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18:56:29 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 --> Using the formula: k=rate of energy flow/area(temp gradient). Where the temp gradient = T1-T2/L Therefore you would first calculate the temp gradient using given values for temp. one and two as well as the given length(thickness). Then plug in the temp gradient into first formula to get value for k.
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18:56:51 06-28-2007 18:56:51 ** 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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18:56:57 ** 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 --> ok
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19:03:52 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 --> The rate of thermal energy is directly proportional to area, and temp gradient. Given the formula Q=kA(T1-T2)/l, where k=thermal conductivity, A=area, T1-T2/L= temp gradient, one can see that this is true.
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19:06:14 ** 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 --> Energy flow is directly proportional to temp gradient, but inversely proportional to thickness. The explanation that the thickness of the material will impede the energy flow makes the inverse relationship very easy to see.
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19:17:35 principles of physics and general college physics 13.8: coeff of expansion .2 * 10^-6 C^-1, length 2.0 m. What is expansion along length if temp increases by 5.0 C?
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RESPONSE --> Given the formula: L=Lo(1 + alpha)(change in T) where L=final length after heating or cooling to a temp T, Lo=initial length, alpha=coefficient of linear expansion, change in temp=T(final)-T(initial). L=2.0(1+.2E-6)(5)=10.0m
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19:27:46 The amount of the expansion is the product of the coefficient of expansion, the original length and the temperature difference: `dL = alpha * L0 * `dT = .2 * 10^-6 C^(-1) * 2.0 m * 5.0 C = 2 * 10^-6 m. This is 2 microns, two-onethousandths of a millimeter. By contrast the coefficient of expansion of steel is 12 * 10^-6 C^(-1);using this for the coefficient of expansion yields a change in length of 24 * 10^-6 m, or 24 microns, which is 240 times as much as for the given alloy.
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RESPONSE --> The formula given in the book adds 1 to the coefficient of linear expansion, where given here, the coeff. of linear expanison is multiplied by -1. Have noted the difference.
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19:34:09 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 --> The coefficient of volume expansion for quartz is 1E-6. Becaues quartz is a solid,the coeff. of linear expanison is approx. equal to 3(alpha). The change in volume=beta(Vo)(change in T) Where beta=coefficient of volume expansion, Vo=initial volume, change in T=T(final)-T(initial).
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19:34:34 ** 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 --> ok
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¯‹¥yõ”Þ‘xÛêš ¶_‰m¯ú踃zîîÜ—ìÍ assignment #011 011. `query 1 Physics II 06-28-2007