#$&* course PHY 232 4/9/2011; 22:30 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 change in the thermal energy of an isolated system is 0. So assuming that the systems are isolated the thermal energy change of one object is equal and opposite to that of the other. For an ideal substance the change in the thermal energy of an object is directly proportional to the change in its temperature and to its mass (i.e., more mass and more change in temperature implies more change in thermal energy). The specific heat is the proportionality constant for the substance. Using `dQ for the change in the thermal energy of an object we can express this as • `dQ = mass * specific heat * `dT. (General College and University Physics students note that most substances do not quite behave in this ideal fashion; for most substances the specific heat is not in fact strictly constant and for most substances changes with temperature.) For two objects combined in a closed system we have `dQ1 + `dQ2 = 0, which gives us the equation • m1 c1 `dT1 + m2 c2 `dT2 = 0 or equivalently • m1 c1 `dT1 = - m2 c2 `dT2. That is, whatever energy one substance loses the other gains. In this situation we know the specific heat of water, the two temperature changes and the two masses. We can therefore solve this equation for specific heat c2 of the unknown substance. ** Your Self-Critique: makes perfect sense! Your Self-Critique Rating:3 ********************************************* Question: query univ phy 17.116 (15.106 10th edition) 1.5 * 10^11 m, 1.5 kW/m^2, sun rad 6.96 * 10^8 m. How did you calculate the total radiation of the Sun and how did you use this result to get the radiation per unit area? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: Part A makes perfect sense: Part A: The power * surface_area_earth / area_from_earth_to_sun: Power * 4Pi *r^2 / (4Pi *r^2) = Power * r^2 / (r^2) This could view from a flux perspective as I*SA: 1.5 kW/m^2 * (1.5 * 10^11 m)^2 / (6.96*10^8 m)^2 = 696.7E 6 W/m^2 This is simply a ratio to determine the rate of radiation over time divided by surface area at the surface of the sun. Part B: It is clear that the transformation of energy is one due to the idea that this is an ideal blackbody! Now I know that rate of conductivity is R = k*gradient (T)*A. And that the Watts at the edge of the atmosphere = 1500 W/m^2 * 4PI* (1.5 * 10^11 m)^2 = 42.4115E 25 W. This is the Energy/sec rewritten as 42.4115E 25 J/s which is equivalent to: (4*`pi * (6.96 x 10 ^ 8 m)^2) * (1) * (5.67051 x 10^-8 W/m^2*K) * T^4 ?????????It made perfect sense to here, but I do not understand where the (5.67051 x 10^-8 W/m^2*K) * T^4 came from the algebra from here is simple resulting in the temp = T = 5934.10766 K on surface of sun. ** confidence rating #$&*: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Outline of solution strategy: If we multiply the number of watts per unit of area by the surface area of the Sun we get the number of watts radiated from the Sun. The energy flows outward in a spherically symmetric manner; at any distance the entire power is distributed over the radius of a sphere concentric with the Sun and of radius equal to the distance. So if we divide that number of watts by the area of a sphere whose radius is equal to that of the Earth’s orbit, we get the number of watts per unit of area at that distance. This strategy is followed in the student solution given below: Good student solution: Surface area of sphere of radius r is 4 pi r^2; if flux intensity is I then flux = 4 pi r^2 I. When r = 1.5 * 10^11 m, I = 1500 W / m^2, so the flux is 4 pi r^2 I = 4 pi * (1.5 * 10^11 m)^2 * 1500 W / m^2 = 4.28 * 10^26 watts. 4.28055 x 10 ^ 26 W / (4*`pi * (6.96 x 10 ^ 8 m)^2) = 4.28055 x 10 ^ 26 W / 6.08735 x 10 ^ 18 m^2 = 70318775.82 J/s/m^2 = 7.03 x 10 ^ 7 J/s/m^2 If the sun is radiating as an ideal blackbody, e = 1, then T would be found as follows: H = `dQ/`dt = 4.28055 x 10 ^ 26 W = (4*`pi * (6.96 x 10 ^ 8 m)^2) * (1) * (5.67051 x 10^-8 W/m^2*K) * T^4 So T^ 4 = 4.28055 x 10 ^ 26 W / 6.087351 x 10 ^ 18 m^2) * 1 * (5.67051 x 10^-8 W/m^2*K) T^4 = 1.240 * 10 ^ 15 K ^4 T = 5934.10766 K on surface of sun. ** Your Self-Critique: I still do not understand where the 5.67051 x 10^-8 W/m^2*K came from, other than that I get it! Your Self-Critique Rating:3
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Given Solution: ** Thermal energy is not radiating in significant quantities from the ice, so only the incoming radiation needs to be considered, and as stated only 70% of that energy is absorbed by the ice.. • 70% of the incoming 600 watts/m^2 is 420 watts / m^2, or 420 Joules/second for every square meter if ice. • Melting takes place at 0 C so there is no thermal exchange with the environment. Thus each square meter absorbs 420 Joules of energy per second. We need to consider the volume of ice corresponding to a square meter. Having found that we can determine the energy required to melt the given thickness: • A 1.2 cm thickness of ice will have a volume of .012 m^3 for every square meter of surface area; the mass will be close to 1000 kg/m^3, so there are about 12 kg of ice for every m^2 of surface (you can obtain a more accurate result by using the a more accurate density; the density of ice (which floats in water) is actually somewhat less than that of water). • It takes about 330,000 Joules to melt a kg of ice at 0 C, so to melt 12 kg requires around 4,000,000 J. At 420 Joules/sec this will require roughly 10,000 seconds, or around 3 hours. All these calculations were done mentally and are therefore approximate. You should check them yourself, using appropriately precise values of the constants, etc. ** " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: univ phy (omitted from 12th edition, but should be worked now) was 17.115 Solar radiation of intensity 600 watts / m^2 is incident on an ice sheet. The temperature above and below the ice sheet is 0 Celsius. Assuming that 70% of the radiation is absorbed at the surface of the ice, how long take to melt a layer 1.2 cm this? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution:e = about .7 => the absorbed energy = 420 J/s We can find the mass of the water by multiplying the density of water by the volume. Rho = 1000 kg/m^3; we know the depth of the ice to be 1.2 cm and we are given the rate of change of energy per area in square meters so let us just use 1 m^2 for the CSA => V = .012 m^3 => The mass of our sheet of ice is about 995 kg/m^3 *.012 m^3 = 11.94 kg To melt one kg of ice it takes roughly 330 kJ/kg [this to change from solid to liquid, not to “heat it up”] 330 kJ/kg*11.94kg = 3.94 Mega-J 3.94E 6 J = 420 J/s * `dt => `dt = 3.94E 6 J / 420 J/s = 9 381 sec /3600sec = 2.6 hours confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** Thermal energy is not radiating in significant quantities from the ice, so only the incoming radiation needs to be considered, and as stated only 70% of that energy is absorbed by the ice.. • 70% of the incoming 600 watts/m^2 is 420 watts / m^2, or 420 Joules/second for every square meter if ice. • Melting takes place at 0 C so there is no thermal exchange with the environment. Thus each square meter absorbs 420 Joules of energy per second. We need to consider the volume of ice corresponding to a square meter. Having found that we can determine the energy required to melt the given thickness: • A 1.2 cm thickness of ice will have a volume of .012 m^3 for every square meter of surface area; the mass will be close to 1000 kg/m^3, so there are about 12 kg of ice for every m^2 of surface (you can obtain a more accurate result by using the a more accurate density; the density of ice (which floats in water) is actually somewhat less than that of water). • It takes about 330,000 Joules to melt a kg of ice at 0 C, so to melt 12 kg requires around 4,000,000 J. At 420 Joules/sec this will require roughly 10,000 seconds, or around 3 hours. All these calculations were done mentally and are therefore approximate. You should check them yourself, using appropriately precise values of the constants, etc. ** " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!
#$&* course PHY 242 4/9/2011; 22:30 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 change in the thermal energy of an isolated system is 0. So assuming that the systems are isolated the thermal energy change of one object is equal and opposite to that of the other. For an ideal substance the change in the thermal energy of an object is directly proportional to the change in its temperature and to its mass (i.e., more mass and more change in temperature implies more change in thermal energy). The specific heat is the proportionality constant for the substance. Using `dQ for the change in the thermal energy of an object we can express this as • `dQ = mass * specific heat * `dT. (General College and University Physics students note that most substances do not quite behave in this ideal fashion; for most substances the specific heat is not in fact strictly constant and for most substances changes with temperature.) For two objects combined in a closed system we have `dQ1 + `dQ2 = 0, which gives us the equation • m1 c1 `dT1 + m2 c2 `dT2 = 0 or equivalently • m1 c1 `dT1 = - m2 c2 `dT2. That is, whatever energy one substance loses the other gains. In this situation we know the specific heat of water, the two temperature changes and the two masses. We can therefore solve this equation for specific heat c2 of the unknown substance. ** Your Self-Critique: makes perfect sense! Your Self-Critique Rating:3 ********************************************* Question: query univ phy 17.116 (15.106 10th edition) 1.5 * 10^11 m, 1.5 kW/m^2, sun rad 6.96 * 10^8 m. How did you calculate the total radiation of the Sun and how did you use this result to get the radiation per unit area? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: Part A makes perfect sense: Part A: The power * surface_area_earth / area_from_earth_to_sun: Power * 4Pi *r^2 / (4Pi *r^2) = Power * r^2 / (r^2) This could view from a flux perspective as I*SA: 1.5 kW/m^2 * (1.5 * 10^11 m)^2 / (6.96*10^8 m)^2 = 696.7E 6 W/m^2 This is simply a ratio to determine the rate of radiation over time divided by surface area at the surface of the sun. Part B: It is clear that the transformation of energy is one due to the idea that this is an ideal blackbody! Now I know that rate of conductivity is R = k*gradient (T)*A. And that the Watts at the edge of the atmosphere = 1500 W/m^2 * 4PI* (1.5 * 10^11 m)^2 = 42.4115E 25 W. This is the Energy/sec rewritten as 42.4115E 25 J/s which is equivalent to: (4*`pi * (6.96 x 10 ^ 8 m)^2) * (1) * (5.67051 x 10^-8 W/m^2*K) * T^4 ?????????It made perfect sense to here, but I do not understand where the (5.67051 x 10^-8 W/m^2*K) * T^4 came from the algebra from here is simple resulting in the temp = T = 5934.10766 K on surface of sun. ** confidence rating #$&*: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Outline of solution strategy: If we multiply the number of watts per unit of area by the surface area of the Sun we get the number of watts radiated from the Sun. The energy flows outward in a spherically symmetric manner; at any distance the entire power is distributed over the radius of a sphere concentric with the Sun and of radius equal to the distance. So if we divide that number of watts by the area of a sphere whose radius is equal to that of the Earth’s orbit, we get the number of watts per unit of area at that distance. This strategy is followed in the student solution given below: Good student solution: Surface area of sphere of radius r is 4 pi r^2; if flux intensity is I then flux = 4 pi r^2 I. When r = 1.5 * 10^11 m, I = 1500 W / m^2, so the flux is 4 pi r^2 I = 4 pi * (1.5 * 10^11 m)^2 * 1500 W / m^2 = 4.28 * 10^26 watts. 4.28055 x 10 ^ 26 W / (4*`pi * (6.96 x 10 ^ 8 m)^2) = 4.28055 x 10 ^ 26 W / 6.08735 x 10 ^ 18 m^2 = 70318775.82 J/s/m^2 = 7.03 x 10 ^ 7 J/s/m^2 If the sun is radiating as an ideal blackbody, e = 1, then T would be found as follows: H = `dQ/`dt = 4.28055 x 10 ^ 26 W = (4*`pi * (6.96 x 10 ^ 8 m)^2) * (1) * (5.67051 x 10^-8 W/m^2*K) * T^4 So T^ 4 = 4.28055 x 10 ^ 26 W / 6.087351 x 10 ^ 18 m^2) * 1 * (5.67051 x 10^-8 W/m^2*K) T^4 = 1.240 * 10 ^ 15 K ^4 T = 5934.10766 K on surface of sun. ** Your Self-Critique: I still do not understand where the 5.67051 x 10^-8 W/m^2*K came from, other than that I get it! Your Self-Critique Rating:3 ********************************************* Question: univ phy (omitted from 12th edition, but should be worked now) was 17.115 Solar radiation of intensity 600 watts / m^2 is incident on an ice sheet. The temperature above and below the ice sheet is 0 Celsius. Assuming that 70% of the radiation is absorbed at the surface of the ice, how long take to melt a layer 1.2 cm this? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution:e = about .7 => the absorbed energy = 420 J/s We can find the mass of the water by multiplying the density of water by the volume. Rho = 1000 kg/m^3; we know the depth of the ice to be 1.2 cm and we are given the rate of change of energy per area in square meters so let us just use 1 m^2 for the CSA => V = .012 m^3 => The mass of our sheet of ice is about 995 kg/m^3 *.012 m^3 = 11.94 kg To melt one kg of ice it takes roughly 330 kJ/kg [this to change from solid to liquid, not to “heat it up”] 330 kJ/kg*11.94kg = 3.94 Mega-J 3.94E 6 J = 420 J/s * `dt => `dt = 3.94E 6 J / 420 J/s = 9 381 sec /3600sec = 2.6 hours confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** Thermal energy is not radiating in significant quantities from the ice, so only the incoming radiation needs to be considered, and as stated only 70% of that energy is absorbed by the ice.. • 70% of the incoming 600 watts/m^2 is 420 watts / m^2, or 420 Joules/second for every square meter if ice. • Melting takes place at 0 C so there is no thermal exchange with the environment. Thus each square meter absorbs 420 Joules of energy per second. We need to consider the volume of ice corresponding to a square meter. Having found that we can determine the energy required to melt the given thickness: • A 1.2 cm thickness of ice will have a volume of .012 m^3 for every square meter of surface area; the mass will be close to 1000 kg/m^3, so there are about 12 kg of ice for every m^2 of surface (you can obtain a more accurate result by using the a more accurate density; the density of ice (which floats in water) is actually somewhat less than that of water). • It takes about 330,000 Joules to melt a kg of ice at 0 C, so to melt 12 kg requires around 4,000,000 J. At 420 Joules/sec this will require roughly 10,000 seconds, or around 3 hours. All these calculations were done mentally and are therefore approximate. You should check them yourself, using appropriately precise values of the constants, etc. ** " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: univ phy (omitted from 12th edition, but should be worked now) was 17.115 Solar radiation of intensity 600 watts / m^2 is incident on an ice sheet. The temperature above and below the ice sheet is 0 Celsius. Assuming that 70% of the radiation is absorbed at the surface of the ice, how long take to melt a layer 1.2 cm this? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution:e = about .7 => the absorbed energy = 420 J/s We can find the mass of the water by multiplying the density of water by the volume. Rho = 1000 kg/m^3; we know the depth of the ice to be 1.2 cm and we are given the rate of change of energy per area in square meters so let us just use 1 m^2 for the CSA => V = .012 m^3 => The mass of our sheet of ice is about 995 kg/m^3 *.012 m^3 = 11.94 kg To melt one kg of ice it takes roughly 330 kJ/kg [this to change from solid to liquid, not to “heat it up”] 330 kJ/kg*11.94kg = 3.94 Mega-J 3.94E 6 J = 420 J/s * `dt => `dt = 3.94E 6 J / 420 J/s = 9 381 sec /3600sec = 2.6 hours confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** Thermal energy is not radiating in significant quantities from the ice, so only the incoming radiation needs to be considered, and as stated only 70% of that energy is absorbed by the ice.. • 70% of the incoming 600 watts/m^2 is 420 watts / m^2, or 420 Joules/second for every square meter if ice. • Melting takes place at 0 C so there is no thermal exchange with the environment. Thus each square meter absorbs 420 Joules of energy per second. We need to consider the volume of ice corresponding to a square meter. Having found that we can determine the energy required to melt the given thickness: • A 1.2 cm thickness of ice will have a volume of .012 m^3 for every square meter of surface area; the mass will be close to 1000 kg/m^3, so there are about 12 kg of ice for every m^2 of surface (you can obtain a more accurate result by using the a more accurate density; the density of ice (which floats in water) is actually somewhat less than that of water). • It takes about 330,000 Joules to melt a kg of ice at 0 C, so to melt 12 kg requires around 4,000,000 J. At 420 Joules/sec this will require roughly 10,000 seconds, or around 3 hours. All these calculations were done mentally and are therefore approximate. You should check them yourself, using appropriately precise values of the constants, etc. ** " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&! ********************************************* Question: univ phy (omitted from 12th edition, but should be worked now) was 17.115 Solar radiation of intensity 600 watts / m^2 is incident on an ice sheet. The temperature above and below the ice sheet is 0 Celsius. Assuming that 70% of the radiation is absorbed at the surface of the ice, how long take to melt a layer 1.2 cm this? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution:e = about .7 => the absorbed energy = 420 J/s We can find the mass of the water by multiplying the density of water by the volume. Rho = 1000 kg/m^3; we know the depth of the ice to be 1.2 cm and we are given the rate of change of energy per area in square meters so let us just use 1 m^2 for the CSA => V = .012 m^3 => The mass of our sheet of ice is about 995 kg/m^3 *.012 m^3 = 11.94 kg To melt one kg of ice it takes roughly 330 kJ/kg [this to change from solid to liquid, not to “heat it up”] 330 kJ/kg*11.94kg = 3.94 Mega-J 3.94E 6 J = 420 J/s * `dt => `dt = 3.94E 6 J / 420 J/s = 9 381 sec /3600sec = 2.6 hours confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** Thermal energy is not radiating in significant quantities from the ice, so only the incoming radiation needs to be considered, and as stated only 70% of that energy is absorbed by the ice.. • 70% of the incoming 600 watts/m^2 is 420 watts / m^2, or 420 Joules/second for every square meter if ice. • Melting takes place at 0 C so there is no thermal exchange with the environment. Thus each square meter absorbs 420 Joules of energy per second. We need to consider the volume of ice corresponding to a square meter. Having found that we can determine the energy required to melt the given thickness: • A 1.2 cm thickness of ice will have a volume of .012 m^3 for every square meter of surface area; the mass will be close to 1000 kg/m^3, so there are about 12 kg of ice for every m^2 of surface (you can obtain a more accurate result by using the a more accurate density; the density of ice (which floats in water) is actually somewhat less than that of water). • It takes about 330,000 Joules to melt a kg of ice at 0 C, so to melt 12 kg requires around 4,000,000 J. At 420 Joules/sec this will require roughly 10,000 seconds, or around 3 hours. All these calculations were done mentally and are therefore approximate. You should check them yourself, using appropriately precise values of the constants, etc. ** " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!#*&!