pavelcherepan Posted April 5, 2015 Share Posted April 5, 2015 (edited) Let's assume that you have some zircon and we estimate that the age is 4.5 My. What does it mean? It just give us an indication for the min age. We have no guarantee that the minerals that we have are the oldest ever made on earth or in the universe. I believe you meant 4.5 by, not my, right? That's why we use meteorites. They are pretty much unchanged from the time of their formation and they all agree on the same age as I've shown in the quote in the previous post. Also your calculation contradicts basic cosmology/astronomy as with that kind of age the Earth must've formed around Population III or Population II star which all have very-very low metallicity and are supposed to be huge and die quickly within millions of years none of which we're observing. Also there wouldn't be enough silicon, iron and oxygen to form a rocky planet at that time. How much contradictory evidence do you need to finally agree that your estimate is wrong-wrong-wrong? Edited April 5, 2015 by pavelcherepan Link to comment Share on other sites More sharing options...
Robittybob1 Posted April 5, 2015 Share Posted April 5, 2015 OK, then. You are massively overthinking the problem I posed. All one needs to do is compare the incoming solar radiation with the radiated power of a blackbody at 1000 K. The context here was whether we could ignore solar radiation. A concept that has been inexplicably missing one one side of this discussion is when terms in equations are important or not. Since I'm getting frustrated that the reticence to do fairly simple analysis, I'll show the numbers myself. We receive about 1.3 kW/m2 from the sun if it's directly overhead, so the average of that over the earth is about 325 W/m2. A blackbody at 1000 K will be radiating 56.7 kW/m2. That's a pretty straightforward calculation. It should not be melting anyone's brain who is interested in discussing science. It's been pointed out that the earth is not a blackbody, so that reduces the emitted radiation, but it also reduces the amount of incoming radiation that's absorbed. So any way you want to slice it, you can ignore the effect of the sun while the earth is very hot. The solar radiation begins to be important at about 500 K (that's when incoming solar is several percent of the radiated power) Remember, in the big picture of this thread, the original scenario was 6000 K, and the question was whether cooling to the point that we could form oceans could occur in around 100 my. Solar radiation was negligible for the initial rapid cooling of such a molten proto-earth. Emissivity is not the problem that it's being portrayed as. Half of a really big number is still a really big number. It doesn't matter if the calculation is "correct" (i.e. did the math right) if the overall analysis is flawed. A proper analysis needs to account for all sources of heating and cooling and all forms of heat transfer. You originally asked a very general question — how the earth could cool in 100 my — and were given a conceptual answer. Over the course of this very long thread you have been given various explanations by a number of people of other details that would have to go into a more detailed analysis, most of which are missing from your "correct" calculation. Can we have a runaway greenhouse gas effect? A runaway greenhouse effect is a process in which a net positive feedback between surface temperature and atmospheric opacity increases the strength of the greenhouse effect on a planet until its oceans boil away.[1][2] An example of this is believed to have happened in the early history of Venus...... Positive feedbacks do not have to lead to a runaway effect, as the gain is not always sufficient. A strong negative feedback always exists (radiation from a planet increases in proportion to the fourth power of temperature, in accordance with the Stefan-Boltzmann law) so the positive feedback effect has to be very strong to cause a runaway effect (see gain). An increase in temperature from greenhouse gases leading to increased water vapor (which is itself a greenhouse gas) causing further warming is a positive feedback, but not a runaway effect, on Earth. So even though the power from the Sun is less compared to the radiation from land surface, the atmospheric gases act as a blanket and the incoming heat gets trapped, and hence hinders radiation and the whole situation just gets hotter and hotter. Radiation from the land is reflected back down but the incoming radiation is absorbed by the gas. This type of situation would resist cooling but did it happen? Link to comment Share on other sites More sharing options...
David Levy Posted April 5, 2015 Author Share Posted April 5, 2015 (edited) Also your calculation contradicts basic cosmology/astronomy as with that kind of age the Earth must've formed around Population III or Population II star which all have very-very low metallicity and are supposed to be huge and die quickly within millions of years none of which we're observing. Also there wouldn't be enough silicon, iron and oxygen to form a rocky planet at that time. How much contradictory evidence do you need to finally agree that your estimate is wrong-wrong-wrong? There is no contradiction with any evidence. None of the above is considered as a proved theory. Therefore, this discovery could potentially contradict with some unproved cosmology/astronomy concepts. Edited April 5, 2015 by David Levy Link to comment Share on other sites More sharing options...
pavelcherepan Posted April 5, 2015 Share Posted April 5, 2015 There is no contradiction with any evidence. None of the above is considered as a proved theory. Therefore, this discovery could potentially contradict with some unproved cosmology/astronomy concepts. Is that so? I think that its in fact quite well established and studied based on spectral data. <Metallicity> Link to comment Share on other sites More sharing options...
David Levy Posted April 5, 2015 Author Share Posted April 5, 2015 (edited) Is that so? I think that its in fact quite well established and studied based on spectral data. <Metallicity> Let me explain you clearly. There were two clever scientists that developed a formula for heat radiation. This formula is called - Stefan-Boltzmann Law http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/stefan.html#c2 In this formula they have added the emissivity factor. If e = 1 then it is consider as ideal radiator or black body object. The science had fully accepted their formula. This formula had been used to develop the frame calculation for Radiative Cooling Time http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/cootime.html#c1 The science used that frame to calculate the requested time for Cooling the Earth to 300K. http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/cootime2.html#c1 Unfortunately, with or without attention, it was assumed that the Earth is an ideal radiator (or blackbody). The science ignores completely the yellow star in the sky, the greenhouse effect and the atmosphere. Based on that assumption – it was proved that the Earth can be cool down from any temp to 300K in just 30 My. So far the science was quite proud with this formula as it meets the main stream concept. Unfortunately, (for some scientists) it is clear now that the Earth shouldn't be considered as a black body. Based on a very simple calculation it is proved that the age of the Earth is much higher than the current expectation. In the past it was believed that the Earth is the center of the Universe. Later on it was discovered that this theory was incorrect. So what? Now we understand that the age of the Earth is longer than our current expectation. So what? With regards to Metallicity or any other unproved theory. The science should be proud with this discovery. It doesn't contradict with any real evidence or observation. In the contrary, it opens a new perspective on the Universe. Please be happy. We are facing a new dawn for the science. Edited April 5, 2015 by David Levy Link to comment Share on other sites More sharing options...
Robittybob1 Posted April 5, 2015 Share Posted April 5, 2015 So remember the Earth is forming and probably without a magnetic field so the greenhouse effect is being reduced all the time by the rapid loss of the atmosphere. So how much longer do you think the cooling was. You don't want to stick to those ridiculous figures in the OP do you? Link to comment Share on other sites More sharing options...
swansont Posted April 5, 2015 Share Posted April 5, 2015 Can we have a runaway greenhouse gas effect? Not with our atmosphere. So even though the power from the Sun is less compared to the radiation from land surface, the atmospheric gases act as a blanket and the incoming heat gets trapped, and hence hinders radiation and the whole situation just gets hotter and hotter. Radiation from the land is reflected back down but the incoming radiation is absorbed by the gas. This type of situation would resist cooling but did it happen? It gets hotter, raising the radiation rate. Rapidly, as the power depends on T4. Let me explain you clearly. There were two clever scientists that developed a formula for heat radiation. This formula is called - Stefan-Boltzmann Law http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/stefan.html#c2 In this formula they have added the emissivity factor. If e = 1 then it is consider as ideal radiator or black body object. The science had fully accepted their formula. "The science" also knows how to properly apply the various equations involved, putting "the science" a leg up on you. This formula had been used to develop the frame calculation for Radiative Cooling Time http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/cootime.html#c1 The science used that frame to calculate the requested time for Cooling the Earth to 300K. http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/cootime2.html#c1 Unfortunately, with or without attention, it was assumed that the Earth is an ideal radiator (or blackbody). The science ignores completely the yellow star in the sky, the greenhouse effect and the atmosphere.[/size] Based on that assumption – it was proved that the Earth can be cool down from any temp to 300K in just 30 My. "The science" here was Kelvin, and we know Kelvin was wrong. That was pointed out to you in the second post in this thread. As in, the very first response to you. You haven't discovered anything new here. "New to you" does not mean new to science. The yellow star does not affect the degree to which the earth can be considered a blackbody. Incoming radiation is part of the equation. We just ran a little exercise (well, I did, because you did not or could not complete the exercise) showing just how (un)important the sun is. The greenhouse effect is an added part of an analysis, as would be the internal heating from radioactive decay. Also, one thing you have apparently missed is that the analysis presented at hyperphysics assumes that the earth completely cools to a uniform temperature. In that sense the Kelvin analysis would represent an overestimate, since you can have a hotter interior (as we do) with conduction and convection from the interior, and that means less energy radiated, which (all else being equal) takes less time. (all else isn't equal, but that's got to be part of a more detailed analysis) So far the science was quite proud with this formula as it meets the main stream concept. Unfortunately, (for some scientists) it is clear now that the Earth shouldn't be considered as a black body. Based on a very simple calculation it is proved that the age of the Earth is much higher than the current expectation. It's been clear for a long, long time. And your very simple calculation is too simple. In the past it was believed that the Earth is the center of the Universe. Later on it was discovered that this theory was incorrect. So what? Now we understand that the age of the Earth is longer than our current expectation. So what? With regards to Metallicity or any other unproved theory. The science should be proud with this discovery. It doesn't contradict with any real evidence or observation. In the contrary, it opens a new perspective on the Universe. Please be happy. We are facing a new dawn for the science. You haven't discovered anything. Discovery means being the first to find something out. You aren't anywhere close to being first to learn about thermodynamics. The sad fact is you've absorbed only a small fraction of the information that people have passed on to you in this thread. I think you need to stop proclaiming "discovery" whenever you finally understand something a little better. And certainly, IMO, drop the "let me explain this to you" attitude. If this were a course on thermodynamics, you'd be getting a failing grade thus far. Link to comment Share on other sites More sharing options...
Mordred Posted April 5, 2015 Share Posted April 5, 2015 (edited) Dear Mordred You are a best teacher. I consider you as my partner. You have significant part of this discovery. I have learned many aspects of modern science from you. Therefore, I have some sort of disappointment from your feedback. I would expect that you would verify the calculation, look at the formulas and ask me about blackbody or graybody. Instead, it seems that you are unhappy as this discovery contradicts some of the current unproved theories. So let's try to distinguish between theory, reality and none relevant issue: Theories It isn't preception.It isn't approximation. It is a real verification. Please read Stefan-Boltzmann Law. I have used the formula exactly as it should be! http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/stefan.html#c2 Let's assume that you have some zircon and we estimate that the age is 4.5 My. What does it mean? It just give us an indication for the min age. We have no guarantee that the minerals that we have are the oldest ever made on earth or in the universe. I have High appriciation for all the scientists. But in this case, it is expected to verify the formula and the calculation. . Thanks for the compliment. All formulas are approximations. The Stefan Boltzmann formula included. Try to stop and ask yourself " where did the material come from to form the Earth?" The answer should be those meteorites. So given that information, then ask yourself in regards to Stefan Boltzmann equation. What is the radiation area while the Earth is forming? During its initial formation. The diameter of the Earth isn't the same as today. Nor is the precollision diameter (Theia) the same after the collision. Then you also need to factor in different the surrounding temperature. Today the Universe is 2.7 Kelvin. However 4 billion years ago it is slightly higher. This is another factor. (Though a minor one). Then you also need to consider time of convection, the surface cools faster than the material beneath the surface, it takes time for the heat to rise to the surface. These two points should be sufficient to show that the Stefan Boltzmann law would at best be an approximation, in this application. This is why one must look for observational evidence and empherical evidence. The other posters covered the other problems. Here is a paper showing Stefan Bolzmann law. The method used by Kelvin, Newton and Perry. Should show some of the problems with each method http://www.arthurstinner.com/stinner/pdfs/2002-ageoftheearth.pdf Here is a decent paper covering several aspects of the Earths age via meteorites, and mineral isotopes dating. http://www.google.ca/url?sa=t&source=web&cd=20&ved=0CD0QFjAJOAo&url=http%3A%2F%2Fwww.researchgate.net%2Fprofile%2FYouxue_Zhang%2Fpublication%2F221947444_The_age_and_accretion_of_the_earth%2Flinks%2F53fbfc120cf2dca8fffee686.pdf&rct=j&q=age%20of%20the%20Earth%20pdf&ei=qZchVfb-NciesAWhuYC4BQ&usg=AFQjCNEFgVbFdqARXb1IMrxOLrjZcgkvYQ&sig2=qOJQ2nz9nnBMj1gl8cf0ww Edited April 5, 2015 by Mordred Link to comment Share on other sites More sharing options...
David Levy Posted April 6, 2015 Author Share Posted April 6, 2015 (edited) All formulas are approximations. The Stefan Boltzmann formula included. Is it real? Stefan Boltzmann is not just one more formula. It isn't a hypothetical idea or theory It's a Law. http://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_law "The Stefan–Boltzmann law, also known as Stefan's law, describes the power radiated from a black body in terms of its temperature. Specifically, the Stefan–Boltzmann law states that the total energy radiated per unit surface area of a black body across all wavelengths per unit time (also known as the black-body radiant exitance or emissive power)," You are the master. How can we claim that Stefan–Boltzmann law is approximation? Shall we also claim that Newton law is approximation? Please – let's set a constructive discussion. Can we use this law? Yes or no Edited April 6, 2015 by David Levy Link to comment Share on other sites More sharing options...
pavelcherepan Posted April 6, 2015 Share Posted April 6, 2015 (edited) Is it real? Stefan Boltzmann is not just one more formula. It isn't a hypothetical idea or theory It's a Law. http://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_law "The StefanBoltzmann law, also known as Stefan's law, describes the power radiated from a black body in terms of its temperature. Specifically, the StefanBoltzmann law states that the total energy radiated per unit surface area of a black body across all wavelengths per unit time (also known as the black-body radiant exitance or emissive power)," You are the master. How can we claim that StefanBoltzmann law is approximation? Shall we also claim that that Newton law is approximation? Please let's set a constructive discussion. Can we use this law? Yes or no Yes, both Newton's laws and Stefan-Boltzman laws are approximations. Doesn't matter if they are called laws, those are still theories. The meaning of "law" is different in science compared to justice. Edited April 6, 2015 by pavelcherepan 2 Link to comment Share on other sites More sharing options...
Klaynos Posted April 6, 2015 Share Posted April 6, 2015 Newtons laws are approximations. In physics, laws are subsets of theories. Theories are the pinnacle of physical endeavour they are well tested and part of that means understanding their domain of applicability, assumptions and limitations. Link to comment Share on other sites More sharing options...
Robittybob1 Posted April 6, 2015 Share Posted April 6, 2015 Yes, both Newton's laws and Stefan-Boltzman laws are approximations. Doesn't matter if they are called laws, those are still theories. The meaning of "law" is different in science compared to justice. Are you sure they are approximations? An example please? Link to comment Share on other sites More sharing options...
pavelcherepan Posted April 6, 2015 Share Posted April 6, 2015 Newton's laws fail to predict properly when velocities are a appreciable fraction of c, that's why it's been largely superseeded by Special relativity. Link to comment Share on other sites More sharing options...
Mordred Posted April 6, 2015 Share Posted April 6, 2015 Is it real? Stefan Boltzmann is not just one more formula. It isn't a hypothetical idea or theory It's a Law. http://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_law "The StefanBoltzmann law, also known as Stefan's law, describes the power radiated from a black body in terms of its temperature. Specifically, the StefanBoltzmann law states that the total energy radiated per unit surface area of a black body across all wavelengths per unit time (also known as the black-body radiant exitance or emissive power)," You are the master. How can we claim that StefanBoltzmann law is approximation? Shall we also claim that Newton law is approximation? Please let's set a constructive discussion. Can we use this law? Yes or no The Stefan Boltzmann law works well in the correct applications. However in terms of the system your modelling several factors change. Surface area, convection rates as the crust develops greater depth, this influences emmisivity rates, your also dealing with different materials with different conductivity values. Iron has a different conductivity than say silicon etc. As the Earth goes through its development stages the system changes. Link to comment Share on other sites More sharing options...
swansont Posted April 6, 2015 Share Posted April 6, 2015 Please – let's set a constructive discussion. Can we use this law? Yes or no Yes, we can use it, as long as it's used properly. All laws have restrictions on their domain of applicability. The problem is that you aren't accounting for all of the heat transfer if all you use is Stefan-Boltzmann. Link to comment Share on other sites More sharing options...
David Levy Posted April 6, 2015 Author Share Posted April 6, 2015 (edited) The Stefan Boltzmann law works well in the correct applications. However in terms of the system your modelling several factors change. Surface area, convection rates as the crust develops greater depth, this influences emmisivity rates, your also dealing with different materials with different conductivity values. Iron has a different conductivity than say silicon etc. As the Earth goes through its development stages the system changes. Thanks Now, we all agree that Stefan Boltzmann law is valid and we can use it. However, we need to verify that we are using it correctly. So, let's look again on Stefan Boltzmann law formula: P = dE/dt = ε·σ·A·(T(hot)4 –T(ambient)4) In this formula there is no requirement for: Surface area, convection rates as the crust develops greater depth or even solar radiation There is only one small sign which covers all of those factors. This is the emissivity - ε As you have stated: "Surface area, convection rates as the crust develops greater depth, this influences emissivity rates" Hence, if we can find the correct values of the emissivity, there is no need to calculate the green house effect, the radiation, the curst of Iron/silicon conductivity or solar radiation. All included in one small sign - emissivity – ε. However, based on this formula, the cooling time is as follow: t(cooling) = Nk / (2·ε·σ·A) · (1/T(final) 3 –1/T(hot)3) Please advice if you agree with that? Edited April 6, 2015 by David Levy -1 Link to comment Share on other sites More sharing options...
swansont Posted April 6, 2015 Share Posted April 6, 2015 Now, we all agree that Stefan Boltzmann law is valid and we can use it. However, we need to verify that we are using it correctly. So, let's look again on Stefan Boltzmann law formula: P = dE/dt = ε·σ·A·(T(hot)4 –T(ambient)4) In this formula there is no requirement for: Surface area A is surface area convection rates as the crust develops greater depth That's a different kind of heat transfer. Convection (and there would also be conduction). The S-B law covers radiation only. It is the heat transfer equation for ONE kind of heat transfer, using ONE body. or even solar radiation That would be an additional term, if one properly applied the equation to a problem (see previous answer). Any proper analysis of this would start with a power balance equation, to find the NET rate of heat transfer. You would include all of the gain and loss terms. There is only one small sign which covers all of those factors. This is the emissivity - ε Absolutely not. As you have stated: "Surface area, convection rates as the crust develops greater depth, this influences emissivity rates" Hence, if we can find the correct values of the emissivity, there is no need to calculate the green house effect, the radiation, the curst of Iron/silicon conductivity or solar radiation. All included in one small sign - emissivity – ε. Nope. It's hard to quantify how wrong this is. However, based on this formula, the cooling time is as follow: t(cooling) = Nk / (2·ε·σ·A) · (1/T(final) 3 –1/T(hot)3) Please advice if you agree with that? Nope. Stop making stuff up. That's the cooling time based on purely radiative cooling with infinite conduction. I know this because I've studied (and passed) physics, and it's also explained on the hyperphysics page you've linked to! To wit: If the heat loss is purely radiative and not limited by heat transfer to the radiating surface, then the cooling time can be modeled for a hot object. The above relationship assumes infinite thermal conductivity so that the temperature of the whole object is equal to the surface temperature. In the real world, the surface will cool faster than the interior. The rate of heat transfer from the interior will be expected to limit the rate of radiative loss from the surface. Please face the fact that you don't understand this and listen to the people who do. Then test your understanding before forging ahead. 2 Link to comment Share on other sites More sharing options...
David Levy Posted April 6, 2015 Author Share Posted April 6, 2015 (edited) The S-B law covers radiation only. It is the heat transfer equation for ONE kind of heat transfer, using ONE body. Thanks You claim that Stefan Boltzmann law covers radiation only. So how can we add the impact of the solar heat to this formula? What is the real meaning of the emissivity in Stefan Boltzmann law? What kind of factors it represent? What is the estimated emissivity value of the Earth? Edited April 6, 2015 by David Levy Link to comment Share on other sites More sharing options...
swansont Posted April 6, 2015 Share Posted April 6, 2015 Thanks You claim that Stefan Boltzmann law covers radiation only. So how can we add the impact of the solar heat to this formula? Add it, i.e. do a power balance. Just maintain the proper sign convention. If you want power emitted to be positive, then you attach a negative sign to terms where power comes in. Power out - Power in = net power out What is the real meaning of the emissivity in Stefan Boltzmann law? What kind of factors it represent? What is the estimated emissivity value of the Earth? It's a deviation from the behavior of an ideal blackbody; it measures how good of an emitter something is, compared to a perfect blackbody. Since blackbodies are also perfect absorbers, it's also related to reflection, since radiation can either be absorbed or reflected. A good reflector will have a low emissivity. A poor reflector will have a high one. There are subtleties to this, too — emissivity will have some wavelength dependence, and some special exceptions since materials can emit non-thermal radiation when there's a resonance. But since we're looking for a kind of order-of-magnitude result, small issues like that don't matter much. here's a table of emissivities http://www.thermoworks.com/emissivity_table.html Soil's emissivity is above 0.9 Lava's emissivity is close to 1 at higher temperatures (600-1000 ºC) but drops down to around 0.7-0.8 at lower temperatures according to https://books.google.com/books?id=86kZBAAAQBAJ&pg=PA89&lpg=PA89&dq=emissivity+lava&source=bl&ots=AVK9TEG2Df&sig=UyvdJmqn92yVhq5eD6M10uPvUDc&hl=en&sa=X&ei=PeEiVfzwLomkNueqgaAI&ved=0CD0Q6AEwBQ#v=onepage&q=emissivity%20lava&f=false see also the previous page of the book Link to comment Share on other sites More sharing options...
Mordred Posted April 6, 2015 Share Posted April 6, 2015 (edited) There is two lines to consider in one of the papers I posted above. "Earths conductivity were not homogeneous, but greater near the centre by a factor of 10, the cooling time would be increased by a factor of 56. In addition, he argued that if some degree of fluidity exists in the Earth, then thermal conductivity must be supplemented by convection." This is the section discussing Perry. "The discovery of radioactivity had two dramatic effects on the debate about the age of the Earth. First, it quickly became clear that since radioactive sources were found everywhere, including deep in the crust of the Earth, the heat budget of the Earth could not be reliably estimated. Secondly, by about 1910 the new methods of radioactive dating held the promise of finding a reliable way to determine the age of the Earth." This section is under radiation These are from http://www.arthurstinner.com/stinner/pdfs/2002-ageoftheearth.pdf Edited April 6, 2015 by Mordred Link to comment Share on other sites More sharing options...
David Levy Posted April 7, 2015 Author Share Posted April 7, 2015 Lava's emissivity is close to 1 at higher temperatures (600-1000 ºC) but drops down to around 0.7-0.8 at lower temperatures according to see also the previous page of the book It's very interesting article. However, it is focusing on the emissivity of volcano/lava activity on the surface of the Earth. However, I have asked specifically about the emissivity of the Earth itself. A good reflector will have a low emissivity. A poor reflector will have a high one. There are subtleties to this, too — emissivity will have some wavelength dependence, and some special exceptions since materials can emit non-thermal radiation when there's a resonance. Yes, I assume that the Earth could be considered as a good reflector. Therefore its emissivity should be low. Hence, what is value of the emissivity which we can apply to Stefan Boltzmann law in order to calculate the real cooling time of the Earth? Link to comment Share on other sites More sharing options...
Mordred Posted April 7, 2015 Share Posted April 7, 2015 You can't exclusively with Stefan Boltzmann. It doesn't factor in a continous heat source such as radiation, convection and conduction. Did you not read the article I posted twice now? http://www.arthurstinner.com/stinner/pdfs/2002-ageoftheearth.pdf Link to comment Share on other sites More sharing options...
swansont Posted April 7, 2015 Share Posted April 7, 2015 It's very interesting article. However, it is focusing on the emissivity of volcano/lava activity on the surface of the Earth. However, I have asked specifically about the emissivity of the Earth itself. But we're talking about a molten earth, right? Since we are discussing how fast it cools off? Yes, I assume that the Earth could be considered as a good reflector. Therefore its emissivity should be low. Again, this is moot. We're talking about rock. Molten, then solid. Not plants, or other things that affect the current earth's emissivity but would have been different 4.5 bya Hence, what is value of the emissivity which we can apply to Stefan Boltzmann law in order to calculate the real cooling time of the Earth? Pick 0.7 for the solid earth It's not going to matter appreciably. 1 Link to comment Share on other sites More sharing options...
David Levy Posted April 7, 2015 Author Share Posted April 7, 2015 (edited) Thanks Mordred Somehow, I'm not sure that I get a consistent feedback with regards to Stefan Boltzmann law. It one had you claim that Stefan Boltzmann law works well in the correct applications: The Stefan Boltzmann law works well in the correct applications. However in terms of the system your modelling several factors change. Surface area, convection rates as the crust develops greater depth, this influences emmisivity rates, your also dealing with different materials with different conductivity values. Also, in the article which you have pointed it is stated: " The cooling time from an initial temperatureTi to a final temperature Tf is given approximately by t = (1/T 3− 1/T 3 )mc/3σA where Ti is the temperature at time t , Tf the ambient temperature (both expressed in kelvins), m is the mass, c the specific heat capacity and A the area of the object (see Block A2). Applying this formula to the small iron sphere, we find that the cooling time is about 47 minutes. For the time of cooling of a globe of the size of the Earth that is made entirely of iron, we obtain about 45 000 years (see Block A2)." Hence, the science is using Stefan Boltzmann law to calculate the cooling time of the Earth. However, in the other hand you claim: You can't exclusively with Stefan Boltzmann. It doesn't factor in a continous heat source such as radiation, convection and conduction.Did you not read the article I posted twice now?http://www.arthurstinner.com/stinner/pdfs/2002-ageoftheearth.pdf The only difference between my calculation and the above one is the emissivity. They assume that the emissivity is 1 and I tried to find the real value. As you can see they do not care about - "continous heat source such as radiation, convection and conduction" They do not care about - "Surface area, convection rates as the crust develops greater depth, this influences emmisivity rates, your also dealing with different materials with different conductivity values" If they can use it, then I can also use it. If you wish to discuss about the emissivity - you are more than welcome. So please - try to take a decision. Can we use the above formola for cooling time? Yes or no (without but..) With regards to emissivity Pick 0.7 for the solid earth Based on my calculation : ε = 0.0013 You claim that it is 0.7. Can you please explain how did you get this value? Edited April 7, 2015 by David Levy Link to comment Share on other sites More sharing options...
Strange Posted April 7, 2015 Share Posted April 7, 2015 Based on my calculation : ε = 0.0013 How did you come up with this value? Are there any materials which have emissivity this low? Some materials which resemble parts of the Earth's surface: Asphalt: 0.88 Concrete: 0.91 Ice: 0.97 Snow: 0.8 to 0.9 Water: 0.96 http://en.wikipedia.org/wiki/Emissivity#Emissivities_of_common_surfaces The lowest value there is polished silver (0.02). So you are saying that the Earth's average emissivity is 10 times less than that. Sounds implausble. Link to comment Share on other sites More sharing options...
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