#$&* course PHY 202 2/8 about 12:15 pm 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:OK ------------------------------------------------ Self-Critique Rating:OK ********************************************* 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 R=k*(‘dT/L)*A, area and temp. gradient are both proportional to the rate of thermal energy flow, and the thickness is inversely proportional. This means that greater area, the greater the rate of flow. If the thickness is smaller, the temperature gradient increases, thus does the rate of flow. 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:OK ------------------------------------------------ Self-Critique Rating:OK ********************************************* 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: ‘dL = ‘alpha*L*’dT = (.2*10^-6 C^-1) * 2.0 m * 5.0 C = 2*10^-6 m confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: This problem is solved using the concept of a coefficient of expansion. The linear coefficient of thermal expansion of a material, denoted alpha, is the amount of expansion per unit of length, per unit of temperature: • expansion per unit of length is just (change in length) / (original length), i.e., • expansion per unit of length = `dL / L0 Thus expansion per unit of length, per unit of temperature is (expansion per unit of length) / `dT. Denoting this quantity alpha we have • alpha = (`dL / L0) / `dT. This is the 'explanatory form' of the coefficient of expansion. In algebraically simplified form this is • alpha = `dL / (L0 * `dT). In this problem we want to find the amount of the expansion. If we understand the concept of the coefficient of expansion, we understand that the amount of the expansion is the product of the coefficient of expansion, the original length and the temperature difference: If we don’t completely understand the idea, or even if we do understand it and want to confirm our understanding, we can solve the formula alpha = `dL / (L0 * `dT) for `dL and plug in our information: • `dL = alpha * L0 * `dT = .2 * 10^-6 C^(-1) * 2.0 m * 5.0 C = 2 * 10^-6 m. This is 2 microns, two one-thousandths of a millimeter. By contrast the coefficient of expansion of steel is 12 * 10^-6 C^(-1); using this coefficient of expansion yields a change in length of 1.2 * 10^-4 m, or 120 microns, which is 60 times as much as for the given alloy. Self-Critique:OK ------------------------------------------------ Self-Critique Rating:OK ********************************************* Question: 5. The surface temperature of the Sun is about 5750 K. What is this temperature on the Fahrenheit scale? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: T(F) = (9/5)*(T(K) - 273.15) + 32 = (9/5)*(5750 - 273.15) + 32 = 9890.33 F confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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):I used the formula given in the book but I understand your reasoning in your solution. ------------------------------------------------ Self-critique Rating:3 #$&* ********************************************* Question: 12. How large an expansion gap should be left between steel railroad rails if they may reach a maximum temperature 35.0ºC greater than when they were laid? Their original length is 10.0 m. Assume that steel has coefficient of linear expansion of 12 * 10^-6 / K. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Since we are dealing with a change in the temperature, and given a coefficient with units k^-1 with a temperature in C, we don’t need to convert to the same units of temp. ‘dL = ‘alpha * L_0 * ‘dT = 12*10^-6 (1/K) * 10.0 m * 35 K = 0.0042 m or 4.2 mm My initial thought was to multiply this by two since the rails were adjoining but then I thought about the fact that the expansion will occur on both ends of the rail so on one side of the rail it will only be 2.1 mm. confidence rating #$&*: 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):OK ------------------------------------------------ Self-critique Rating:OK #$&* ********************************************* Question: (optional for Principles of Physics students) (a) Suppose a meter stick made of steel and one made of invar (an alloy of iron and nickel) are the same length at 0ºC . What is their difference in length at 22.0ºC ? (b) Repeat the calculation for two 30.0-m-long surveyor’s tapes. Assume coefficient of expansion for invar to be 1.2 * 10^-6 / K. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: ‘dL = ‘alpha * L * ‘dT ‘dL_s_ms = 12*10^-6 K^-1 * 1.0 m * 22 K = 0.000264 m = 0.26 mm ‘dL_i_ms = 1.2*10^-6 K^-1 * 1.0 m * 22 K = 0.0000264 m = 0.026 mm The difference would be 0.26 mm - 0.026 mm = .234 mm ‘dL_s_st = 12*10^-6 K^-1 * 30 m * 22 K = 0.00792 m = 7.9 mm ‘dL_i_st = 122*10^-6 K^-1 * 30 m * 22 K = 0.000792 m = 0.79 mm The difference would be 7.9 mm - 0.79 mm = 7.11 mm confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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):OK ------------------------------------------------ Self-critique Rating:OK #$&* ********************************************* Question: query general phy 13.12: what is the coefficient of volume expansion for quartz, and by how much does the volume change? (Note that Principles of Physics and University Physics students do not do General Physics problems) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: ‘beta = 1*10^-6 C^-1 ‘dT = 200C - 30C = 170C Diameter = 8.75 cm Radius = 4.375 cm V = (4/3)*pi*(4.375 cm)^3 = 350.77 cm^3 ‘dV = ‘beta * V * ‘dT = 1*10^-6 C^-1 * 351 cm^3 * 170 C = 0.06 cm^3 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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:OK ------------------------------------------------ Self-Critique Rating:OK ********************************************* 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: L=15 cm = 0.15 m, A=5 m^2, R = 1200 W, ‘dT = 20 C - 10 C = 10 C ‘dT/L = 10 C / 0.15 m = 66.7 C/m k = R/((‘dT/L)*A) = (1200 W) / (66.7 C/m * 5 m^2) = 3.60 W/(C*m) L=15 cm = 0.15 m, A = 3000 cm^2 = 0.3 m^2, k = 3.60 W/(C*m) ‘dT=20 C - 10C = 10 C ‘dT/L = 20 C / 0.15 m = 133 C/m R = k * (‘dT/L) * A = 3.6 W/(C*m) * 133 C/m * 0.3 m^2 = 144 W confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-Critique Rating:OK ********************************************* Question: `q002. What is the specific heat of a material if it requires 5000 Joules to raise the temperature of half a kilogram of the material from 20 Celsius to 30 Celsius? By how much would the temperature of 100 grams of the same material change if it absorbed 200 Joules of heat? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: Q = 5000 J, m = 0.5 kg, ‘dT = 30 C - 20 C = 10 C, c = ? Q=m*c*’dT C = Q / (m*’dT) = 5000 J / (0.5 kg * 10 C) = 1000 J/(kg*C) Q = 2000 J, m = 100 g = 0.1 kg, c = 1000 J/(kg*C), ‘dT = ? ‘dT = Q / (m*c) = 200 J / (0.1 kg * 1000 J/(kg*C)) = 2 C confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-Critique Rating:OK ********************************************* Question: query univ 17.101 / 17.103 (15.93 10th edition) (Note that Principles of Physics and General College Physics students don't do University Physics problems). A copper calorimeter of mass .446 kg contains .095 kg of ice, all at 0 C. .035 kg of steam at 100 C and 1 atm pressure is added. What is the final state of the system? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: 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) 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., 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. ** 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: ???I assume this and the following question are for University Physics. If not let me know and I will resubmit.???