۔˘ YoxV assignment #001 yO̳aÉGɜ Physics II 09-11-2005
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22:07:50 09-11-2005 22:07:50 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?......!!!!!!!!...................................
NOTES -------> Conductivity is the rate at which thermal energy is equal to the conductivity of the object*the area of the wall*the temperature gradient.
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22:10:21 ** 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). **......!!!!!!!!...................................
RESPONSE --> i was correct.
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22:18:53 09-11-2005 22:18:53 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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NOTES -------> we find the conductivity by multiplying the area of the wall by the rate of energy flow by the temp. gradient. Thus the conductivity is proportional to the area because any change in the area can be multiplied by the conductivity of the original to arive at the answer if the area had been that size when the equation was initially ran.
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22:20:09 09-11-2005 22:20:09 ** 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. **......!!!!!!!!...................................
NOTES -------> I did not get this answer correct.
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22:20:46 query video experiment 1 (summary not needed)
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RESPONSE --> I will send this in an attachment email.
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22:20:54 ok
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RESPONSE --> ok
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22:22:55 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 --> We saw a large amount of energy lost by the water through the drop in the water temperatur. This energy must be used where the work is being done and the main work being done was the ice melting. The temperature of the ice did not change a lot, the temperature of the water was what changed most.
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22:23:49 ** 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. **......!!!!!!!!...................................
RESPONSE --> I was correct.
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22:34:16 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 --> unfortunatly I do not have the correct book so I do not have the means of answering this question at this time.
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22:35:39 ** 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 **......!!!!!!!!...................................
RESPONSE --> I am sure this is right, I just cant prove anything with out the book. I am going to end this here because there is no need in going on if I was given the University Physics book.
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˧ڏyxԼ~ assignment #002 yO̳aÉGɜ Physics II 09-11-2005
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22:41:14 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......!!!!!!!!...................................
RESPONSE --> With the change in temperature we can find out what amount of energy was lost or gained by each element. We take the difference in the beginning water temp and final water temp and multiply this number by a fixed constant 4180 J/kg c and the amount of water present this gives us the amount of energy lost by the water. The energy has no place to go but to the other element. We can take this amount of energy and divide it by the amount of the seccond element present to find the amount per j/kg in this case. We can then take the amount of temperature change in the substance and divide the amount of energy per j/kg by the change in temp to get the temp. gradient in j/kg c.
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22:42:52 09-11-2005 22:42:52 ** 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. **......!!!!!!!!...................................
NOTES -------> I was correct, i did not however include the equations.
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22:43:15 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 --> again I do not have the correct book.
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22:44:24 ** 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. **......!!!!!!!!...................................
RESPONSE --> looks like good notes to me.
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22:45:09 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?......!!!!!!!!...................................
RESPONSE --> not for me
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22:45:14 09-11-2005 22:45:14 ** 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......!!!!!!!!...................................
NOTES -------> na
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22:45:27 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. **......!!!!!!!!...................................
RESPONSE --> na
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22:45:32 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 --> na
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22:45:37 ** 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. **......!!!!!!!!...................................
RESPONSE --> na
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Video Experiment 1