#$&* course Phy 122 1/27 3pm If your solution to stated problem does not match the given solution, you should self-critique per instructions athttp://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm
.............................................
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:OK ------------------------------------------------ Self-Critique Rating:OK ********************************************* Question: Explain in terms of proportionalities how thermal energy flow, for a given material, is affected by area (e.g., is it proportional to area, inversely proportional, etc.), thickness and temperature gradient. your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv It is directly proportional to area, inversely proportional to thickness, and directly proportional to temperature gradient. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution:3 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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:OK ------------------------------------------------ Self-Critique Rating:OK ********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: I have looked at the formulas, but I'm still not sure how to get the correct answer. I'm trying to memorize what each abbreviation stands for, so I am semi confused about it thus far.
.............................................
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: Seeing how it is supposed to be worked helps me grasp what I am supposed to do. Still a little confusing, but I am starting to catch on. ------------------------------------------------ Self-Critique Rating:3 ********************************************* Question: 5. The surface temperature of the Sun is about 5750 K. What is this temperature on the Fahrenheit scale? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: A Kelvin is 1.8 Fahrenheit degrees, so we use this conversion factor to find what 5750 K is in Fahrenheit degrees. 5750 K * (1.8 Fahrenheit degree / 1 K) = 10350 Fahrenheit degrees confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: 5750 K means 5750 Kelvin degrees above absolute zero. A Kelvin degree is 1.8 Fahrenheit degrees, so this temperature is 5750 K * 1.8 (F / K) = 10350 Fahrenheit degrees above absolute zero. 0 on the Fahrenheit scale is about 460 Fahrenheit degrees above absolute zero, so 10350 Fahrenheit degrees above absolute zero is about (10350 - 460) Fahrenheit = 9890 Fahrenheit. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Ohhhhhh I did not realize that 5750 K was 5750 K degrees above absolute zero, which effected the answer. I got it right to a certain step, but did not continue enough to ge the right answer. ------------------------------------------------ Self-critique Rating:3
.............................................
Given Solution: The expansion of a 10.0 meter steel rail when temperature increases by 35 C is `dL = 10.0 meter * (35 C * 12 * 10^-6 / K) = 0.042 meter, or a little over 4 millimeters. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I got the wrong answer. I got one too many zeros in my answer compared to the correct answer. I suppose I found my number incorrectly when finding what 12 * 10^-6 was, but I thought for sure I did that correctly. ------------------------------------------------ Self-critique Rating:3 #$&* " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #$&* " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!