9B-C
Assignment Physics #10
Cole
[20Q.1] Irreversible, because for a reversible process it must be possible for all things involved in the process must be returned to their initial states. Although the pot has been returned to it's initial Q Q < 0 and entropy gained + by the kitchen at state there is the entropy lost by the burner - TH TC TC. Note the kitchen gains more entropy than the burner loses because the denomintor is smaller for the kitchen. [20Q.4] The coils on the low pressure side of the compressor get cold and absorb heat from the room the coils on the high pressure side get hot and exhaust heat to the room. No airconditioner is perfectly efficient (second law), so the amount of heat exhausted is larger than the amount absorbed. Thus, if both the cold coils and hot coils are inside the room, the room will actually get hotter, although there will be a local area near the cold coils that will be cooler. The only way to effect a net cooling of the room is to stick the hot coils outside. The cooler of ice would make the room cooler as heat flows from the room into the ice. The net entropy change is positive because Q < 0 , but the ice gains heat at a lower the room loses heat at a higher temperature Sroom = - TH Q temperature, so Sice = + > |Sroom | Stotal > 0. Thus, this process would not violate the TC second law. [20Q.6] If a human being were created by all of the molecules coming together and one instant and spontaneously creating the person, yes, it would be most improbable! This is not the way it happens. The method is called stratified stability. Atoms bang into each other and at first form simple molecules. When the simple molecules become dense enough (they form a more complex stable level), it becomes probable that some will bang into eachother forming complex molecules (the next level). When these become dense enough, it becomes probable that the next higher level will evolve, ad infinitum. Hence biological evolution does not violate the statistical interpretation of entropy. Note that the thermodynamic version of the second law does not really apply because living biological systems are "far" from being in equilibrium. Biological processes are highly irreversible and produce far more entropy than the ordering would negate. [20.17] a) From the first law: QC + QH = U + W . Because the device is run in a cycle, U = 0, and Qc = 5.00 kJ. W = QC + QH . Use the second law. The entropy change of the device for Q Q T 298 one cycle is zero H + c = 0 QH = - H Qc = - (5 kJ) = -5.67 kJ . TH Tc Tc 263 273 W = 5.00 kJ - 5.67 kJ = - 0.67 kJ and it acts as a refrigerator b) QH = - (5 kJ) = -5.19 kJ 263 W = 5.00 kJ - 5.19 kJ = - 0.19 kJ. The work is negative so it acts as a refrigerator. c) Now the -10C reservoir is the hot reservoir! W = Qc + 5.0 kJ where T 248 QC = - C QH = - (5 kJ) = -4.71 kJ W = -4.71 kJ + 5kJ = 0.29 kJ The device is TH 263 functioning as an engine.
Assigment #10, 12/9/05
Page 1 of 5
UC Davis
[20.34] a) On average each half of the box will contain half the particles so 250 nitrogen and 50 oxygen. b) Let's calculate the entropy change for the oxygen first and use the fact that the entropy is additive. Initially all of the particles are on the same side of the box so W = 1. The number of 100! ways to get an equal mix of 100 particles into two groups of 50 each is W = (50!)2 100! Soxy = klnW - klnWo = kln - kln1 Using the properties of n, (50!)2 Soxy = kln100! -2 kln50!. Use Stirling's formula: lnN!= NlnN - N 100 = 100 kln2 Similarly for nitrogen: Soxy = k (100)ln100 - 2(50)kln50 = 100 kln 50 Sn = 500 kln2 The total is: Soxy + Sn = 100 kln2 + 500 kln2 = 600 kln2 = 5.74 10-21 J/K. c) There are 2N possible combinations of N particles in two partitions (see solution for 18.30). Wo 1 Whereas, there is just one way of putting all N particles on a given side. p = N = 100 for 2 2 1 oxygen and 500 for nitrogen. To so both at once pTot = poxy pn = 2-100 2-500 = 2-600. This is too 2 small for a calculator. Take the log: log p = log 2-600 = -600 log2 = -180.62 p = 10-18010-0.62 = 2.4 10-181. [20.36] Make a table of all the states. The states are: 4 heads, 3 heads, 2 heads, 1 head, 0 heads. Note there are 2N = 24 = 16 possible combinations = Wi. We have been 1 {H H H H interpreting the probability as the frequency of occurance. T H H H W4 H W0 H W3 H 1 1 4 H T H H b) p3 H = 4 = ; a) p4 H = 4 = ; p0 H = 4 = 4 2 16 2 16 2 16 W W 4 6 H H T H p1H = 14H = c) p2 H = 24H = H H H T 2 16 2 16 d) pi = 1. The probabilites sum to 1 because we have counted all the T T H H T H T H possibilities. 4! 4! N! N! = = 6 , W3 H = = =4 Note that W2 H = T H H T 6 (nH !)(nT !) (2!)(2!) (nH !)(nT !) (3!)(1!) H T H T H H T T H T T H T T T H T T H T 4 T H T T H T T T [20.59] a) For the T-S diagram of a Carnot cycle, the T 1 {T T T T isotherms are hoizontal lines and because Q = T dS = 0 S = constant, the adiabats are vertical TH lines. The carnot cycle forms a rectangle in T-S space. b) Note that Q = T dS = area. QH = TH S and QC = -TCS c) From the first law, QH + QC = U + W, but for a cycle U = 0 W = QH + QC , so TC W QH + QC QC e= = =1+ . From our diagram S QH QH QH S Assigment #10, 12/9/05 Page 2 of 5 UC Davis
- TC S T = 1 - C which is the Carnot efficiency. T TH S TH d) An isochoric heating increases the entropy T T S = Cv ln H so the isochors are lines that have slope. The H TC T-S diagram has the shape shown. Along the isochors, the heats exchanged are equal but opposite in sign so they make no TC net contribution to heat; hence QH is the area under the top isothermal which has the same area underneath it as the Carnot S cycle. In the Carnot cycle the net work done only depends on the work done during the isothermal expansion compression. and The work done along the adiabatic expansion and compressions cancel out exactly. In the Stirling cycle, the isochoric legs W is the same do no work, so the Stirling Cycle does the same work as the Carnot cycle so e = QH for both cycles. e =1+ [20.60] a) The boiling water loses heat equal to that the ice gains. Qw = - mI Lf, so the entropy - mi L f -(0.16 kg)(334 10 3 J/kg) change is Sw = = = - 143 J/K. Tw 373 K mi L f (0.16 kg)(334 10 3 J/kg) = = 196 J/ K The ice gains heat so its entropy change is b) Si = Ti 273 K positive. c) During the time interval, the rod is in a steady state and nothing changes. It's thermodynamic state does not change (actually it is not characterized by one state because it has a range of temperatures). Srod = 0. d) The entropy is additive: ST = Sw + Si + Sr = 196 J/K - 143 J/K = + 53 J/K Because the total entropy change is positive, this is an irreversible process. [14Q.25] The velocity at a point in space is constant. As the particle moves to a different point, it's velocity changes and so it undergoes an acceleration. [14Q.27] A tornado is a spinning air mass. To conserve angular momentum, as the air moves inward, it must spin faster (just like an ice skater). By Bernoulli's principle, the pressure is least where the speed is greatest, so near the core of the tornado the pressure is least. It acts as a large vacuum sucking in everything in it's path. dV dx 1 dV 1.20 m3 / s =A = Av v1 = = 2 = 17.0 m/s dt dt A1 dt (0.15 m) 17m/s v b) By the continuity equation: v2A2 = v1 A1 r2 2 = 1 r12 r2 = (0.15 m) = 0.32 m 3.8 m/s v2 [14.35] If the diameter of the mains is very large, then the velocity of the flow in the mains is approximately zero. When the water reaches the highest point it can go, the velocity will again be zero. Using Bernoulli's equation, P2 + gh = P1 P1 atmospheric pressure, the gauge pressure in the mains must be P2 - P1 = gh =(1000 kg/m3)(9.8 m/s2 )(15 m) = 1.47 105 Pa. [14P.31] a) The volume flow rate is [14.37] The net upward force is Fy = PL A - PT A - mg Using Bernoulli's equation and Assigment #10, 12/9/05 Page 3 of 5 UC Davis
assuming the thickness of the wing is negligible: PL + 1 vL2 = PT + 1 vT2 2 2 Fy = ( 1 vT2 - 1 vL2 )A - mg 2 2 1 Fy = 2 (1.2 kg/m3)((70 m/s)2 - (60 m/s)2)(16.2 m2 ) - 1340 kg(9.8 m/s2) = -496 N [14.76] a) The forces must sum to create the centripetal acceleration. This acceleration is caused by a pressure gradient. Here it is easier to use the force per volume. From chapter 1 in the notes: P Fr = - = - r 2 where the force is radially inward and the pressure gradient points radially r Pr r out. Separating variables and integrating: dP = 2 r dr b) dP = 2 r dr
Pa 0
Pr - Pa = + 1 2 r2 . c) Let Pr be the pressure at r and the depth h. Using equation (14-5) 2 Pr - Pa = gh 1 2r2 = gh h = 2r2 /(2g). 2 [14.80] (a) Apply Bernoulli's equation to points 1 at 2: P + gH = P + g(H-h) + 1 v2 2 v = 2gh Now it becomes a problem in kinematics. x = vt where t is the time to fall a height H-h H-h =
1 2
1 h H 2 v
gt2 t =
2 H -h g
(
)
x = 2 h H -h
(
)
x
b) Solve for h x2 = 4hH - 4h2 4h2 - 4Hh + x2 = 0 h = H H 2 - x2 So there are two solutions, equidistant from H/2. 2 2
[14.84] a) For an incremental piece of air m: L = mvr Conserving L: m v1 r1 = m v2 r2 30 km r v2 = 1 v1 = 200 km/h = 17 km/h. 350 km r2 b) Using Bernoull's equation, the pressure will be greater out near the rim where the air is moving more slowly. P1 + 1 v1 2 = P2 + 1 v22 2 2 1m / s 1 3 p = (1.2 kg / m3 )((200 km / h)2 - (17 km / h)2 ) = 1.8 10 Pa. 2 3.6 km / h 2 v c) 1 mv 2 = mgh h = = 160 m. 2 2g d) In part (c) we assumed the pressure is constant, but at higher altitude the pressure is lower providing a lifting which can be several kilometers in a hurricane. dV = v1 A1 = v2 A2 [14.86] a) Conserve mass: dt 1 dV 6.00 10 -3 m3 / s 6.00 10 -3 m3 / s v= v1 = = 6.00 m / s & v2 = = 1.50 m / s A dt 10.0 10 -4 m 2 40.0 10 -4 m 2 1 2 b) Use Bernoulli's equation p = (v12 - v2 ) = 1.69 10 4 Pa 2 c) The air pressure difference must be able to support the Ahg weight of the excess mercury in p (1.69 10 4 Pa ) the tube p = gh h = = = 12.7 cm . g (13.6 10 3 kg / m3 )(9.80 m / s2 ) Assigment #10, 12/9/05 Page 4 of 5 UC Davis
2
Additional Problems [1] One way to view the pressure is as two hills next to one another. Pressure is plotted in the vertical direction versus the x-y plane. The pressure dies out from high pressure P to one standard atmosphere a long distance from the two high pressure cells:
[2] a) Let f be the force per unit volume F/V; f = - P Because the pressure only depends on y, we only need to consider the component of the gradient. ^ ^ d C = j 2Cy The force on the air will point in the positive y direction f = - j 2 2 2 dy y + a ( y + a) (north) b) The maximum in the pressure can be found by setting the gradient (first derivative) to 2Cy zero and solve for y. 2 2 = 0 y = 0. c) High pressure is at y = 0, so for y > 0 the ( y + a) wind will blow north-away from high pressure. For y < 0, the wind will blow south because f points in the negative y direction. [3] This problem is far easier if we use the polar form of the gradient: U U U 1 1 + + = =- 2 U = r r r sin r r r [4] The net flux is: dm/dt = TM v dA. From the figure on the problem set, the number of lines entering the entire surface equals the number leaving so dm/dt = 0. We can think of the flux as composed of two parts: (dm/dt)s through the sock, and (dm/dt)c through the circle. The flux through the circle is trivial: v dA = - v dA , so that, (dm/dt)c = circlev dA = - vcirdA = -a2v. Because (dm/dt)s = - (dm/dt)c, (dm/dt)s = + a2v.
Assigment #10, 12/9/05
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UC Davis