#$&* course Phy 232 Question: query intro set problem 14 substance, water, both temperatures and masses known, final temperature known, find spec ht
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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: OK Your Self-Critique Rating: OK ********************************************* 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: Rate energy from sun reaches earth is 1.50 kW/m^2. Distance from sun to earth = 1.50*10^11. Radius of sun = 6.96 * 10^8 m. We can calculate the number of watts being emitted from the sun by multiplying the rate energy by the surface area of the sun. Thus, (4*pi*r^2)*(1.50 kW/m^2) = kW (4*pi*(1.5*10^11)^2)*(1.50 kW/m^2) = 4.24*10^23 kW = 4.24*10^26 W. 4.24115*10 ^ 26 W / (4*`pi * (6.96 x 10 ^ 8 m)^2) = 4.24115*10^26/6087345845760000000 = 69671579.4939 watts/m^2 = 6.967*10^7 Watts/m^2 I do not understand how to solve for the temperature of the surface. I’m sure you use dQ somehow, but I don’t know how. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: After looking at the solution, I am confused as to what they did to solve for the temperature. Why is it raised to the 4th power in the equation? And where did the value, 5.67051 x 10^-8 W/m^2*K come from? Your Self-Critique Rating: 1
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Given Solution: .012 ** 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. ** Your Self-Critique: I used more accurate numbers so that is why my answer is different than that of the solutions. Your Self-Critique Rating: 3 " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!