Physics 231 Lecture 32 Problem and review

Transcription

1 Physics 231 Lecture 32 Problem and review

2 Checking Understanding: Pressure and Forces The two identical cylinders contain samples of gas. Each cylinder has a lightweight piston on top that is free to move, so the pressure inside each cylinder is equal to atmospheric pressure. One cylinder contains hydrogen, the other nitrogen. The mass of gas in each cylinder is the same. The temperature of the hydrogen gas is A. greater than the temperature of the nitrogen. B. equal to the temperature of the nitrogen. C. less than the temperature of the nitrogen. PV = nrt PV T = nr m = n H i 2g = n N i 28g n H = n i 28g N 2g PV n H R PV n N R n H = 14n N T H = = T N T H /T N =? V H =V N =V P H =P N =P m H =m N =m 1 n H 1 n N = n N n H = 1 14 Slide 12-26

3 Example of equilibration and energy conservation At a fabrication plant, a hot metal forging has a mass of 75 kg and a specific heat capacity of 430 J/(kg C). To harden it, the forging is quenched by immersion in 710 kg of oil that has a temperature of 32 C and a specific heat capacity of 645 cal/(kg C). The final temperature of the oil and forging at thermal equilibrium is 47 C. Assuming that heat flows only between the forging and the oil, determine the initial temperature of the metal forging. Energy conservation: Q c m T T Δ Q = c m T T Δ Q +Δ Q = 0 metal oil Δ = ( ) oil oil oil ( f oil ) 4.19J cal metal metal metal f metal o o coil = 645cal/(kg C) = 2700kJ/(kg C) ( ) = ( ) c m T T c m T T T metal metal f metal oil oil f oil f T = metal T = T + metal f ( ) c m T T oil oil f oil c metal m metal ( ) c m T T oil oil f oil c metal m Δ Q metal metal = ΔQ oil m metal m oil c metal 75 kg 710 kg ( )( ) o C) ( 75kg) 430J/kg/ o C o o o 2700J / (kg C) 710kg C o 47 C 939 C c oil 645cal/kg/ o C = + = 430J / (kg T f T oil,0 T metal,0? 47 o C 32 o C

4 example Two moles of the monatomic gas helium (C V =3/2R) are initially at a temperature of 300K. The gas is cooled at constant volume. How much heat must be removed to decrease the temperature (in Kelvin) by a factor of two? a) 1.7 kj b) 2.7 kj c) 3.7 kj d) 4.7 kj e) 5.7 kj 3 Eth,i 2 E = E + Q th,f E 3 nrt 2 = nrt i th,f f th,i Q E E = th,f th,i = nrtf nrt i 3 Q nr Tf Ti 2 = = 3 2moles 8.31J / mole K 150K 2 = ( ) ( ) ( ) ( ) Q = 3.7kJ

5 m f X X RMS Speed revisited (3 gases) Gas X = 3 n N X X m Gas Y Temperatures, volumes, masses and number of moles all equal Differing degrees of freedom By the ideal gas law, PV=nRT or PV=Nk B T f Y Y = 5 n N Y Y m f Z Z = Gas Z 6 n N Y Z P X =P Y =P Z Same RMS speeds too, since Different Internal Energies: U=f/2 nrt v RMS 3kT B 3RT = = m M f Δ U = nrδ T = Q 2 Molar specific heat at constant volume C V = Q nδt ΔU = = nδt f R 2 Caution: If gas is heated at constant pressure, not constant volume, the heat capacity C p is larger, reflecting the work done during expansion.

6 Checking Understanding Rank the following in terms of the number of moles, from greatest number of moles to least: g of He (A = 4) g of Ne (A = 20) g of O 2 (atomic oxygen, O, has A = 16) g of Ar (A = 40) g of Pb (A = 207) A. 5 > 4 > 3 > 2 > 1 B. 5 > 4 > 2 > 3 > 1 C. 3 > 1 > 4 > 2 > 5 D. 1 > 4 > 3 > 2 > 5 n He = 20 / 4 = 5 n Ne = 60 / 20 = 3 N O2 = 120 / 32 = 3.75 N Ar = 160 / 40 = 4 N Pb = 207 / 208 =.995 Slide 12-14

7 Conceptual question The parameters that describe a gas are n: the number of moles, P: the pressure, T: the temperature, and V: the volume. An ideal gas is contained within a cylinder sealed at the top with an immovable plug. When heat flows into the gas, what are all the parameters that change? a) n, V b) n, T c) P, T d) T e) P, T, V n is constant, V is constant U increases by Q U=f/2nRT so T changes P=nRT/V so P changes

8 56 v rms,ne = 3RT M Ne v rms,he = 3RT M He 3RT / / / v rms,he v rms,ne = M He 3RT / / / = 1/ M He 1/ M Ne M He M Ne M He M Ne M Ne = M Ne M He = 20 4 = 2.2

9 mv 2 rms,0 = 3 2 k B T 0 If all velocities increase by 4, the rms velocity increases by mv 2 rms,f = 1 2 m(4v rms,0 )2 = mv 2 rms,0 = k B T 0 T f = 16T 0

10 1 2 mv 2 GmM 0 r 0 = 1 2 mv 2 GmM f 1 r f 2 v 2 = GM f r f GM r 0 = GM R GM 2.1R v f 2 = 2 2.1RGM RGM 2.1R 2 v f = 2.2GM 2.1R = 8.09km / s

11 4. T E 2 = 4π 2 3 M sun r E 2 T biggie = 4π 2 3 r M biggie sun 2 T biggie = 2 T E 4π 2 3 r M biggie sun 4π 2 3 r M E sun T 2 biggie 2 T E T biggie T E = r 3 biggie 3 r E = r 3/2 biggie = 2.86E11 3/2 r E 1.50E = 2.63

12 Buoyancy considerations Are there other forces active besides gravity and the buoyant force? If there are no other forces, Is the object s density less than that of the liquid? If so, it floats. If not, it sinks. If there are other forces, Do other forces pull the object partly out of the liquid decreasing the displaced volume? Do other forces push the object into the liquid increasing the displaced volume? If the object is submerged, the displaced volume is the volume of the object. If the object is not submerged, the displaced volume is determined by assuming that sum of the forces, including the buoyant force, equals zero to get the buoyant force and use that to get the displaced volume. If the displaced volume increases, the liquid level rises If the displaced volume decreases, the liquid level falls.

13 49 F B mg F down = 0 F down = 5500N F B = ρ w V elsie g mg = mg = 688.7kg g = 6749N F B = N = 1.221x10 4 N V elsie = F B ρ w g ρ elsie = m = ρ gm w V elsie F B = 1000kg / m3 6749N 1.221x10 4 N = 553kg / m 3

14 Conceptual question A 200-ton ship enters the lock of a canal. The fit between the sides of the lock and the ship is tight so that the weight of the water left in the lock after it closes is much less than 200 tons. Can the ship still float if the weight of the quantity of water left in the lock is much less than the ship s weight? a) Yes, as long as the water gets up to the ship s waterline. b) No, the ship touches bottom because it weighs more than the water in the lock. If the water lever reaches the waterline, the weight of displaced water is the weight of the water that would be need to occupy the volume that the boat displaces below the water line. When that occurs the buoyant force equals the weight of the boat, and the boat floats.

15 Conceptual question A lead weight is fastened on top of a large solid piece of Styrofoam that floats in a container of water. Because of the weight of the lead, the water line is flush with the top surface of the Styrofoam. If the piece of Styrofoam is turned upside down so that the weight is now suspended underneath it, a) the arrangement sinks. b) the water line is below the top surface of the Styrofoam. c) the water line is still flush with the top surface of the Styrofoam. Same displaced water in both figures!

16 Conceptual quiz A lead weight is fastened to a large solid piece of Styrofoam that floats in a container of water. Because of the weight of the lead, the water line is flush with the top surface of the Styrofoam. If the piece of Styrofoam is turned upside down, so that the weight is now suspended underneath it, the water level in the container a) rises. b) drops. c) remains the same. If an object floats: F B =ρ w gv displaced =weight of object. The weight doesn t change. V displaced doesn t change. The level doesn t change.

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18 11. P bottom = P top + ρgh ρgh = 1025*9.8*11000Pa = 1.105x10 8 Pa F = A P bottom = π (.15m) x10 8 Pa = 7.8e6N

19 47. An organ pipe, open at one end, produces the middle A note (440 Hz) when sustaining a standing wave at its third harmonic. What is the length of the pipe? 1) 0.22 m 6) 1.07 m 2) 0.37 m 7) 1.62 m 3) 0.58 m 8) 2.56 m 4) 0.78 m 9) 3.19 m 5) 0.97 m 10) 5.22 m L = 3 4 λ = 3 4 v f = m / s 440Hz =.58m

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22 ƒ' = ƒ v + v o v v s 200Hz = 182Hz 340m / s + v 0 340m / s 340m / s + v o = *340m / s v o = *340m / s 340m / s = 33.6m / s

23 45. A wave has a frequency of 60 Hz and a speed of 120 m/s. What are its period and wavelength? 1) 60 s, 2.0 m 6) s, 0.50 m 2) 60 s, 0.50 m 7) s, 2.0 m 3) 2.0 s, 5.0 m 8) 7,200 s, m 4) 2.0 s, 6.0 m 9) 7,200 s, 60 m 5) s, 16 m 10) s, 6.0 m λ = 120m / s = 2m 60Hz 1 T = 60Hz = 16.6ms

24 44. If two people talk simultaneously and each creates an intensity level of 45 db at a certain point, what is the total intensity level at this point? 1) 46 db 6) 65 db 2) 48 db 7) 70 db 3) 51 db 8) 80 db 4) 55 db 9) 90 db 5) 60 db 10) 100 db β 1 = 10log 10 β 2 = 10log 10 β 2 = 10log 10 I I 0 2I I 0 2I I 0 = 45db = 10log 10 I I 0 = 45db +10log log 2 10 ( ) ( ) = 45db + 3db = 48db