#$&* course PHY 202 1/28/15 at 10:15PM 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: I used a different format in my equation, but the solution is the same. ------------------------------------------------ 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 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: Since the equation for thermal energy flow is Thermal energy flow = (conductivity)(temperature gradient)(area) Then the rate of energy flow is directly proportional to conductivity, temperature gradient, and area. 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: I suppose I should have analyzed my response more, as opposed to just basing my solution solely off of the formula. The edited student’s response makes sense to me, because a material with greater area will be able to pass a greater energy flow through it. Then, as the conductivity of the material increases, it makes sense that the rate of energy able to flow through would be higher as well due to this characteristic. From what I understand, rate of energy flow is proportional to the temperature gradient because if the gradient is higher, then energy will flow through it more quickly as if to restore equilibrium between the two (such as two objects in contact with different temperatures that eventually come into equilibrium). ------------------------------------------------ Self-Critique Rating:3 ********************************************* 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: Change in L = Coeff expansion * Initial length * Temp change = (0.2*10^-6 C^-1) (2m) (5.0 C) = 2.0*10^-6 m confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: This problem is solved using the concept of a coefficient of expansion. The linear coefficient of thermal expansion of a material, denoted alpha, is the amount of expansion per unit of length, per unit of temperature: • expansion per unit of length is just (change in length) / (original length), i.e., • expansion per unit of length = `dL / L0 Thus expansion per unit of length, per unit of temperature is (expansion per unit of length) / `dT. Denoting this quantity alpha we have • alpha = (`dL / L0) / `dT. This is the 'explanatory form' of the coefficient of expansion. In algebraically simplified form this is • alpha = `dL / (L0 * `dT). In this problem we want to find the amount of the expansion. If we understand the concept of the coefficient of expansion, we understand that the amount of the expansion is the product of the coefficient of expansion, the original length and the temperature difference: If we don’t completely understand the idea, or even if we do understand it and want to confirm ourunderstanding, we can solve the formula alpha = `dL / (L0 * `dT) for `dL and plug in our information: • `dL = alpha * L0 * `dT = .2 * 10^-6 C^(-1) * 2.0 m * 5.0 C = 2 * 10^-6 m. This is 2 microns, two one-thousandths of a millimeter. By contrast the coefficient of expansion of steel is 12 * 10^-6 C^(-1); using this coefficient of expansion yields a change in length of 1.2 * 10^-4 m, or 120 microns, which is 60 times as much as for the given alloy. Self-Critique: OK ------------------------------------------------ Self-Critique Rating: ********************************************* Question: 5. The surface temperature of the Sun is about 5750 K. What is this temperature on the Fahrenheit scale? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: F = (K - 273) * (9/5) + 32 F = (5750 - 273) *(9/5) + 32 = 9890.6 F confidence rating #$&*: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): OK ------------------------------------------------ Self-critique Rating: #$&* ********************************************* Question: 12. How large an expansion gap should be left between steel railroad rails if they may reach a maximum temperature 35.0ºC greater than when they were laid? Their original length is 10.0 m. Assume that steel has coefficient of linear expansion of 12 * 10^-6 / K. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Change in L = Coeff expansion * Initial length * Change in temp Change in L = (12*10^-6/K) (10m) (35C+273) Change in L = 0.03696m If each steel rail may expand 0.03696m, then the gap between two adjacent rails should be 0.03696m. confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): I added 273 to 35C in my solution in order to convert to temperature in Kelvin. I did this because the expansion coefficient was given in terms of K. Is this incorrect?? ------------------------------------------------ Self-critique Rating:
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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): While I understand how to calculate the change in length based on the coefficient of expansion of metals, I don’t understand the second part of the problem with regards to the two 30-meter tapes. What is being compared - the two tapes to each other or to the invar or to the steel? - and how would a tape have a coefficient of expansion? ------------------------------------------------ Self-critique 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: I am at a total loss as to where the sphere diameter in your solution came from. I found myself unable to answer the second part of the question because I know where to find the information for initial volume or temperature change. ------------------------------------------------ Self-Critique Rating:
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Given Solution: ** Let Tf be the final temperature of the system. The ice doesn't change temperature until it's melted. It melts at 0 Celsius, and is in the form of water as its temperature rises from 0 C to Tf. If all the ice melts, then the melting process requires .0950 kg * (3.3 * 10^5 J / kg) = 30 000 J of energy, very approximately, from the rest of the system. If all the steam condenses, it releases .0350 kg * 2.256 * 10^6 J / kg = 80 000 Joules of thermal energy, very approximately, into the rest of the system. We can conclude that all the ice melts. We aren't yet sure whether all the steam condenses. If the temperature of all the melted ice increases to 100 C, the additional thermal energy required is (.0950 kg) * (4186 J / (kg C) ) * 100 C = 40 000 Joules, very approximately. The container is also initially at 0 C, so to raise it to 100 C would require .446 kg * (390 J / (kg C) ) * 100 C = 16 000 Joules of energy, very approximately. Thus to melt the ice and raise the water and the container to 100 C would require about 30 000 J + 40 000 J + 16 000 Joules = 86 000 Joules of energy. The numbers are approximate but are calculated closely enough to determine that the energy required to achieve this exceeds the energy available from condensing the steam. We conclude that all the steam condenses, so that the system will come to equilibrium at a temperature which exceeds 0 C (since all the ice melts) and is less than 100 C (since all the steam will condense). We need to determine this temperature. The system will then come to temperature Tf so its change in thermal energy after being condensing to water will be 4186 J / (kg K) * .035 kg * (Tf - 100 C). The sum of all the thermal energy changes is zero, so we have the equation m_ice * L_f + m_ice * c_water * (Tf - 0 C) + m_container * c_container * ( Tf - 0 C) - m_steam * L_v - + m_steam * c_water * ( Tf - 100 C ) = 0. The equation could be solved for T_f in terms of the symbols, but since we have already calculated many of these quantities we will go ahead and substitute before solving: [ 0.0950 kg * 3.3 * 10^5 J / kg ] + [0.0950 kg * 4186 J/kg*K *(Tf - 0 C)] + [.446 kg * 390 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 we get 170 J/C * Tf + 390 J/C * Tf - 79000 J - 14000 J + 31 000 J + 140 J / C * Tf = 0 or 700 J / C * Tf = 62 000 J, approx. or Tf = 90 C (again very approximately) Self-Critique: ------------------------------------------------ Self-Critique Rating: ********************************************* Question: query univ phy 17.98 / 17.100 (90 in 10th edition): C = 29.5 J/mol K + (8.2 + 10^-3 J/mol K^2) T . How much energy is required to change the temperature of 3 moles from 27 C to 227 C? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** In this case the specific heat is not constant but varies with temperature. The energy required to raise the temperature of 3 moles by `dT degrees (where `dT is considered to be small enough that the change in specific heat is insignificant) while at average temperature T is `dQ = 3 mol * C * dT = 3 mol * (29.5 J/mol K + (8.2 * 10^-3 J/mol K^2) T) * `dT. To get the energy required for the given large change in temperature (which does involve a significant change in specific heat) we integrate this expression from T= 27 C to T = 227 C, i.e., from300 K to 500 K. An antiderivative of f(t) = (29.5 J/mol K + (8.2 + 10^-3 J/mol K^2) T) is F(T) = 29.5 J / (mol K) * T + (8.2 + 10^-3 J/mol K^2) * T^2 / 2. We simplify and apply the Fundamental Theorem of Calculus and obtain F(500 K) - F(300 K). This result is then multiplied by the constant 3 moles. The result for Kelvin temperatures is about 3 moles * (F(500 K) - F(300 K) = 20,000 Joules. ** Self-Critique: ------------------------------------------------ Self-Critique Rating: ********************************************* Question: University Physics Problem 17.106 (10th edition 15.96): Steam at 100 Celsius is bubbled through a .150 kg calorimeter initially containing .340 kg of water at 15 Celsius. The system ends up with a mass of .525 kg at 71 Celsius. From these data, what do we conclude is the heat of fusion of water? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: **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 which is easily solved to give us • Hf = 79,000 J / (.035 kg) = 2,257,000 J / kg. ** ------------------------------------------------ Self-Critique Rating: ********************************************* Question: `q003. A container with negligible mass holds 500 grams of water, and 100 grams of ice and 800 grams of a substance whose specific heat is 1800 Joules / (kilogram * Celsius), all at 0 Celsius. How much steam at 100 Celsius must be bubbled through the water to raise the temperature of the system to 20 Celsius? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: Phase Change of Ice to Water - Q = mL Q = 0.1kg * 333kJ/kg = 33.3 kJ to change ice to water at 0C = 33300 J Heating Water to 20C 0.1kg ice has now melted to water at 0C 0.6kg water total Increasing from 0C to 20C gives a temp change of 20C Q = m * c * change in temp Q = (0.6kg) (4186 J/kg*C) (20C) Q = 50232 J Increasing temp of substance by 20C Q = m * c * change in temp Q = (0.8kg) (1800J/kg*C) (20C) Q = 28800 J Total Energy Required = 33300 J + 50232 J + 28800 J = 112332 J required to raise the temperature of the entire system from 0C to 20C Amount of Steam Required 112.332 kJ = m * c * change in temp 112.332 kJ = m * 2010 J/kg*C * 20C m = 0.00279 kg confidence rating #$&*:’m not entirely sure that my choice for the last equation is correct)