course Phy 202 Xm壾ַ콲C~assignment #011
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14:38:30 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 --> the rate of conductivity is greater as the cross sectional area increases and is greater as the temperature dfference is greater. It is less as the thickness becomes greater.
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14:39:21 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 --> ok
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14:40:45 ** 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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14:44:50 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 --> 0.000002m
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14:45:17 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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14:50:39 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 --> 1*10^-6 /C is the coefficient of volume expansion of quartz
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14:51:14 ** 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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14:51:24 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 --> ok
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14:52:05 ** 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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14:52:11 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 --> ok
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14:52:18 ** 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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14:52:25 University Physics Problem 17.106 (10th edition 15.96): Give your solution.
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RESPONSE --> ok
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14:52:30 **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 --> ok
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Æ䑊J\陸ٙ assignment #012 012. `query 2 Physics II 07-20-2008
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14:56:56 query intro set problem 14 substance, water, both temperatures and masses known, final temperature known, find spec ht Explain how the final temperature of the system is combined with the known initial temperatures and masses of both substances to obtain the unknown specific heat
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RESPONSE --> the temperature change, the final temperature minus the inital temperature times the mass times the specific heat gives the energy change.
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14:57:57 ** The change in the thermal energy of an isolated system is 0. So assuming that the systems are isolated the thermal energy change of one object is equal and opposite to that of the other. For an ideal substance the change in the thermal energy of an object is proportional to the change in its temperature and to its mass (i.e., more mass and more change in temperature implies more change in thermal energy). The specific heat is the proportionality constant for the substance. Using `dQ for the change in the thermal energy of an object we can express this as `dQ = mass * specific heat * `dT. (General College and University Physics students note that specific heat is not in fact strictly constant and for most substances changes with temperature.) We know that what energy one substance loses, the other gains, and we know the specific heat of water , so we can set the two quantities equal and solve for specific heat of the unknown substance. For two objects combined in a closed system we have `dQ1 + `dQ2 = 0, which gives us the equation m1 c1 `dT1 + m2 c2 `dT2 = 0 or equivalently m1 c1 `dT1 = - m2 c2 `dT2. If we know all the quantities except, say, the specific heat c2 of the second substance we can easily solve for c2. **
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RESPONSE --> ok
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14:58:08 prin phy Ch 13.26. Kelvin temperatures corresponding to 86 C, 78 F, -100 C, 5500 C and -459 F.
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RESPONSE -->
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14:58:34 The Kelvin temperature is 273 above the Celsius temperature (actually 273.15 below, but the degree of precision in the given temperatures is not sufficient to merit consideration of tenths or hundredths of a degree). 86 C, -100 C and 5500 C are therefore equivalent to ( 86 + 273 ) K = 359 K, -(100 + 273) K = 373 K, (5500 + 273) K = 5773 K. 78 F is (78 F - 32 F) = 46 F above the freezing point of water. 46 Fahrenheit degrees is the same as (5/9 C / F ) * 46 F = 26 C above freezing, or since freezing is at 0 C, just 26 C. The Kelvin temperature is therefore (26 + 273) K = 299 K. -459 F is (459 + 32) F = 491 F below freezing, or (5/9 C / F) * (-491 F) = 273 C below freezing, which is -273 C or (-273 + 273) K = 0 K. This is absolute zero, to the nearest degree.
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RESPONSE --> ok
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15:05:14 prin phy and gen phy Ch 13.30 air at 20 C is compressed to 1/9 of its original volume. Estimate the temperature of the compressed air assuming the pressure reaches 40 atm.
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RESPONSE --> 930C
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15:07:34 query gen phy ch 13.38 fraction of air released after tire temp increases from 15 to 38 C at 220 kPa gauge
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RESPONSE --> 7.4%
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15:08:01 ** T1 = 288 K, T2 = 311 K so T2 / T1 = 311 / 288 = 1.08, approx. This is approx. an 8% increase in temperature. The pressure will end up at 3ll / 288 * 321 kPa = 346 kPa, approx (note that we have to use actual rather than gauge pressure so init pressure is 220 kPa + 101 kPa = 321 kPa, approx. You then have to change the number n of moles of gas to get back to 331 kPa, so n3 / n2 = P3 / P2 = 321 kPa / 346 kPa or approximately .93, which would be about a 7% decrease. Note that the results here are mental estimates, which might not be particularly accurate. Work out the process to see how the accurate numbers work out. **
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RESPONSE --> ok
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15:08:07 query univ phy 17.116 (15.106 10th edition) 1.5 * 10^11 m, 1.5 kW/m^2, sun rad 6.96 * 10^8 m. How did you calculate the total radiation of the Sun and how did you use this result to get the radiation per unit area?
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RESPONSE --> ok
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15:08:12 ** GOOD STUDENT SOLUTION The total radiation of the sun was the rate it reaches earth times the imaginary surface of the sphere from the sun center to earth atmosphere, or 1500 W/m^2 * (4`pir^2) = 1500W/m^2 * 2.8537 x10^23 m^2 = 4.28055 x 10 ^ 26 W. } Radiation per unit of area surface of the sun would be
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RESPONSE --> ok
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15:08:25 4.28055 x 10 ^ 26 W / (4*`pi * (6.96 x 10 ^ 8 m)^2) = 4.28055 x 10 ^ 26 W / 6.08735 x 10 ^ 18 m^2 = 70318775.82 J/s/m^2 = 7.03 x 10 ^ 7 J/s/m^2 If the sun is radiating as an ideal blackbody, e = 1, the T would be found: H = `dQ/`dt = 4.28055 x 10 ^ 26 W = (4*`pi * (6.96 x 10 ^ 8 m)^2) * (1) * (5.67051 x 10^-8 W/m^2*K) * T^4 So T^ 4 = 4.28055 x 10 ^ 26 W / 6.087351 x 10 ^ 18 m^2) * 1 * (5.67051 x 10^-8 W/m^2*K) T^4 = 1.240 * 10 ^ 15 K ^4 T = 5934.10766 K on surface of sun. **
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RESPONSE --> ok
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15:08:31 univ phy 17.115 time to melt 1.2 cm ice by solar radiation 600 w/m^2, 70% absorption, environment at 0 C.
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RESPONSE --> ok
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15:08:45 ** Thermal energy is not radiating in significant quantities from the ice, so you use 70% of the incoming 600 watts/m^2, which gives you 420 watts / m^2, or 420 Joules/second for every square meter if ice. Melting takes place at 0 C so there is no thermal exchange with the environment. A 1.2 cm thickness of ice will have a volume of .012 m^3 for every square meter of surface area; the mass will be close to 1000 kg/m^3, so there are about 12 kg of ice for every m^2 of surface (you can correct this by using the correct density of ice). It takes about 330,000 Joules to melt a kg of ice, so to melt 12 kg requires around 4,000,000 J. At 420 Joules/sec the time required will be about 10,000 seconds, or around 3 hours. All my calculations are approximate and done mentally so you should check them yourself, using more precise values of the constants, etc. **
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RESPONSE --> ok
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