#$&* course phy122 Your comments definitely helped me. I still need some help figuring out the last few steps on a couple of these, but overall I think that I understand it better than I did. Thank you so much for all of your help! I greatly appreciate it! Query2#$&*
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Given Solution: ** 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 the proportionality equation • rate of thermal energy conduction = conductivity * temperature gradient * area, or in symbols • R = k * (`dT/`dx) * A. (note: R is the rate at which thermal energy Q is transferred with respect to clock time t. Using the definition of rate of change, we see that the average rate over a time interval is `dQ / `dt, and the instantaneous rate is dQ / dt. Either expression may be used in place of R, as appropriate to the situation.) For an object of uniform cross-section, `dT is the temperature difference across the object and `dx is the distance between the faces of the object. The distance `dx is often denoted L. Using L instead of `dx, the preceding proportionality can be written • R = k * `dT / L * A We can solve this equation for the proportionality constant k to get • k = R * L / (`dT * A). (alternatively this may be expressed as k = `dQ / `dt * L / (`dT * A), or as k = dQ/dt * L / (`dT * A)). STUDENT COMMENT I really cannot tell anything from this given solution. I don’t see where the single, solitary answer is. INSTRUCTOR RESPONSE The key is the explanation of the reasoning, more than the final answer, though both are important. However the final answer is given as k = R * L / (`dT * A), where as indicated in the given solution we use L instead of `dx. Two alternative answers are also given. Your solution was 'Well, according to the information given in the Introductory Problem Set 5, finding thermal conductivity (k) can be determined by using k = (‘dQ / ‘dt) / [A(‘dT / ‘dx)].' The given expressions are equivalent to your answer. If you replace `dx by L, as in the given solution, and simplify you will get one of the three given forms of the final expression. However note that you simply quoted and equation here (which you did solve correctly, so you didn't do badly), and gave no explanation or indication of your understanding of the reasoning process. Self-Critique: Oh my goodness. This is over my head. I looked at the videos on the DVD. Perhaps I should watch these again. I should also watch some you tube videos for solving this problem. Professor, I know that you are busy and I have 5000 questions. Should I enlist the help of a tutor? &&&&& I feel better about my answer now. &&&& ------------------------------------------------ ------------------------------------------------ Self-Critique Rating: @& R=k*(dT/dx)*A First, can you identify the symbols for conductivity, rate of energy flow, area, change in temperature, and thickness of the wall? &&&& The R stands for rate. The k stands for the conductivity of the material. The dT stands for the temperature difference across the object. The dx stands for the distance between the faces of the object. The A stands for area. &&&&& Second, does it make sense to you that you get more energy flow if you have a bigger wall area, or a bigger temperature difference between inside and outside? &&&&&Yes, it does now. &&&& Also does it make sense that a thinner wall will result in more energy flow? *@ @& Now can you see that multiplying by dT and by A will result in a greater result if dT and A are greater, and that dividing by a lesser dx will also make the result greater? If so you can then see how the equation could represent the situation. *@ @& The k in the equation is a property of the material, what we call the conductivity of the material. It should be clear that a material with a larger k will produce a greater rate of energy flow, given the same conditions, than a material with a smaller k. *@ @& Let me know if this helps you make sense of the situation. And if it doesn't keep the questions coming until we've resolved this. &&&&&& So, if I think of k like the insulation in my walls for example, the higher the number, the better insulation it will provide, correct?
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Given Solution: ** CORRECT STUDENT ANSWER WITHOUT EXPLANATION: Energy flow is: • directly proportional to area • inversely proportional to thickness and • directly proportional to temperature gradient Good student answer, slightly edited by instructor: The energy flow for a given object increases if the cross-sectional area (i.e., the area perpendicular to the direction of energy flow) increases. Intuitively, this is because the more area you have the wider the path available so more stuff can move through it. By analogy a 4 lane highway will carry more cars in a given time interval than will a two lane highway. In a similar manner, energy flow is directly proportional to cross-sectional area. Temperature gradient is the rate at which temperature changes with respect to position as we move from one side of the material to the other. That is, temperature gradient is the difference in temperature per unit of distance across the material: • temperature gradient is `dT / `dx. (a common error is to interpret temperature gradient just as difference in temperatures, rather than temperature difference per unit of distance). For a given cross-sectional area, energy flow is proportional to the temperature gradient. If the difference in the two temperatures is greater then the energy will move more quickly from one side to the other. For a given temperature difference, greater thickness `dx implies smaller temperature gradient `dT / `dx. The temperature gradient is what 'drives' the energy flow. Thus greater thickness implies a lesser temperature gradient the lesser temperature gradient implies less energy flow (per unit of cross-sectional area) per unit of time and we can say that the rate of energy flow (with respect to time) is inversely proportional to the thickness. Self-Critique: Wow, I need review. When I read the text this information does not seem to jump out at me. I had to google the answers above. ------------------------------------------------ ------------------------------------------------ Self-Critique Rating: ********************************************* ********************************************* Question: 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? &&&&& dL=alpha *LO*dT dL=.2*10^-6C^-1*2.0m*5.0C Professor, How would I calculate a change in temperature, if there is no initial temperature given? &&&&& YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: I would use an equation to solve for this problem. If I only knew which one to use. Professor, how will I ever memorize all of these formulas? confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Poor. I really don’t understand.
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Given Solution: This problem is solved using the concept of a coefficient of expansion. The linear coefficient of thermal expansion of a material, denoted alpha, is the amount of expansion per unit of length, per unit of temperature: • expansion per unit of length is just (change in length) / (original length), i.e., • expansion per unit of length = `dL / L0 Thus expansion per unit of length, per unit of temperature is (expansion per unit of length) / `dT. Denoting this quantity alpha we have • alpha = (`dL / L0) / `dT. This is the 'explanatory form' of the coefficient of expansion. In algebraically simplified form this is • alpha = `dL / (L0 * `dT). In this problem we want to find the amount of the expansion. If we understand the concept of the coefficient of expansion, we understand that the amount of the expansion is the product of the coefficient of expansion, the original length and the temperature difference: If we don’t completely understand the idea, or even if we do understand it and want to confirm our understanding, we can solve the formula alpha = `dL / (L0 * `dT) for `dL and plug in our information: • `dL = alpha * L0 * `dT = .2 * 10^-6 C^(-1) * 2.0 m * 5.0 C = 2 * 10^-6 m. This is 2 microns, two one-thousandths of a millimeter. By contrast the coefficient of expansion of steel is 12 * 10^-6 C^(-1); using this coefficient of expansion yields a change in length of 1.2 * 10^-4 m, or 120 microns, which is 60 times as much as for the given alloy. Self-Critique: I feel like I need step by step instruction as to where the numbers came from. Perhaps I would just write out the formula instead of using abbreviations. This could be more helpful: so if the formula is dL=alpha*LO*dT=2 * 10^-6 C^(-1) * 2.0 m * 5.0 C = 2 * 10^-6 m. Would you mind telling me what each letter in the formula denotes? I will go online and do more research for myself. ------------------------------------------------ ------------------------------------------------ Self-Critique Rating: &&&& I see that I don’t need to have an initial temperature, I would just plug in the temperature given above. I think that I read too deeply into these problems sometimes. I think that I make them much more complicated than they are. &&&& @& It should make sense that, for a given temperature change, the amount of expansion is greater for a longer object. Also that the expansion is greater for a greater change in temperature. So it makes sense that the amount of expansion should be proportional to the length of the object, multiplied by the change in its temperature. *@ @& Of course we need a little more than this. What we need is a number to multiply by this product. That number is alpha, the coefficient of linear expansion. Its value will depend on the material (some materials expand more readily than others). *@ @& So the amount of expansion involves the original length L0, temperature change `dT, and coefficient of expansion alpha. The amount of expansion will be alpha multiplied by the product of L0 and `dT: amount of expansion = alpha * L0 * `dT We will use `dL, which stands for change in length, to stand for the amount of expansion, so our final rule is `dL = alpha * L0 * `dT *@ ********************************************* ********************************************* Question: 5. The surface temperature of the Sun is about 5750 K. What is this temperature on the Fahrenheit scale? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Temperature conversions, I should be able to do this! If 1 Kelvin degree is 1.8 F degrees, I should be able to take 5750K and multiply it by 1.8 F to get 10350 degrees F. &&&& After reviewing, I see that I need to subtract 460 degrees from the 10350 degrees F to arrive at 9890 degrees F because absolute zero on the Farenheit scale is 460 degrees. &&&&&
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Given Solution: 5750 K means 5750 Kelvin degrees above absolute zero. A Kelvin degree is 1.8 Fahrenheit degrees, so this temperature is 5750 K * 1.8 (F / K) = 10350 Fahrenheit degrees above absolute zero. 0 on the Fahrenheit scale is about 460 Fahrenheit degrees above absolute zero, so 10350 Fahrenheit degrees above absolute zero is about (10350 - 460) Fahrenheit = 9890 Fahrenheit. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Ok, I see. I need to subtract 460 degrees F because 0 on the F scale is 460 degrees above absolute zero. ------------------------------------------------ ------------------------------------------------ Self-critique Rating: @& Good. *@ I’m starting to feel like a failure at this. @& It doesn't look that way from here. &&&& Thank you, your comments are really helping me. I am actually starting to understand! *@ #$&* ********************************************* ********************************************* Question: 12. How large an expansion gap should be left between steel railroad rails if they may reach a maximum temperature 35.0ºC greater than when they were laid? Their original length is 10.0 m. Assume that steel has coefficient of linear expansion of 12 * 10^-6 / K. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: dL= 10 meters*(35.0C*12*10^-6/K)=0.042m @& Good. *@ confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The expansion of a 10.0 meter steel rail when temperature increases by 35 C is `dL = 10.0 meter * (35 C * 12 * 10^-6 / K) = 0.042 meter, or a little over 4 millimeters. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ ------------------------------------------------ Self-critique Rating: #$&* ********************************************* ********************************************* Question: (optional for Principles of Physics students) (a) Suppose a meter stick made of steel and one made of invar (an alloy of iron and nickel) are the same length at 0ºC . What is their difference in length at 22.0ºC ? (b) Repeat the calculation for two 30.0-m-long surveyor’s tapes. Assume coefficient of expansion for invar to be 1.2 * 10^-6 / K. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&& Steel: dL=12*10^-6/K*30m*22Degrees C=0.00026m
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Given Solution: ** 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.. The coefficient of volume expansion is the proportionality constant beta in the relationship `dV = beta * V0 * `dT (completely analogous to the concept of a coefficient of linear expansion). We therefore have `dV = beta* V0*dT = 3 x 10^(-6) C^ (-1) * 352 cm^3 * (200C - 30 C) = 0.06 cm^3 ** STUDENT COMMENT: Similar to length an increase in temp. causes the molecules that make up this substance to move faster and that is the cause of expansion? INSTRUCTOR RESPONSE: At the level of this course, I believe that's the best way to think of it. There is a deeper reason, which comes from to quantum mechanics, but that's is way beyond the scope of this course. STUDENT COMMENT I found it difficult to express this problem because I was unable to type a lot of my steps into word, as they involved integration. However, I will take from this exercise that I should be more specific about where I got my numbers from and what I was doing for each of the steps I am unable to write out. INSTRUCTOR RESPONSE Your explanation was OK, though an indication of how that integral is constructed would be desirable. I understood what you integrated and your result was correct. For future reference: The integral of f(x) with respect to x, between x = a and x = b, can be notated int(f(x) dx, a, b). A common notation in computer algebra systems, equivalent to the above, is int(f(x), x, a, b). Either notation is easily typed in, and I'll understand either. Self-Critique: ------------------------------------------------ ------------------------------------------------ Self-Critique Rating: ********************************************* ********************************************* Question: `q001. A wall of a certain material is 15 cm thick and has cross-sectional area 5 m^2. It requires 1200 watts to maintain a temperature of 20 Celsius on one side of the wall when the other side is held at 10 Celsius. What is the thermal conductivity of the material? How many watts would be required to maintain a wall of the same material at 20 Celsius when the other is at 0 Celsius, if the cross-sectional area of the wall was 3000 cm^2? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: &&&& R=kAdT/dx R=k5m^2*10/15cm I got this far, now how do I get k by itself. I’m sure that the 1200 watts plays a part in this equation, but I don’t know what exactly. I went online and was looking for hints. The search reported that I needed to have a heat flux, so I found a formula for heat flux: W/m^2 which gives me 1200W/5m^2=240 so I will now plug the 240 into my equation. R=240w/m2*10/15cm Is that right? The latter part of the equation: R= kAdT/dx R=k3000cm^2*20/15cm Now, how do I calculate for the watts needed? I’m not sure if I have completed this correctly and I know that there has to be a step missing. @& Note that you are given the wall thickness `dx, the cross-sectional area A, the rate of energy flow R and the two temperatures, from which you can find the temperature difference. So you know everything except k in the equation R = k A `dT / `dx Can you solve that equation for k? (Be sure to let me know if not). Can you then plug the quantities into your expression and simplify to get k? *@ confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ ------------------------------------------------ Self-Critique Rating: ********************************************* ********************************************* Question: `q002. What is the specific heat of a material if it requires 5000 Joules to raise the temperature of half a kilogram of the material from 20 Celsius to 30 Celsius? By how much would the temperature of 100 grams of the same material change if it absorbed 200 Joules of heat? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: &&&& To find specific heat I need to take the joules and divide by the kg. So I will take 5000 joules/.5kg=10000 I will then multiply 10000 by the 10 degree temperature difference in C to give me 10000*10=100000J/kg/C I would have to convert grams to kg which would be .1kg if it was originally 100 grams. Gosh, I don’t know if I have been looking at this for too long or if I am just too dense to figure it out right now. I’m thinking that I need to somehow figure out how many joules it takes to raise the temperature one degree. So to get that I would need to divide the 10000 by the 10 degree temperature to give me 1000. So it takes 1000 joules to raise the temperature one degree.