course Phy202 åz߽
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20:23:47 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 --> We can find conductivity given the rate of energy flow, temperatures, and thickness of the wall by finding the specific SI units that relate to each of these variable and applying them into an equation that will give us the right electrical conductivity.
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20:31:33 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 --> By using the ideal gas law of PV = NkT for kinetic energy, each of these variables of area, thickness, and temp. can be measured and determined here based on other constants and variables that can be seen in equilibrium in this equation.
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20:38:24 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 --> By multiplying (.2*10^-6 C^-1) (2.0m) (5.0 C) the expansion would be 2*10^-6 N/m^2.
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20:41:37 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 volume expansion for quartz is (C^-1) and changes by 1* 10^-6 (C^-1).
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20:44:08 ** 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 I understand
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20:48:16 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 --> By dividing (29.5)/ (8.2 +10^-3), T= 3.6 C
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20:49:56 ** 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 I understand now what I did.
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20:50:29 University Physics Problem 17.106 (10th edition 15.96): Give your solution.
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RESPONSE --> I have the General College Physics text....
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ВGץ~tӓ U assignment #002 002. `query 1 Physics II 10-22-2008"