course Phy 202
......!!!!!!!!...................................
11:06:10 query introset 5 # 12: Finding the conductivity given rate of energy flow, area, temperatures, thickness of wall. Describe how we find the conductivigy given the rate of energy flow, area, temperatures, and thickness of the wall?
......!!!!!!!!...................................
RESPONSE --> The definition of conductivity will help us in the process of finding the conductivity. The rate at which thermal energy is conducted is equal to the product of the conductivity, area of the wall, and temperature gradient. So, conductivity is equal to the rate of conduction divided by the area and the temperature gradient. The area and rate of conduction is usually given, but to figure out the temperature gradient, you just divide the difference in the temperature change by the thickness of the wall. So, to find the conductivity the generalized formula would be: thermal conductivity (k) = rate of thermal energy flow/(cross sectional area * temperature gradient)
.................................................
......!!!!!!!!...................................
11:06:45 ** 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 Rate of thermal energy condction = conductivity * temperature gradient * area, or R = k * `dT/`dx * A. For an object of uniform cross-section `dT is the temperature difference across the object and `dx is the distance L between the faces of the object. In this case the equation is R = k * `dT / L * A and we can solve to get k = R * L / (`dT * A). **
......!!!!!!!!...................................
RESPONSE --> Okay
.................................................
......!!!!!!!!...................................
11:12:45 Explain how energy flow, for a given material, is affected by area (e.g., is it proportional to area, inversely proportional, etc.), thickness and temperature gradient.
......!!!!!!!!...................................
RESPONSE --> The rate at which thermal energy is transferred is proportional to the cross sectional area of the wall and to the temperature gradient. The energy flow is going to be lower if a wall is thicker, because there is just more wall for the energy to go through. It makes sense. The temperature gradient will be The temperature gradient is proportional to the temperature difference, and the gradient will change by the same factor.
.................................................
......!!!!!!!!...................................
11:18:44 ** CORRECT STUDENT ANSWER WITHOUT EXPLANATION: Energy flow is directly proportional to area inversely propportional to thickness and directly proportional to temperature gradient GOOD STUDENT ANSWER WITH EXPLANATIONS, PLUS INSTRUCTOR COMMENTARY: The energy flow for a given material increases if the area increases. This is because the more area you have the wider a path something has to go through so more of it can move through it. Just like a 4 lane highway will carry more cars in a given time interval than a two lane highway will. So the relationship of energy flow to area is proportional. Energy flow, however is inversely proportional to thickness. This is because although the thermal energy flows through the material, the material impedes it. So if the thickness increases the thermal energy will have to travel farther through the resistance and be impeded more. ** Also for given temperature difference, greater thickness `dx implies smaller temperature gradient `dT / `dx. Small temperature gradient doesn't 'drive' the energy flow as much. Energy flow is also proportional to the temperature gradient. Meaning if the difference in the two temperatures is greater then the energy will move faster from one side to the other. Temperature gradient is not difference in temperatures, it's difference in temperature per unit of distance across the material. Temperature gradient is `dT / `dx, not just `dT. Greater temperature gradient means greater difference in temperature over any given small distance increment. The greater the temperature difference across this increment the more energy will flow. **
......!!!!!!!!...................................
RESPONSE --> So, the more area you have the greater the energy flow. A bigger wall means more energy movement. The highway explanation helped a lot to make that clear. I understand the wall thickness, because it is like insulation. The more insulation in the walls the less cold air gets through to your house, so it makes sense that a thicker wall decreases energy flow. For temperature gradient, I understand that if the difference in temperatures is greater that more energy will be passing through the wall. I like to think of the house example again. The colder it gets outside, the more you have to turn up the heat inside, because the colder air is getting into the house faster which means that the energy is transferring through the wall faster.
.................................................
......!!!!!!!!...................................
11:22:28 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?
......!!!!!!!!...................................
RESPONSE --> L = Lo(1 + a * delta T) L = 2m [1 + (0.2 x 10^-6 (C)-1) 5C] L = 2.000002 m The length of the table increases by .000002 meters
.................................................
......!!!!!!!!...................................
11:43:05 The amount of the expansion is the product of the coefficient of expansion, the original length and the temperature difference: `dL = alpha * L0 * `dT = .2 * 10^-6 C^(-1) * 2.0 m * 5.0 C = 2 * 10^-6 m. This is 2 microns, two-onethousandths of a millimeter. By contrast the coefficient of expansion of steel is 12 * 10^-6 C^(-1);using this for the coefficient of expansion yields a change in length of 24 * 10^-6 m, or 24 microns, which is 240 times as much as for the given alloy.
......!!!!!!!!...................................
RESPONSE --> Iron and Nickel have a low coefficient of linear expansion [0.2 x 10-6 (C)-1]. If the table is initially 2 meters, how much will the table expand if the temperature is raised by 5C? L = Lo(1 + a*delta T) L = 2m [1 + (0.2 x 10-6 (C)-1) 5C] L = 2.000002 m So, the change was .000002 meters of 2 microns, because you have to subtract the initial 2 meters from the 2.000002 to get how much the table actually expanded. That makes sense. Also, I can see why it is easier to use the (`dL = alpha * L0 * `dT) formula, because I am having to subtract the initial 2 meters from my answer to get the difference and it is an extra step that would not be necessary if I used the right formula. If the table was steel: L = Lo(1 + a * delta T) L = 2m [1 + (12 x 10-6 (C)-1) 5C] L = 2.00012 m Okay I am confused for this one because my difference for the steal was 12 x 10^-5. Were we not supposed to multiply by the 5C? Because I do get 24 * 10^-6 if I just multiply alpha and Lo, but that is not what the formula says to do.
.................................................
......!!!!!!!!...................................
11:46:40 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)
......!!!!!!!!...................................
RESPONSE --> A quartz sphere is 8.75 cm in diameter. What will be its change in volume if it is heated from 30C to 200C? First, the initial volume of the sphere needs to be found. Volume of a sphere = (4/3)pr^3 = (4/3) p 4.3753 =350.77cm^3 Then, you have to convert the cm3 into L. 350.77 cm^3 x 1 Liter/1000 cm^3 = .35077 Liters Finally you solve for 'dV = * Vo * 'dT with = (1 x 10^-6) 'dV = [1 x 10^-6 (C)-1] (.35077Liters) (200C -30C) 'dV = 5.96 x 10^-5 Liters
.................................................
......!!!!!!!!...................................
11:54:03 ** 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.. We therefore have dV = beta* V0*dT = 3 x 10^(-6) C^ (-1) * 352 cm^3 * (200C - 30 C) = 0.06 cm^3 **
......!!!!!!!!...................................
RESPONSE --> So I did not have to change my cm^3 to Liters. Let me re-work the problem and see if I get the same answer. Volume of a sphere = (4/3)pr^3 = (4/3) p 4.3753 =350.77cm3 So, I got approximately 351 cm^3 Now solve for 'dV = * Vo * 'dT with = (1 x 10-6) 'dV = [1 x 10-6 (C)-1] (351 cm^3) (200C -30C) 'dV = .06 cm^3 Okay, I got the same answer, but how come in the solution, beta was changed from 1 x 10^-6 to 3 x 10^-6 ? Is that just a typo, because the answer would be different if you had actually used that number.
.................................................
......!!!!!!!!...................................
11:55:02 query univ 17.103 (15.93 10th edition) (Note that Principles of Physics and General College Physics students don't do University Physics problems).
......!!!!!!!!...................................
RESPONSE --> I am not in University Physics, I am a general college physics student.
.................................................
......!!!!!!!!...................................
11:55:20 ** The ice doesn't change temperature until it's melted, at which time it is in the form of water with the specific heat of water. Also the steam will come to temperature Tf so its change in thermal energy after being condensted will be 4186 J / (kg K) * .035 kg * (Tf - 100 C). I prefer to say that the sum of all the thermal energy changes is zero, so that we don't have to worry about taking a negative of a negative (which you should have done on your right-hand side, and which would have avoided the negative result). I would write the equation as follows: [.446 kg * 390 J/kg*K * (Tf - 0 C)] + [0.0950 kg * 4186 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 this gives you 170 J/C * Tf + 390 J/C * Tf - 79000 J - 14000 J + 140 J / C * Tf = 0 or 700 J / C * Tf = 93000 J, approx. or Tf = 130 C. This isn't possible--we can't end up warmer than the original temperature of the steam. We conclude that not all the steam condenses and that the system therefore reaches equilibrium at 100 C, with a mixture of water and steam. Our energy conservation equation will therefore be [.446 kg * 390 J/kg*C * (100 C - 0 C)] + [0.0950 kg * 4186 J/kg*C *(100 C - 0 C)] - mCondensed * 2.256 x 10^6 J/kg = 0 where mCondensed is the mass of the condensed steam. This gives us 17000 J + 39000 J - mCondensed * 2.3 * 10^6 J/kg = 0 or mCondensed = 56000 J / (2.3 * 10^6 J/kg) = .023 kg. We end up with .095 kg * .023 kg = .118 kg of water and .035 kg - .023 kg = .012 kg of steam. **
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
11:55:30 query univ phy 17.100 (90 in 10th edition): C = 29.5 J/mol K + (8.2 + 10^-3 J/mol K^2) T . Give your solution to this problem.
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
11:55:33 ** Specific heat is not constant but varies with temperature. The energy required to raise the temperature of 3 moles by dT degrees while at temperature T is 3 mol * C * dT = 3 mol * (29.5 J/mol K + (8.2 + 10^-3 J/mol K^2) T) * dT. You have to integrate this expression from T= 27 C to T = 227 C, which is from 300 K to 500 K. Antiderivative of (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. Simplify and apply Fundamental Theorem of Calculus (find F(500) - F(300) if you think the temperature T is in Kelvin or F(227) - F(27) if you think it's in Celsius; this isn't specified in the problem and while the units tend to imply Kelvin temperature the resulting specific heats would be unrealistic for most real substances), then multiply by the constant 3 moles. The result for Kelvin temperatures is about 20,000 Joules. **
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
11:55:35 ** Specific heat is not constant but varies with temperature. The energy required to raise the temperature of 3 moles by dT degrees while at temperature T is 3 mol * C * dT = 3 mol * (29.5 J/mol K + (8.2 + 10^-3 J/mol K^2) T) * dT. You have to integrate this expression from T= 27 C to T = 227 C, which is from 300 K to 500 K. Antiderivative of (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. Simplify and apply Fundamental Theorem of Calculus (find F(500) - F(300) if you think the temperature T is in Kelvin or F(227) - F(27) if you think it's in Celsius; this isn't specified in the problem and while the units tend to imply Kelvin temperature the resulting specific heats would be unrealistic for most real substances), then multiply by the constant 3 moles. The result for Kelvin temperatures is about 20,000 Joules. **
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
11:55:40 University Physics Problem 17.106 (10th edition 15.96): Give your solution.
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
11:55:43 **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 so that Hf = 79,000 J / (.035 kg) = 2,257,000 J / kg. **
......!!!!!!!!...................................
RESPONSE -->
................................................."