course PHY232 Professor Smith,I did not see Assingment one (1) being received by you. I may have missed the use of the work submission form. The assignment was completed on 2 Jan.
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16:26:58 Most queries in this course will ask you questions about class notes, readings, text problems and experiments. Since the first two assignments have been experiments, the first two queries are related to the experiments. While the remaining queries in this course are in question-answer format, the first two will be in the form of open-ended questions. Interpret these questions and answer them as best you can.
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RESPONSE -->
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16:28:30 Suppose you measure the length of a pencil. You use both a triply-reduced ruler and the original ruler itself, and you make your measurements accurate to the smallest mark on each. You then multiply the reading on the triply-reduced ruler by the appropriate scale factor. Which result is likely to be closer to the actual length of the pencil? What factors do you have to consider in order to answer this question and how do they weigh into your final answer?
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RESPONSE --> The orignal ruller itself would give the most accurate measurment. The triply-reduced ruler scale factor may not be accurate. confidence assessment: 2
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16:29:57 Answer the same question as before, except assume that the triply-reduced ruler has no optical distortion and you know the scale factor accurate to 4 significant figures.
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RESPONSE --> With no optical distortion and a scale facotor accurate to 4 significant fiures would depend on how accurate my measurment need to be. If the measurment need to be accurate to within 5 signifcant figures, then the triply-reduce ruler would be problematic. confidence assessment: 2
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16:31:15 Suppose you are to measure the length of a rubber band whose original length is around 10 cm, measuring once while the rubber band supports the weight of a small apple and again when it supports the weight of two small apples. You are asked to report as accurately as possible the difference in the two lengths, which is somewhere between 1 cm and 2 cm. You have available the singly-reduced copy and the triply-reduced copy, and your data from the optical distortion experiment. Which ruler will be likely to give you the more accurate difference in the lengths? Explain what factors you considered and how they influence your final answer.
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RESPONSE --> The singly -reduced copy will give the most accurat measurment since the error is not propogated. confidence assessment: 2
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16:35:11 Later in the course you will observe how the depth of water in a uniform cylinder changes as a function of time, when water flows from a hole near the bottom of the cylinder. Suppose these measurements are made by taping a triply-reduced ruler to the side of a transparent cylinder, and observing the depth of the water at regular 3-second intervals. {}{}The resulting data would consist of a table of water depth vs. clock times, with clock times 0, 3, 6, 9, 12, ... seconds. As depth decreases the water flows from the hole more and more slowly, so the depth changes less and less quickly with respect to clock time. {}{}Experimental uncertainties would occur due to the optical distortion of the copied rulers, due to the the spacing between marks on the rulers, due to limitations on your ability to read the ruler (your eyes are only so good), due to timing errors, and due to other possible factors. {}{}Suppose that depth changes vary from 5 cm to 2 cm over the first six 3-second intervals. {}{}Assume also that the timing was very precise, so that there were no significant uncertainties due to timing. Based on what you have learned in experiments done in Assignments 0 and 1, without doing extensive mathematical analysis, estimate how much uncertainty would be expected in the observed depths, and briefly explain the basis for your estimates. Speculate also on how much uncertainty would result in first-difference calculations done with the depth vs. clock time data, and how much in second-difference calculations. {}{}How would these uncertainties affect a graph of first difference vs. midpoint clock time, and on a graph of second difference vs. midpoint clock time? {}How reliably do you think the first-difference graph would predict the actual behavior of the first difference? {}Answer the same for the second-difference graph. {}{}What do you think the first difference tells you about the system? What about the second difference?
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RESPONSE --> confidence assessment: 0
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ɆؗԄ݅x assignment #002 002. `query 1 Physics II 02-01-2009
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22:09:48 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 --> H= ka * (tf - ti)/L So, if we know the flow (H), tempertures (tf-ti), and the length, we can then solve for (a) the thermal conductivity
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22:10:45 ** 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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22:12:17 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 energy flow is directly proportional to the area and inversly proportional to the length. H = ka *(tf-ti)/L
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22:13:21 ** 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 --> I need to give a more consice answer
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22:21:10 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 --> Linear expansion Change in the length is equal to the coefficent of linerar expansion for the material times the initial length times the change in temperature. The change would be equal to 2 x 10^-6
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22:21:37 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 --> I understand
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22:23:13 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 coeffient of volume expansion for quartz is equal to 0.12 x 10^-5 Do not need to do this problem
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22:23:21 ** 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 -->
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22:24:09 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 --> I understand
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22:25:03 ** 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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22:28:32 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 --> I did not finish this problem in the homework
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22:32:54 ** 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 --> I will rework the problem
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22:33:18 University Physics Problem 17.106 (10th edition 15.96): Give your solution.
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RESPONSE --> I did not finish the assignment
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22:39:01 **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 --> I will rework the problem
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~zznԌרOg assignment #001 001. Physics II 02-01-2009
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22:46:17 Most queries in this course will ask you questions about class notes, readings, text problems and experiments. Since the first two assignments have been experiments, the first two queries are related to the experiments. While the remaining queries in this course are in question-answer format, the first two will be in the form of open-ended questions. Interpret these questions and answer them as best you can.
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RESPONSE --> I have some confusion over the expierments since they are covered in chapter 14 and the University Physics assignments were in chapter 17?
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22:47:49 Suppose you measure the length of a pencil. You use both a triply-reduced ruler and the original ruler itself, and you make your measurements accurate to the smallest mark on each. You then multiply the reading on the triply-reduced ruler by the appropriate scale factor. Which result is likely to be closer to the actual length of the pencil? What factors do you have to consider in order to answer this question and how do they weigh into your final answer?
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RESPONSE --> I googled a triply-reduced ruler and did not get a hit? So, I am going to assume it is 1/3 scale. The conversion factor of the scale would be most important and the optical illusion. confidence assessment: 2
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22:48:30 Answer the same question as before, except assume that the triply-reduced ruler has no optical distortion and you know the scale factor accurate to 4 significant figures.
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RESPONSE --> This would be ok if the precison required 4 significant figures. confidence assessment: 2
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22:49:38 Suppose you are to measure the length of a rubber band whose original length is around 10 cm, measuring once while the rubber band supports the weight of a small apple and again when it supports the weight of two small apples. You are asked to report as accurately as possible the difference in the two lengths, which is somewhere between 1 cm and 2 cm. You have available the singly-reduced copy and the triply-reduced copy, and your data from the optical distortion experiment. Which ruler will be likely to give you the more accurate difference in the lengths? Explain what factors you considered and how they influence your final answer.
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RESPONSE --> The singly-reduced copy would give the most accurate results since there is no correction factor that must be applied. confidence assessment: 2
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22:52:39 Later in the course you will observe how the depth of water in a uniform cylinder changes as a function of time, when water flows from a hole near the bottom of the cylinder. Suppose these measurements are made by taping a triply-reduced ruler to the side of a transparent cylinder, and observing the depth of the water at regular 3-second intervals. {}{}The resulting data would consist of a table of water depth vs. clock times, with clock times 0, 3, 6, 9, 12, ... seconds. As depth decreases the water flows from the hole more and more slowly, so the depth changes less and less quickly with respect to clock time. {}{}Experimental uncertainties would occur due to the optical distortion of the copied rulers, due to the the spacing between marks on the rulers, due to limitations on your ability to read the ruler (your eyes are only so good), due to timing errors, and due to other possible factors. {}{}Suppose that depth changes vary from 5 cm to 2 cm over the first six 3-second intervals. {}{}Assume also that the timing was very precise, so that there were no significant uncertainties due to timing. Based on what you have learned in experiments done in Assignments 0 and 1, without doing extensive mathematical analysis, estimate how much uncertainty would be expected in the observed depths, and briefly explain the basis for your estimates. Speculate also on how much uncertainty would result in first-difference calculations done with the depth vs. clock time data, and how much in second-difference calculations. {}{}How would these uncertainties affect a graph of first difference vs. midpoint clock time, and on a graph of second difference vs. midpoint clock time? {}How reliably do you think the first-difference graph would predict the actual behavior of the first difference? {}Answer the same for the second-difference graph. {}{}What do you think the first difference tells you about the system? What about the second difference?
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RESPONSE --> 3cm / 18 secons = 0.16cm/second Alot of uncertainity exist. The uncertainity would be highest on the second differenc vs. midpoint clock time. The first graph would be more reliabel than the second graph due to the smoothing of the velocity versus the acceleration. confidence assessment: 2
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22:53:27 Suppose the actual second-difference behavior of the depth vs. clock time is in fact linear. How nearly do you think you could estimate the slope of that graph from data taken as indicated above (e.g., within 1% of the correct slope, within 10%, within 30%, or would no slope be apparent in the second-difference graph)? Again no extensive analysis is expected, but give a brief synopsis of how you considered various effects in arriving at your estimate.
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RESPONSE --> The slope would be approximatly 10% if it was linear. confidence assessment: 1
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ԯʡPˍ assignment #003 003. `query 2 Physics II 02-01-2009
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22:54:32 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 --> Do not need to do this problem
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22:55:30 ** 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 --> I understand
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22:55:45 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 --> Do not need to do this problem
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22:56:19 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 --> Do not need to do this problem
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22:56:36 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 --> Do not need to do this problem
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22:56:47 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 --> Do not need to do this problem
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22:57:01 ** 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 --> Do not need to do this problem
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22:57:46 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 --> I did not work this problem
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22:59:11 ** 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 --> I will go back and rework this problem
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23:03:17 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 , I will go back and work this prolem
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23:03:39 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 --> I did not work this problem
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23:07:15 ** 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 --> Oki
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