course PHY 232 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
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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. Your Self-Critique: My answer looks different from the given solution, but can’t my answer also be used to find conductivity???
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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 didn’t explain my answer in the detail to which I should have explained what I meant by direct and inversely. Your Self-Critique rating #$&*3 ********************************************* Question: principles of physics and general college physics 13.8: coeff of expansion .2 * 10^-6 C^-1, length 2.0 m. What is expansion along length if temp increases by 5.0 C? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: Change in length = alpha*L0*change in temp Alpha = coefficient of linear expansion L0 = original length Change in length =(.2*10^-6 C^-1) * 2.0m * 5.0 C =2*10^-6 m confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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., • 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. Your Self-Critique:OK Your Self-Critique rating #$&* OK ********************************************* Question: query general phy 13.12: what is the coefficient of volume expansion for quartz, and by how much does the volume change? (Note that Principles of Physics and University Physics students do not do General Physics problems) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. Your Self-Critique: Your Self-Critique rating #$&* ********************************************* Question: query univ 17.103 (15.93 10th edition) (Note that Principles of Physics and General College Physics students don't do University Physics problems). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: Melting energy: 0950 kg * (3.3 * 10^5 J / kg) = 30 000 J Thermal energy to melt ice: .0950 kg * 4186 J / kg C * 100 C = 40 000 J Thermal energy needed for container: .446 kg * 390 J / kg C * 100 C = 16 000 J So, The energy it takes to melt ice, heat the water and container to 100C = 30000J + 40000J + 16000J = 86,000J Since the change in thermal energies add to become 0 you can find the final temperature with the following set up (using numbers we previously obtained)… (0.0950 kg * 3.3 * 10^5 J / kg ) + (0.0950 kg * 4186 J/kg*K *(Tf - 0 C)) + (.446 kg * 390 J/kg*K * (Tf - 0 C)) - .0350 kg * 2.256 x 10^6 J/kg + 4186 J / kg K * .035 kg * (Tf - 100 C) = 0 So Tf = 90 c confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: This problem took me a long time and many attempts for me to feel like I did it correct. I should’ve put my intermittent steps but that would have even taken longer so I hope that just giving my final answer is okay. I only put my confidence at 1 because it did take me many times and by the time I came up with the answer I had confused myself. Your Self-Critique rating #$&*3 ********************************************* Question: query univ phy 17.100 (90 in 10th edition): C = 29.5 J/mol K + (8.2 + 10^-3 J/mol K^2) T . YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: DQ = nCdT dQ = (3 mol) * (29.5j/mol k + (8.2 + 10^-3 j/mol k^2)T) * dt 27 + 276 = 303 K 227 + 276 = 503 K dQ = 3 [29.5k + ((8.2 * 10^-3k^2)/2)] = 3 [(29.5(503) + (8.2 * 10^-3(503)^2/2)) – (29.5(303) + ((8.2* 10^-3(303)^2)/2)] = 3 [15875.84 – 9314.92] = 47627.52 – 27944.75 = 19682.77 J confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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., 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: My answer didn’t come out to an even 20,000 Joules but since it was so close I’m guessing it was because I used 503K and 303k instead of 500K and 300k.
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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: My numbers vary slightly due to rounding. "