#$&* course Phy 232 Question: Describe how we find the conductivity given the rate of energy flow, area, temperatures, and thickness of the wall.
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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 dont 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. Your Self-Critique: I feel my answer is sufficient but may not have gone into detail enough. Your Self-Critique Rating: 2 ********************************************* 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: Thermal energy flow is proportional to area. We know this because in the equation H = k*A*T, the Area is in the numerator. Thermal energy flow is inversely proportional to the thickness. We know this because in the equation, the thickness is in the denominator. Thermal energy flow is proportional to the temperature gradient. We know this because in the equation, the temperature gradient is in the numerator. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 pathavailable 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: My answer is correct, I may not have gone into enough detail, however. Your Self-Critique Rating: 2 ********************************************* 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). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: Can mass = 0.446 kg, contains 0.0950 kg of ice. Initial Temp = 0 degrees C 0.0350kg of steam is added at 100 degrees C 1 atm pressure is added What is the final temperature of the can and its contents? For our purposes, it is safe to assume all of the ice melts, since it begins to melt at 0 degrees Celsius. We can also conclude the final temperature will be somewhere between 0 degrees Celsius and 100 degrees Celsius. I do not understand how to obtain this temperature. This is all I could come up with. confidence 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) Your Self-Critique: I do not understand how a lot of those numbers were obtained, such as 3.3 * 10^5 J / kg and 4186 J / (kg C). I could not figure out where they came from. Your Self-Critique Rating: 1
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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., from 300 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. ** Your Self-Critique: I do not understand where the value of C, (29.5 J/mol K + (8.2 * 10^-3 J/mol K^2) T), is obtained from. Your Self-Critique Rating: 2
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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. ** Your Self-Critique: I do not understand where the some of your values came from. I think I am missing something very important. I will go back and read the section. Your Self-Critique Rating: 1 " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!