course Phy 202
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19:16:34 query introset 5 # 12: Finding the conductivity given rate of energy flow, area, temperatures, thickness of wall. Describe how we find the conductivigy given the rate of energy flow, area, temperatures, and thickness of the wall?
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RESPONSE --> Average temperature gradient = `dT / `dx. If the thickness of a wall is `dx and the inside and outside temperatures of the wall surface are T2 and T1, then `dT = T2 - T1. So this means that we must then use the formula to get temperature gradient: `dT / `dx = (T2 - T1) / `dx For both the inside and outside of the walls.
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19:17:24 ** The rate at which thermal energy is conducted across for a object made of a given substance is proportional to the temperature gradient (the rate at which temperature changes across the object), and to the cross-sectional area of the object. The conductivity is the constant of proportionality for the given substance. So we have Rate of thermal energy condction = conductivity * temperature gradient * area, or R = k * `dT/`dx * A. For an object of uniform cross-section `dT is the temperature difference across the object and `dx is the distance L between the faces of the object. In this case the equation is R = k * `dT / L * A and we can solve to get k = R * L / (`dT * A). **
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RESPONSE --> I don't thnk that I did that question right. I had it confused with another one. Please let me know if I need to change something around??
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19:18:57 Explain how energy flow, for a given material, is affected by area (e.g., is it proportional to area, inversely proportional, etc.), thickness and temperature gradient.
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RESPONSE --> It has to do with the thickness of the wall on the inside and outside. Also the temperature will either increase or decrease depending on the walls.
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D՛yі assignment #001 CߗDƶgUJ Physics II 06-05-2006
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19:26:16 query introset 5 # 12: Finding the conductivity given rate of energy flow, area, temperatures, thickness of wall. Describe how we find the conductivigy given the rate of energy flow, area, temperatures, and thickness of the wall?
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RESPONSE --> Average temperature gradient = `dT / `dx. by using this equation and the thickness of the wall we are able to find the conductivity b/c we know the celsius temperature, the flow of energy, and area
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19:27:04 ** The rate at which thermal energy is conducted across for a object made of a given substance is proportional to the temperature gradient (the rate at which temperature changes across the object), and to the cross-sectional area of the object. The conductivity is the constant of proportionality for the given substance. So we have Rate of thermal energy condction = conductivity * temperature gradient * area, or R = k * `dT/`dx * A. For an object of uniform cross-section `dT is the temperature difference across the object and `dx is the distance L between the faces of the object. In this case the equation is R = k * `dT / L * A and we can solve to get k = R * L / (`dT * A). **
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RESPONSE --> I don't know how well I did on that but the reasoning I have behind it might make some sense.
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19:28:00 Explain how energy flow, for a given material, is affected by area (e.g., is it proportional to area, inversely proportional, etc.), thickness and temperature gradient.
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RESPONSE --> Temperature gradient = `dT / `dx = (T2 - T1) / `dx thickness of a wall is `dx and the inside and outside temperatures of the wall surface are T2 and T1, then `dT = T2 - T1.
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19:28:07 ** CORRECT STUDENT ANSWER WITHOUT EXPLANATION: Energy flow is directly proportional to area inversely propportional to thickness and directly proportional to temperature gradient GOOD STUDENT ANSWER WITH EXPLANATIONS, PLUS INSTRUCTOR COMMENTARY: The energy flow for a given material increases if the area increases. This is because the more area you have the wider a path something has to go through so more of it can move through it. Just like a 4 lane highway will carry more cars in a given time interval than a two lane highway will. So the relationship of energy flow to area is proportional. Energy flow, however is inversely proportional to thickness. This is because although the thermal energy flows through the material, the material impedes it. So if the thickness increases the thermal energy will have to travel farther through the resistance and be impeded more. ** Also for given temperature difference, greater thickness `dx implies smaller temperature gradient `dT / `dx. Small temperature gradient doesn't 'drive' the energy flow as much. Energy flow is also proportional to the temperature gradient. Meaning if the difference in the two temperatures is greater then the energy will move faster from one side to the other. Temperature gradient is not difference in temperatures, it's difference in temperature per unit of distance across the material. Temperature gradient is `dT / `dx, not just `dT. Greater temperature gradient means greater difference in temperature over any given small distance increment. The greater the temperature difference across this increment the more energy will flow. **
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RESPONSE --> ok
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19:28:27 query video experiment 1 (summary not needed)
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RESPONSE --> Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK. Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems ok.
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19:28:33 ok
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RESPONSE -->
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19:29:42 Explain how the experiment shows that thermal energy must go into ice to melt it, even if its temperature doesn't change
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RESPONSE --> b/c it must have cold in order to be able to change. therefore it has to be in the ice for something to have a chance to melt
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19:30:19 ** GOOD STUDENT EXPLANATION: The experiment demonstrates that thermal energy must go into the ice to melt it. The way it is demonstrated, is the water temperature dropped from 18.4 degrees to 0 degrees when the snow was added. This means that the water lost an amount of thermal energy. The snow began the experiment at 0 degrees and after it was melted, it and the other water was still at 0 degrees. So the snow did not gain any temperature but it had to gain some thermal energy. We know this because the other quantity of water lost some thermal energy and it had to go somewhere. So since the snow's gain in thermal energy was not used to raise the temperature of the snow, it had to be used to do the phase change. GOOD STUDENT QUESTION AND INSTRUCTOR RESPONSE: The ice is absorbing the energy lost by the water, which is therefore causing it to melt. If the ice did not absorb the energy lost by the water then it would not melt??? Exactly. Put ice at 0 deg in water at 0 deg and no energy is exchanged. No ice melts, no water freezes. Of course if that 0 deg water is warmed, as by a surrounding room, things start to change. **
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RESPONSE --> Well do I always have to come up with an answer like that one b/c it seems pretty complex to be able to know all of that?
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19:32:41 query general phy 13.12: what is the coefficient of volume expansion for quartz, and by how much does the volume change? (Note that Principles of Physics and University Physics students do not do General Physics problems)
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RESPONSE --> Volumetric thermal expansion
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19:33:10 ** The coefficient of volume expansion for quartz is 1 x 10^(-6) C^(-1). The sphere has diameter 8.75 cm, so its volume is 4/3 pi r^3 = 4/3 pi ( 4.38 cm)^3 = 352 cm^3, approx.. We therefore have dV = beta* V0*dT = 3 x 10^(-6) C^ (-1) * 352 cm^3 * (200C - 30 C) = 0.06 cm^3 **
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RESPONSE --> got that one wrong I was doing the units instead
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19:33:22 query univ 17.103 (15.93 10th edition) (Note that Principles of Physics and General College Physics students don't do University Physics problems).
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RESPONSE --> alrighty
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19:33:36 ** The ice doesn't change temperature until it's melted, at which time it is in the form of water with the specific heat of water. Also the steam will come to temperature Tf so its change in thermal energy after being condensted will be 4186 J / (kg K) * .035 kg * (Tf - 100 C). I prefer to say that the sum of all the thermal energy changes is zero, so that we don't have to worry about taking a negative of a negative (which you should have done on your right-hand side, and which would have avoided the negative result). I would write the equation as follows: [.446 kg * 390 J/kg*K * (Tf - 0 C)] + [0.0950 kg * 4186 J/kg*K *(Tf - 0 C)] - .0350 kg * 2.256 x 10^6 J/kg + 4186 J / (kg K) * .035 kg * (Tf - 100 C) = 0. Noting that change in temperature of a Kelvin degree is identical to a change of a Celsius degree this gives you 170 J/C * Tf + 390 J/C * Tf - 79000 J - 14000 J + 140 J / C * Tf = 0 or 700 J / C * Tf = 93000 J, approx. or Tf = 130 C. This isn't possible--we can't end up warmer than the original temperature of the steam. We conclude that not all the steam condenses and that the system therefore reaches equilibrium at 100 C, with a mixture of water and steam. Our energy conservation equation will therefore be [.446 kg * 390 J/kg*C * (100 C - 0 C)] + [0.0950 kg * 4186 J/kg*C *(100 C - 0 C)] - mCondensed * 2.256 x 10^6 J/kg = 0 where mCondensed is the mass of the condensed steam. This gives us 17000 J + 39000 J - mCondensed * 2.3 * 10^6 J/kg = 0 or mCondensed = 56000 J / (2.3 * 10^6 J/kg) = .023 kg. We end up with .095 kg * .023 kg = .118 kg of water and .035 kg - .023 kg = .012 kg of steam. **
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RESPONSE --> Ok
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19:43:55 query univ phy 17.100 (90 in 10th edition): C = 29.5 J/mol K + (8.2 + 10^-3 J/mol K^2) T . Give your solution to this problem.
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RESPONSE -->
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19:44:09 ** Specific heat is not constant but varies with temperature. The energy required to raise the temperature of 3 moles by dT degrees while at temperature T is 3 mol * C * dT = 3 mol * (29.5 J/mol K + (8.2 + 10^-3 J/mol K^2) T) * dT. You have to integrate this expression from T= 27 C to T = 227 C, which is from 300 K to 500 K. Antiderivative of (29.5 J/mol K + (8.2 + 10^-3 J/mol K^2) T) is F(T) = 29.5 J / (mol K) * T + (8.2 + 10^-3 J/mol K^2) * T^2 / 2. Simplify and apply Fundamental Theorem of Calculus (find F(500) - F(300) if you think the temperature T is in Kelvin or F(227) - F(27) if you think it's in Celsius; this isn't specified in the problem and while the units tend to imply Kelvin temperature the resulting specific heats would be unrealistic for most real substances), then multiply by the constant 3 moles. The result for Kelvin temperatures is about 20,000 Joules. **
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RESPONSE --> ok
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19:44:21 University Physics Problem 17.106 (10th edition 15.96): Give your solution.
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RESPONSE --> don't do
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19:44:28 **The final mass of the system is .525 kg, meaning that .525 kg - (.340 kg + .150 kg) = .035 kg of steam condensed then cooled to 71 C. The thermal energy change of the calorimeter plus the water is .150 kg * 420 J/(kg C) * 56 C + .34 kg * 4187 J / (kg C) * 56 C = 83,250 J, approx. The thermal energy change of the condensed water is -Hf * .035 kg + .035 kg * 4187 J / (kg C) * (-29 C) = -Hf * .035 kg - 2930 J, approx. Net thermal energy change is zero, so we have 83,250 J - Hf * .035 kg - 4930 J = 0 so that Hf = 79,000 J / (.035 kg) = 2,257,000 J / kg. **
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RESPONSE --> k
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