 )g=V displaced,before  w g With 11 people: F g =(M raft +M 10 +M 1 )g B=(V displaced,before +V extra )  w g stationary if F g =B (M raft +M 10 )g+M 1 g=(V displaced,before +V extra )  w g M 1 g=V extra  w g so V extra =(M 1 /  w ) V extra =60/1.0x10 3 =0.06m 3 V extra =LxWxH=8H=0.06 so H=0.75cm so there is still 0.25cm to spare!

11. PHY 231 11 Bernoulli A=5cm 2 A=2cm 2, 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: A 1 v 1 =A 2 v 2 5*v top =2*1 v top =0.4 m/s B) Use Bernoulli. P top +½  v top 2 +  gh top = P bot +½  v bot 2 +  gh bot P top +0.5*(1E+03)*0.4 2 +(1E+3)*9.8*2=(1E+05)+0.5*(1E+03)1 2 P top =80820 Pa.  =1.0x10 3 kg/m 3

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

13. PHY 231 13 Thermal expansion  L=  L o  T L0L0 LL T=T 0 T=T 0 +  T  A=  A o  T  =2   V=  V o  T  =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

14. PHY 231 14 Boyle & Charles & Gay-Lussac IDEAL 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:

15. 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)

16. PHY 231 16 Ch. 10 Metal hoop A metal (thermal expansion coefficient  =17x10 -6 1/ 0 C) hoop of radius 0.10 m is heated from 20 0 C to 100 0 C. By how much does its radius change? 0.1m  L=  L 0  T =17x10 -6 (2  r 0 )80=8.5x10 -4 m r new =(L 0 +  L)/2  =L 0 /2  +  L/2  =r 0 +1.35x10 -4 m

17. PHY 231 17 10: Moles Two moles of Nitrogen gas (N 2 ) are enclosed in a cylinder with a moveable piston. A) If the temperature is 298 K and the pressure is 1.01x10 6 Pa, what is the volume (R=8.31 J/molK)? b) What is the average kinetic energy of the molecules? k B =1.38x10 -23 J/K A)PV=nRT V=nRT/P =2*8.31*298/1.01x10 6 =4.9E-03 m 3 B) E kin,average =½mv 2 =3/2k B T=3/2*1.38x10 -23 *298=6.2x10 -21 J

18. PHY 231 18 Gas law One way to heat a gas is to compress it. A gas at 1.00 atm at 25 0 C is compressed to one tenth of its original volume and it reaches 40.0 atm pressure. What is its new temperature? use P 1 V 1 /T 1 =P 2 V 2 /T 2 P 1 =1.00 atm P 2 =40.0 atm V 2 =V 1 /10 T 1 =273+25=298 K T 2 =P 2 V 2 T 1 /(P 1 V 1 )=40.0*(0.1*V 1 )*298/(1.00*V 1 )= =1192 K=919 o C

19. 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: Q cold =-Q hot m cold c cold (T final -T cold )=-m hot c hot (T final -T hot ) the final temperature is: T final = m cold c cold T cold +m hot c hot T hot m cold c cold +m hot c hot

20. PHY 231 20 Phase Change GAS(high T) liquid (medium T) Solid (low T) Q=c gas m  T Q=c liquid m  T Q=c solid m  T Gas  liquid liquid  solid Q=mL f Q=mL v make sure you can calculate cases like water and gold shown in earlier lectures

21. 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(T h -T c )/  x=kA  T/  x k: thermal conductivity Unit:J/(ms o C) metal k~300 J/(ms o C) gases k~0.1 J/(ms o C) nonmetals~1 J/(ms o C) more than 1 layer:

22. PHY 231 22 Radiation Nearly all objects emit energy through radiation: P=  AeT 4 : Stefan’s law (J/s)  =5.6696x10 -8 W/m 2 K 4 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(T 4 -T 0 4 ) where T: temperature of object T 0 : temperature of surroundings.

23. PHY 231 23 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 most efficient way (fastest) to transfer heat from hot to cold and what are the relative rates of transfer? Area: A Length L 0.1A 0.2L 4A 5L A: cross section surface of black wire,L:its length 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 P 1 :P 2 :P 3 = 1:0.5:0.8 First case is most efficient. 1 2

24. PHY 231 24 Thermal equilibrium 20g of a solid at 70 0 C is placed in 100g of a fluid at 20 o C. After waiting a while the temperature of the whole system is 30 o C and stays that way. The specific heat of the solid is: a)Equal to that of the fluid b)Less than that of the fluid c)Larger than that of the fluid d)Unknown; different phases cannot be compared e)Unknown; different materials cannot be compared Q fluid =-Q solid m fluid c fluid (T final -T fluid )=-m solid c solid (T final -T solid ) C fluid -m solid (T final -T solid ) C solid m fluid (T final -T fluid ) == -20(30-70) 100(30-20) = 0.8 C solid >C fluid

25. PHY 231 25 radiation An object at 27 0 C has its temperature increased to 37 0 C. The power than radiated by this object increases by how many percent? P=  AeT 4 T i =273+27=300 K T f =273+37=310 K P~T 4 P i ~300 4 P f =310 4 P f /P i =1.14 increase by 14%

26. PHY 231 26 First Law of thermodynamics  U=U f -U i =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

27. 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

28. 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 m 3 P=1 atm, how much work has been done on the gas? P 1 V 1 /T 1 =P 2 V 2 /T 2 P 1 =P 2 T 1 =270 K T 2 =300 K V 2 =V 1 T 2 /T 1 =1*400/300=1.33 m 3 Isobaric, so: W=-P  V=-1x10 5 *0.33=-3.3x10 4 J 1 atm=1x10 5 Pa. The work done on the gas is negative, so the gas has done work (positive).