course Phy 232 Msv̑ݏٞassignment #002
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13:20:12 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 --> conductivity, k Temperature gradient, dT/dx energy flow, dQ/dt Temperature, T thickness, delta Area, A To find conductivity we are looking for the rate at which thermal energy is conducted that is equal to the product of the conductivity, the area of the wall, and the temperature gradient. It is thus equal to the rate of conduction divided by the area and the gradient. k = (dQ/dt)/[(dT/dx)*A]
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13:20:17 ** 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 --> okay
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13:21:13 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 equation for energy flow is R = k*A*dT/dx R is proportional to area R is proportional to temperature gradient R is inversely proportional to thickness If two numbers are proportional they are written a=b*c, a/b is a constant and is directly proportional. If two numbers are inversely proportional they are written a=c/b, a*b is a constant and is inversely proportional. Energy flow and area are directly proportional because energy flow increases with increasing area. Energy flow and thickness are inversely proportional because as thickness increases the energy flow will be reduced. Energy flow is proportional to the temperature gradient because heat flows from hot to cold and the larger the temperature difference the faster the heat will travel.
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13:21:18 ** 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 --> okay
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13:24:02 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 --> dL=alpha*Linitial*dT dL = 0.2*10^-6 C^-1 * 2 m* 5 C dL = 2*10^-6 m
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13:24:06 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 --> ok
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13:24:14 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 --> okay
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13:24:17 ** 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 --> okay
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13:42:23 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 --> given: mc = 0.446 kg cc = 390 J/kg*K cw = 4190 J/kg*K Tinitial = 0 C mice = 0.0950 kg msteam = 0.0350 kg Tsteam = 100 C cice = 2.1*10^3 J/kg*K Lf = 3.34*10^5 J/kg Lv = 2.256*10^6 J/kg part a: Qsteam = msteam*cw*(Tf-100 C) = 0.0350 kg*4190 J/kg*K*(Tf-100 C) Qcopper = mc*cc*(Tf-0 C) = 0.446 kg*390 J/kg*K*(Tf-0 C) Qwater =mice*cwater*(Tf-0 C) = 0.0950 kg*4190 J/kg*K*(Tf-0) Qsteam2 = msteam*Lv = 0.035 kg*2.256*10^6 J/kg All of these heats added up are going to come out as zero. Qsteam+Qcopper+Qwater-Qsteam2 = 0 Solve for Tf: Tf = 130.3 C The final temperature can not be higher than the steam because the steam started at 100 C and would be cooled somewhat. Therefore, I am going to assume that not all the steam turned to water and therefore, the final temperature would have to be 100 C. part b: We are going to solve for the mass of steam condensed by using: Qcopper+Qice-Qsteam2 0.446 kg*390 J/kg*C + 0.095*4186*(100-0) - mc*2.256*10^6 = 0 mc = 0.025 kg mwater = 0.095+0.025 = 0.12 kg msteam = 0.035-0.025 = 0.01 kg
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13:42:32 ** 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 --> our rounding was a little different
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13:48:17 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 --> dQ=nCdT T=27C + 273 = 300 K T = 227C + 273 = 500 K integrate from T = 300 to T = 500 dQ=n*29.5 J/mol K + (8.2 + 10^-3 J/mol K^2)* T* dT Q = 19668 J
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13:48:21 ** 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 --> okay
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14:15:46 University Physics Problem 17.106 (10th edition 15.96): Give your solution.
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RESPONSE --> 0.525 kg - (0.34 kg+0.15 kg) = 0.035 kg of steam were condensed and cooled so I am going to use this as the mass of steam Qcal + Qwater + Qsteam = 0.15 kg*420*(71-15) + 0.34*4190*(71-15)-Hf*.35+0.035*4190*(71-100) = 0 Hf = 2.26*10^6 J/kg
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14:15:51 **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 --> okay
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