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Earth $ surface absorbs a averaqe about 960 W/m? from the Sun'$ Irradlance The power absorbed (960 WImf 2)(Adsch wnere Aasc ARe Is Earth $ projected 4n equa am...

Question

Earth $ surface absorbs a averaqe about 960 W/m? from the Sun'$ Irradlance The power absorbed (960 WImf 2)(Adsch wnere Aasc ARe Is Earth $ projected 4n equa amaunt power radiated 50 that Earth remains thermal equilibnum with its environment at nearly Estimate Earth'$ surface temperature by setting the radiated power from Stefan' Iawa equal absorded power ano solving for the temperature Kelvir Stefan's Iaw, aesume take the area 4RE the surface area spherical Earth; (Note: Eart

Earth $ surface absorbs a averaqe about 960 W/m? from the Sun'$ Irradlance The power absorbed (960 WImf 2)(Adsch wnere Aasc ARe Is Earth $ projected 4n equa amaunt power radiated 50 that Earth remains thermal equilibnum with its environment at nearly Estimate Earth'$ surface temperature by setting the radiated power from Stefan' Iawa equal absorded power ano solving for the temperature Kelvir Stefan's Iaw, aesume take the area 4RE the surface area spherical Earth; (Note: Earth's atmosphere acts like blanket and warms the planet to global averade about 30 adove the value calculated here:)



Answers

Earth's surface absorbs an average of about $960 . \mathrm{W} / \mathrm{m}^{2}$ from the Sun's irradiance. The power absorbed is $P_{\mathrm{abs}}=\left(960 . \mathrm{W} / \mathrm{m}^{2}\right)$ $\left(A_{\mathrm{disc}}\right),$ where $A_{\mathrm{disc}}=\pi R_{E}^{2}$ is Earth's projected area. An equal amount of power is radiated so that Earth remains in thermal equilibrium with its environment at nearly 0 K. Estimate Earth's surface temperature by setting the radiated power from Stefan's law equal to the absorbed power and solving for the temperature in Kelvin. In Stefan's law, assume $e=1$ and take the area to be $A=4 \pi R_{E}^{2}$ , the surface area of a spherical Earth. (Note: Earth's atmosphere acts like a blanket and warms the planet to a global average about 30 $\mathrm{K}$ above the value calculated here.)

This exercise. We're going to talk about this stuff on both One constant. Remember that this stuff in Baltimore council talks about black body radiation, black body radiation and how we calculate the power that's radiated by a black body. So the power p is equal to are suffering waterman constant sigma times the area of the black body times the temperature of the black body to the fourth, where seven bowls. One constant sigma is 5.67 times sent to the minors. Eight watts per meter square, Calvin to the fourth. And what we have to do is to compare the power that is observed, absorbed by the earth coming from the sun. I'm gonna call it P s and the power that's emitted by the Earth. I'm gonna call it be due to the fact that it's a bad body, considering that the temperature of the Earth the average temperature is 15 degrees Celsius. So first, let's calculate the power that received by the Earth from the sun so he would have a son. And here we have the earth. The distance between the sun and the Earth D is equal to 1.5 times 10 to the 11th meters. Okay, let's go. Okay, So the intensity off the the sun radiation is equal to the power emitted by the sun divided by the area of a sphere that has a radius equal to the distance between the sun and the earth. This is the intensity of life at leery. Yeah. Okay, so the power p I'm going to write it down here. The power emitted by the sun is sigma times, the area of the sun times the temperature of listening to the fourth. And in order to calculate it, we're gonna need the following. Constance, you need the radius of the sun, which is 6.96 times 10 to the 8 m and the temperature of the surface of the sun, which is 5800 Calvin. So the power radiated by the sun is 5.67 times 10 to the minus eight watts meters squared Calvin to the fourth times, four pi times, radius of the sun squared 6.96. I'm 78th meters squared times the temperature which is five. 5800 Calvin to the four. So the power radiated by the sun is 4.4 times 10 to 26 watts. Okay. Eso The intensity of light is equal to the power divided by the area, which is four pi d squared. D is the distance between the earth and, uh, the we observe it in the earth and, uh, listen, and this is equal to the power received by the earth. That's what I'm calling PS, divided by the area of the earth that received receives, um, solar radiation. And that's half the total area of the Earth because the sun only shines on half the earth at a time. So it's two pi times The radius of the earth squared, so PS is equal to p, divided by B times, the radius of the earth squared. I'm sorry. Divided by two times the square. So P is equal to 4.4 times 10 to the 26 watts. The rate is of the earth is 6.38 times 10 to the sixth meters and went to divided by two times 1.5 times 10 to the 11th meters. So the power coming from the sun, that which is the earth, is four point. Actually, I'm sorry. This is actually a total power. The power that reaches the earth is 3.65 times 10 to the 17. What? This is a power coming from the sun. Now let's calculate the power emitted by the earth due to the fact that it's a black body. So the power emitted from the earth is sigma times. The area of the earth which is 45 times the radius of the earth, squared times the temperature of the earth to the fourth. So this is 5.67 times 10 to the minus 11 star, minus eight lots per meter squared kelvin to the fourth times four pi times the radius of the earth squared that 6.38 times 10 to the sixth meters squared times the temperature of the earth, the temperature It is 15 degrees Celsius and I have to some 273 Calvin in order to obtain 288 Kelly. So this is the temperature of the earth. So the power is three. Actually, I'm sorry. Power is 2.4 times 10 to the 12 watts. So notice that PS, divided by P, is of the order of magnitude of 10 to the fifth. So the power that the Earth receives from the sun is about 10 to 5 times greater than the power that knee earth emit in the form of black body radiation it

If we have the art. And here we have all the atmosphere the Sunday coming in gets reflected, are scattered or radiated by the art, and then it gets recited by the atmosphere it and comes back. This is called greenhouse gas, and the skips the art warm. So without the atmosphere, the art will get colder and colder without the atmosphere. The an attitude that is coming in with be radiated out. So basically, this is the read off a mission which becomes, it was, do the rate off absorption. So right off a mission becomes close to rate of absolution means nothing is getting absorb, and that is why option A is the correct answer.

Hi there. So for this problem, we are told that the average rage at which the heat flows out through the surface of the earth is about you. So that value is the heat over the story. Um and that value is 54 million wops hurt meter squared. And the average thermal conductivity that we do not take with the letter T is equal to 2.5 whoops per meter per kelvin. And assuming a surfaced Imparato award, that is equal to 10 cells. Use degrees. Why should be the temperature at a death? That is equal to this change in the death is well, we call this change in the death is equal to 33 kilometers. No, what we need to obtain from this. Its first the change and the temperature or And with that value we can obtain the final temperature tour. So we're going to start with the equation for the heat. We are told that the heat transfer is equal to Indeed, thermal conductivity ok. Times the arian times the change in temperature or over the change in this case of the death now. And the rate at which the heat flows out is given as a power per story. Um So the quantity given is precisely age over a. That is the value given in here This one right here. And the temperature difference we just need to solve from this equation. We will find quite easily and that is equal to age over a heat over the area times the change in the death of over the thermal conductivity. So we just need to simply substitute all of these values in here. So we will have that this is equal to the These terms right here. Age over a which has about you of 0.0 54 webs per meter square. We transform this because the value that we are given is in millie and we know that millie it's 10 to the minus three now and the change in the death is 33 kilometers. But we know that kilo means tens to the three so we will obtain 33,000 m. And this over the The thermal conductivity that in this case is 2.5 what's per mature percolating. So from this using our calculator we obtained about you off seven 110 Kelvin. Now we know that this is the change in the temperature for And we know that the changes in that territory is defined as the final temperature minus the initial temperature. Or we know that initial temperature is 10 Celsius degrees. So we can transform that To obtain the final temperature or we must do to the value that we obtain plus the initial value. So we need to transform this into Kelvin. We know that we need just to add to that value 273. So we'll obtain that this is 200 and 83 killed in So the final temperature or is the one that we just have obtained which is 710 Kelvin plus 283 kelvin. So from this, we obtained upon you off 90 993 kill me. So this is a solution for this problem. Thank.

So we're gonna suppose that we used to cut a cover an area and we have 1.7 meter by a 0.3 meter section of beach blanket. But the elevation angle of the sun is 60 degrees. So we can say that the target area would then be equal in 1.7 meters times 0.3 meters and this would be just average dimension ends of a any beach blanket and then times co sign of 30 degrees. This is giving us 0.4 square meters. And so the intensity of the radiation at Earth's surface we could say intensity at the surface would be equaling 0.6 or 60% of the incoming intensity. And on Lee, 50% of this is absorbed. And so we can say that here, we know that the current is simply the average. So we're gonna suppose that we used to cut a cover an area and we have 1.7 meter by a 0.3 meter section of beach blanket. But the elevation angle of the sun is 60 degrees, so we can say that the target area would then be equal in 1.7 meters times 0.3 meters and this would be just average dimensions or there occurred. That intensity would be the average power divided by the area. And so this would be equaling the change in the energy with respect to time essentially power multiplied by a that area. This would be equal in the intensity, so the absorbed energy would be equaling 0.5 times the intensity at the surface multiplied by the area multiplied by doubt, the tea. And so this would be equaling points of a any beach blanket and then times co sign of 30 degrees. This is giving us 0.4 square meters, and so the intensity of the radiation at Earth's surface we could say intensity at the surface would be equaling 0.6 or 60% of the incoming intensity. And on Lee, 50% of this is absorbed. And so we can say that here we know that the current is simply the average five times 0.6 times I incoming multiplied by a delta T and so we can solve. The energy absorbed would then be equaling 0.5 times 0.6 times 1370 watts per square meter multiplied by the area of 0.4 square meters. There occurred that intensity would be the average power divided by the area. And so this would be equaling the change in the energy with respect to time essentially power multiplied by a that area, this would be equal in the intensity. So the absorbed energy would be equaling 0.5 times the intensity at the surface multiplied by the area multiplied by Delta T. And so this would be equaling point multiplied by t of an hour or 3600 seconds. And this is giving us six times 10 to the fifth jewels. So this would be our answer. That is the end of the solution. Thank you for watching.


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