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The Sun radiates like a perfect black body with an emissivity of exactly 1 . (a) Calculate the surface temperature of the Sun, given that it is a sphere with a $7.00 \times 10^{8}-\mathrm{m}$radius that radiates $3.80 \times 10^{26} \mathrm{W}$ into 3 -K space. (b) How much power does the Sun radiate per square meter of its surface? (c) How much power in watts per square meter isthat value at the distance of Earth, $1.50 \times 10^{11} \mathrm{m}$ away? (This number is called the solar constant.)

Part $(\mathrm{a}) : T=5761^{\circ} \mathrm{K}$Part $(\mathrm{b}) : \frac{P_{w}}{A}=\frac{Q / t}{A}=6.17 \times 10^{7} \mathrm{W} / \mathrm{m}^{2}$Part $(\mathrm{c}) : \frac{P_{0}}{A}(\text { at a distance d } \mathrm{d})=2.74 \times 10^{-15} \mathrm{W} / \mathrm{m}^{2}$

Physics 101 Mechanics

Chapter 14

Heat and Heat Transfer Methods

Thermal Properties of Matter

Simon Fraser University

University of Sheffield

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number 62. It's about the sun. Um, we're given that the radius of the sun is seven times 10 to the commuters, and we're told that it radiates heat with this at this rate. So basically of power come and the temperature spaces three Kelvin and whereas to find was the temperature of the surface, well, this is the rate of which is the radiance. It's not the net rate, and it would be about the same anyway, because this is close to zero. That's pretty much inexcusable. So that's why I don't have to do the thing where I subtract the temperatures. I'm just going to use this equation. Um, so the heat per time is the read radiating. So that's this and wants Ah, bolts Mons constant. That's always the same. 5.67 times, 10 to the negative. We're told that it's a perfect black box radiator. So my impulsivity is one area I'm giving the radius. Um, so that's the surface area. Surface area of sphere is for pi r squared. So for pie, let me give him my radius seven times in the eighth like it's screwed times. My temperature raised the fourth car, and that's what I'm solving for. So from here it's just a calculator. I'm taking this. I'm dividing by all of this. And then I am taking the fourth root of that. I think the best way to do that in the calculator, then, is a few rays that to the fourth power, you know, take the fourth root of it. It's the same as raising to the fourth power. So after divide all this, I use that button, um X to the white button on the calculator and I d'oh, my answer raised to the 0.25 and I get what my temperature is. 5744 and that would be in Kelvin, Part B. They wouldn't know what is the power per square meter at the surface of the sun Here at the surface of the sun. I want exactly that. I want the power per square meter, said power per area. I was given my power 3.8 times, 10 to 26 divided by this Venus Surface area of the Sun, just like I figured it out up here. So for pie times seven times 10 to the eighth squared for that, I get 6.17 times, 10 to the seventh. And that was in Watts over meter squared. What? Her meter squared and then part See, they won't know what is the power per square meter when you're 1.5 times in the 11th meters away. So that's the distance of the earth is away. So just imagine, Now you know the surface is out here. So power per square area power still saying so. Same amount of energy per second coming out of out of the sun here. So that's still 3.8 times 10 to the 26. Now it's divided by surface area, as if this was this year. So for pie, my radius is gonna be that 1.50 times 10 of the 11 and like it squared. And for this leg only 1344. What per meter squared

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