#$&* course Phy 242 002. `query 2*********************************************
.............................................
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 didn't go into nearly as much detail but the point was the same. ------------------------------------------------ Self-Critique Rating: 3 ********************************************* 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 Thermal energy flow is proportional to area and temperature gradient. This is easy to tell by looking at the equation used in the question above: Rate of thermal energy flow=temperature gradient*area of the wall*conductivity If you increase the area or the temperature gradient, the rate of thermal energy flow will increase. If you decrease the area of the temperature gradient, the rate of thermal energy flow will also decrease. This is why I said they are proportional to each other. However, this is not the case for the thickness. This is harder to see since in the formula above, thickness is not in plain sight. You have to remember that when you solved for the temperature gradient, you took the difference in temperatuer and divided by thickness. So if I rewrite the equation above showing this, this is how it looks: Rate of thermal energy flow=[(T2-T1)/thickness]*area of the wall*conductivity Now, it is easier to see the relationship thickness has with rate of thermal energy flow. Since the thickness is in the denominator, as it increases, this will decrease the rate of thermal energy and when it decreases, this will increase the rate of thermal energy flow. This means that thickness is inversely proportional to the rate of thermal energy flow. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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: ???I'm assuming this doesn't apply to me since I am university physics???
.............................................
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: ------------------------------------------------ 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: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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: The thermal conductivity is 1.8 watts/(C m). I found this by using the equation 'dQ/'dt=kA*'dT/'dx.
.............................................
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: I didn't solve for Tf because the question asked for the state of the system. However, I read through the steps of solving for the Tf and I understand them. ------------------------------------------------ Self-Critique Rating: 3 ********************************************* 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: The energy required is 21,363. Since the specific heat is dependent on temperature, I took the integral of the equation given for the specific heat from the temperature 300 K (27 C) to 500 K (227 C) and multiplied this by 3 moles. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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: OK ------------------------------------------------ Self-Critique Rating: OK ********************************************* 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: The specific heat of water is 2,156,798 J/kg. I found this by multiplying the required energy to change the temperature of the water from 15 C to 71 C plus the energy required to bring the calorimeter from a temperature of 15 C to 71 C. I set these calculations equal to the energy required to condense the steam plus the energy required to bring the temperature of the steam to 71 C. NOTE: I HAD TO LOOK AT THE SOLUTION TO FIND THE SPECIFIC HEAT FOR THE CALORIMETER. IN THE PROBLEM, IT DOES NOT SPECIFY WHAT TYPE OF MATERIAL THE CALORIMETER IS MADE OUT OF. IT WOULD'VE BEEN VERY HELPFUL TO HAVE KNOWN THIS RATHER THAN HAVING TO LOOK AT THE SOLUTION ONCE I REALIZED I HAD TWO UNKNOWNS (THE SPECIFIC HEAT OF THE CALORIMETER AND THE HEAT OF VAPORIZATION OF THE WATER).
.............................................
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: 3 ********************************************* 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 waterto raise the temperature of the system to 20 Celsius? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: .043 kg of steam must be bubbled through. I calculated this by finding the energy required to melt the ice, plus the energy required to raise the temperature of the melted ice, plus the energy required to raise the temperature of the existing water, plus the energy required to raise the temperature of the calorimeter. I set these values equal to the energy required to condense the steam plus the energy required to lower the temperature of the steam. Once I did this, my only unknown was the mass of the steam. I solved and got .043. My equation looked like this: .1*3.3*10^5+.1*20*4186+.5*20*4186+.8*1800*20=x*2.256*10^6+x*4186*80 Note: x is the unknown value of the mass of the steam. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-Critique Rating: ********************************************* Question: `q004. The specific heat of a certain substance increases linearly from 1200 Joules / (kg C) at 150 C to 1400 Joules / (kg C) at 350 C. How much heat would be required to increase the temperature of a 5 kg sample from 200 C to 300 C? Show how this problem could be solved without using an integral. Show how this problem could be solved using an integral. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: The heat that would be required to increase the temperature of a 5 kg sample from 200 C to 300 C would be 650,000 J. I found this without using an integral by first by graphing the line that represents the change in specific heat. I did this by using my two points and the slope between them. Once I drew this, I calculated the specific heats at 200 C and at 300 C using the slope of the line (1). The two points I found were (200, 1250) and (300, 1350). I calculated the area under the line between those two points using a rectangle and a triangle. Once I found the area, I multiplied by 5 kg. To solve this problem using an integral I found the equation of the line and took the integral between the two temperatures (200 and 300). Then I multiplied this by 5 kg. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^