#$&* course Phy 202 002. `query 2*********************************************
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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: ------------------------------------------------ Self-Critique Rating: ********************************************* 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 The less area there is the higer the thermal enrgy flow will be. Thermal energy flow is proportional to area. Thermal energy flow is incersly proportional to thickness. Thermal energy flow is proporitonal to the temperature gradient. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: energy flow increases if the area increases. The greater the area the more heat can move across the space. The rate of energy flow is proportional to the temperature gradient. The greater the difference in the two temperatures the faster energy will move from one side to the other. Greater thickness will affect the temperature gradient, more thickness = less heat transfer Lower temperature gradient = lower rate energy flow. Therefore, rate of energy flow is inversely proportional to the thickness of the wall. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: ------------------------------------------------ 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: alpha = (`dL / L0) / `dT dL = alpha * L0 * `dT dL = .2 * 10^-6 C^(-1) * 2.0 m * 5.0 C dL = 2 * 10^-6 m confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: This problem is solved using the concept of a coefficient of expansion. The linear coefficient of thermal expansion of a material, denoted alpha, is the amount of expansion per unit of length, per unit of temperature: · expansion per unit of length is just (change in length) / (original length), i.e. alpha = (`dL / L0) / `dT, · expansion per unit of length = `dL / L0 Thus expansion per unit of length, per unit of temperature is (expansion per unit of length) / `dT. Denoting this quantity alpha we have · alpha = (`dL / L0) / `dT. This is the 'explanatory form' of the coefficient of expansion. In algebraically simplified form this is · alpha = `dL / (L0 * `dT). In this problem we want to find the amount of the expansion. If we understand the concept of the coefficient of expansion, we understand that the amount of the expansion is the product of the coefficient of expansion, the original length and the temperature difference: If we don’t completely understand the idea, or even if we do understand it and want to confirm our understanding, we can solve the formula alpha = `dL / (L0 * `dT) for `dL and plug in our information: · `dL = alpha * L0 * `dT = .2 * 10^-6 C^(-1) * 2.0 m * 5.0 C = 2 * 10^-6 m. This is 2 microns, two one-thousandths of a millimeter. By contrast the coefficient of expansion of steel is 12 * 10^-6 C^(-1); using this coefficient of expansion yields a change in length of 1.2 * 10^-4 m, or 120 microns, which is 60 times as much as for the given alloy. Self-Critique: ------------------------------------------------ Self-Critique Rating: ********************************************* Question: 5. The surface temperature of the Sun is about 5750 K. What is this temperature on the Fahrenheit scale? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: F = 9/5(K-273)+32 F= 9/5(5750-273)+32 F = 9890.33 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ------------------------------------------------ Self-critique Rating: #$&* ********************************************* Question: 12. How large an expansion gap should be left between steel railroad rails if they may reach a maximum temperature 35.0ºC greater than when they were laid? Their original length is 10.0 m. Assume that steel has coefficient of linear expansion of 12 * 10^-6 / K. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: `dL = 10.0 meter * (35 C * 12 * 10^-6 / K) 'dL = 0.042 meter confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: #$&* ********************************************* Question: (optional for Principles of Physics students) (a) Suppose a meter stick made of steel and one made of invar (an alloy of iron and nickel) are the same length at 0ºC . What is their difference in length at 22.0ºC ? (b) Repeat the calculation for two 30.0-m-long surveyor’s tapes. Assume coefficient of expansion for invar to be 1.2 * 10^-6 / K. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: `dL = 1 m * 22 Celsius * 12 * 10^-6 / K = .00026 m inar: `dL = 1 m * 22 Celsius * 12 * 10^-6 K = .000026 m .00026 m - .000026 m = .00023 m The difference in the 30.0 m long tapses would be 30 times as much expansion. confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: When is temperature is increased by 22 Celsius, a one-meter length of steel will experience change in length `dL = 1 meter * 22 Celsius * 12 * 10^-6 / K = .00026 meters (about a quarter of a millimeter) and a one-meter length of invar will experience change in length `dL = 1 meter * 22 Celsius * 12 * 10^-6 K = .000026 meters. The difference in the lengths of the meter sticks will therefore be about .00026 m - .000026 m = .00023 m. The difference for two 30-meter tapes would be 30 times as great. This difference would be close to a centimeter. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: #$&* ********************************************* Question: query general phy 13.12: what is the coefficient of volume expansion for quartz, and by how much does the volume change? (Note that Principles of Physics and University Physics students do not do General Physics problems) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: coefficient expansion: 1 x 10^(-6) C^(-1) sphere volume: 4/3 pi r^3 = 4/3 pi ( 4.38 cm)^3 = 352 cm^3 `dV = beta * V0 * `dT `dV = = 3 x 10^(-6) C^ (-1) * 352 cm^3 * (200C - 30 C) 'dv = 0.06 cm^3 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 Your Solution: k = R * L / (`dT * A) k = 1200W * 15cm / (10C * 3000 cm^2) k = .75 W/(cm*C) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-Critique Rating: ********************************************* Question: `q002. What is the specific heat of a material if it requires 5000 Joules to raise the temperature of half a kilogram of the material from 20 Celsius to 30 Celsius? By how much would the temperature of 100 grams of the same material change if it absorbed 200 Joules of heat? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: Heat added = specific heat * mass * change in temp 5000J = specific heat * .5kg * 10C 1000 J/kg*C = specific heat confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-Critique Rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: