I do not have direct access to my webs this week, which is causing delays in my responses. I do have access to my email. If you want quicker feedback, you may use email; however continue to also submit your work by the form. The work included here will be posted by the first of next week. assignment #017 017. `query 7 Physics II 03-14-2007
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12:28:24 Query set 5 problems 16-20 explain how we calculate the average force exerted by a particle of mass m moving at speed v which collides at a time interval `dt elastically and at a right angle with a solid wall
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RESPONSE --> When the particles collide with the wall the change in velocity is 2* the velocity, because they are considered elastic,The momentum change is found by the total mass * the velocity change. Once we have the momentum change we can determine the average force by dividing the momentum change by the time interval.
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12:29:32 ** The impulse exerted on a particle in a collision is the change in the momentum of that particle during a collision. The impulse-momentum theorem says that the change in momentum in a collision is equal to the impulse, the average force * the time interval between collisions. The average force is thus change in momentum / time interval; the time interval is the round-trip distance divided by the velocity, or 2L / v so the average force is -2 m v / ( 2L / v) = m v^2 / L If there were N such particles the total average force would be N * m v^2 / L If the directions are random we distribute the force equally over the 3 dimensions of space and for one direction we get get 1/3 the force found above, or 1/3 N * m v^2 / L. This 3-way distribution of force is related to the fact that for the average velocity vector we have v^2 = vx^2 + vy^2 + vz^2, where v is average magnitude of velocity and vx, vy and vz the x, y and z components of the velocity (more specifically the rms averages--the square root of the average of the squared components). **
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RESPONSE --> It looks like I may have left a few steps out , but I understand what they are.
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12:33:20 Summarize the relationship between the thermal energy that goes into the system during a cycle, the work done by the system during a cycle, and the thermal energy removed or dissipated during the cycle.
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RESPONSE --> The work done by the energy in, is always less than the energy in, due to energy dissapated through the cycle. The thermal energy removed during the cycle is a measure of the effeciency of the system.
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12:34:05 ** Work-energy is conserved within an isolated system. So the thermal energy that goes into the system must equal the total of the work done by the system and the thermal energy removed from the system. What goes in must come out, either in the form of work or thermal energy. **
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RESPONSE --> I think I was close with this one. I just have different wording.
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12:36:28 If you know the work done by a thermodynamic system during a cycle and the thermal energy removed or dissipated during the cycle, how would you calculate the efficiency of the cycle?
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RESPONSE --> You can add the work and the dissipated energy and find the total energy. Then you can find the effeciency by dividing the work done by the total energy into the system.
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12:36:39 ** STUDENT SOLUTION: Efficiency is work done / energy input. Add the amount thermal energy removed to the amount of work done to get the input. Then, divide work by the energy input. **
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RESPONSE --> ok
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12:43:25 prin phy and gen phy problem 15.2, cylinder with light frictionless piston atm pressure, 1400 kcal added, volume increases slowly from 12.0 m^3 to 18.2 m^3. Find work and chagne in internal energy.
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RESPONSE --> I have the old version of the book, that one is not in mine.
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12:51:11 Work done at constant pressure is P `dV, so the work done in this situation is `dW = P `dV = 1 atm * (18.2 m^3 - 12 m^3) = (101.3 * 10^3 N/m^2) * (6.2 m^3) = 630 * 10^3 N * m = 6.3 * 10^5 J. A total of 1400 kcal = 1400 * 4200 J = 5.9 * 10^6 J of thermal energy is added to the system, the change in internal energy is `dU = `dQ - `dW = 5.9*10^6 J - 6.3 * 10^5 J = 5.9 * 10^6 J - .63 * 10^6 J = 5.3 * 10^6 J. It is worth thinking about the P vs. V graph of this process. The pressure P remains constant at 101.3 * 10^3 J as the volume changes from 12 m^3 to 18.2 m^3, so the graph will be a straight line segment from the point (12 m^3, 101.3 * 10^3 Pa) to the point (18.2 m^3, 101.3 * 10^3 Pa). This line segment is horizontaland the region above the horizontal axis and beneath the segment is a rectangle whose width is 6.2 * 10^3 m^3 and whose altitude is 101.3 * 10^3 N/m^2; its area is therefore the product of its altitude and width, which is 6.3 * 10^5 N m, or 6.3 * 10^5 J, the same as the word we calculated above.
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RESPONSE --> I followed along with the mathematical process, but I lost you on the graph.
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13:02:30 prin phy and gen phy problem 15.5, 1.0 L at 4.5 atm isothermally expanded until pressure is 1 atm then compressed at const pressure to init volume, final heated to return to original volume. Sketch and label graph.
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RESPONSE --> I'm not sure how to go about this.
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13:06:05 When a confined ideal gas is expanded isothermally its pressure and volume change, while the number of moled and the temperature remain constant. Since PV = n R T, it follows that P V remains constant. In the initial state P = 4.5 atm and V = 1 liter, so P V = 4.5 atm * 1 liter = 4.5 atm * liter (this could be expressed in standard units since 1 atm = 101.3 kPa = 101.3 * 10^3 N/m^2 and 1 liters = .001 m^3, but it's more convenient to first sketch and label the graph in units of atm and liters). During the isothermal expansion, therefore, since P V remains constant we have P V = 4.5 atm liters. At a pressure of 1 atm, therefore, the volume will be V = 4.5 atm liter / P = 4.5 atm liter / (1 atm) = 4.5 liters. The graph follows a curved path from (1 liter, 4.5 atm) to (4.5 liters, 1 atm).
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RESPONSE --> for some reason I'm just not getting this.
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12:32:26 prin phy and gen phy problem 15.19 What is the maximum efficiency of a heat engine operating between temperatures of 380 C and 580 C?
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RESPONSE --> Change temps to K degrees 380+273=653, 580+273=853 (th-tl)/th=effeciency (853-653)/853=.2345
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12:32:59 The maximum possible efficiency is (T_h - T_c) / T_h, where T_h and T_c are the absolute max and min operating temperatures. T_h is (580 + 273)K = 853 K and T_c is (380 + 273) K = 653 K, so the maximum theoretical efficiency is max efficiency = (T_h - T_c) / T_h = (853 K - 653 K) / (853 K) = .23, approx. This means that the work done by this engine will be not greater than about 23% of the thermal energy that goes into it.
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RESPONSE --> ok
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12:48:12 query gen phy problem 15.26 source 550 C -> Carnot eff. 28%; source temp for Carnot eff. 35%?
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RESPONSE --> change temp to K degrees 550+273=823K (th-tl)/th=effeciency so (823-tl)/823=.28 solvinh for tl we get 592.56K for effecincy of 35% (823-tl)/823=.35 solving for tl we get 534.95
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12:50:09 ** Carnot efficiency is eff = (Th - Tc) / Th. Solving this for Tc we multiply both sides by Th to get eff * Th = Th - Tc so that Tc = Th - eff * Th = Th ( 1 - eff). We note that all temperatures must be absolute so we need to work with the Kelvin scale (adding 273 C to the Celsius temperature to get the Kelvin temperature) If Th = 550 C = 823 K and efficiency is 30% then we have Tc =823 K * ( 1 - .28) = 592 K. Now we want Carnot efficiency to be 35% for this Tc. We solve eff = (Th - Tc) / Th for Th: Tc we multiply both sides by Th to get eff * Th = Th - Tc so that eff * Th - Th = -Tc and Tc = Th - eff * Th or Tc = Th ( 1 - eff) and Th = Tc / (1 - eff). If Tc = 576 K and eff = .35 we get Th = 592 K / ( 1 - .35 ) = 592 C / .6 = 912 K, approx. This is (912 - 273) C = 639 C. **
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RESPONSE --> I guess I misunderstood the second part of the question, I solved it for the same th, but a 35% effecincy
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