course Phys 202 Your solution, attempt at solution:If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
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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. 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). Your Self-Critique: I showed = k = (`dQ / `dt) / (A ( `dT / `dx) ) = rate of thermal energy flow / (c.s. area * temperature gradient). Instead of stating what conductivity was. Your Self-Critique Rating: 3 ********************************************* 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 Thermal energy flow is proportional to area. If the area is larger the flow will be larger. Thermal energy flow is inversely proportional to thickness. If the thickness increases the flow decreases. Thermal energy flow is directly proportional to the temperature gradient. If the gradient increases the flow increases. Confidence Rating: 3
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Given Solution: ** CORRECT STUDENT ANSWER WITHOUT EXPLANATION: Energy flow is: directly proportional to area inversely propportional to thicknessand 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. Your Self-Critique: I could have used more detailed explanations such as the examples listed in the solution. Your Self-Critique Rating: 2 ********************************************* 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: delta L = alpha(original L) (Delta T) delta L = (0.2 * 10^-6 C^-1) (2 m) (5.0 C) 2 * 10 ^-6 m Confidence Rating: 3
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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 dont 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. Your Self-Critique: OK Your Self-Critique Rating: OK ********************************************* Question: 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) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: quartz has a coefficient of volume expansion of 1*10^-6 C^-1 (Beta) volume = 4/3 pi R^3 = 4/3 pi (4.375)^3 = 350.8 cm^3 delta V = Beta (Vo)(delta T) = 1*10^-6 C^-1 (350.8) (170) =0.0596 cm^3 Confidence Rating: 3
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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 ** Your Self-Critique: OK Your Self-Critique Rating: OK ********************************************* Question: query univ 17.103 (15.93 10th edition) (Note that Principles of Physics and General College Physics students don't do University Physics problems). ********************************************* Question: query univ phy 17.100 (90 in 10th edition): C = 29.5 J/mol K + (8.2 + 10^-3 J/mol K^2) T . ********************************************* Question: University Physics Problem 17.106 (10th edition 15.96): "