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July 27

Can a gas shield the Earth?

Can we release a gas which will shield the Earth from sun rays and reduce global warming? (That would be the equivalent of CO2 but reflecting heat instead of absorbing.)--Mr.K. (talk) 11:17, 27 July 2010 (UTC)[reply]

Solar radiation management discusses several schemes. -- Finlay McWalter • Talk 11:20, 27 July 2010 (UTC)[reply]

The sun emits most of its energy in the visible range of the spectrum. I don't know of too many gases that significantly reflect in the visible, are stable in the atmosphere, and are mostly harmless to the ecosystem. Also, of course, it would seem to be more rational to do less meddling with a large, complex, critical system, rather than more. --Stephan Schulz (talk) 11:34, 27 July 2010 (UTC)[reply]

(Actually, you have it wrong - the whole problem with CO2 is that it DOES "reflect heat" - thereby trapping the heat in our atmosphere instead of letting it radiate harmlessly out into space!)

The tricky part is that you need to find something that's able to reflect visible and UV light - but be transparent to infrared. That's because it would have to bounce away the frequencies of light where the sun is injecting energy into the planet - but not prevent the planet from radiating away waste heat as infrared light. The Greenhouse gasses like CO2 that are causing all of the problems are quite the opposite - they are transparent to visible light (letting sunlight into the planetary atmosphere) and reflect infrared (preventing waste heat from escaping).

So the material you're looking for wouldn't be a transparent substance like most gasses - because those aren't reflecting visible light - they are transmitting it. The material you're imagining would have to be bright white to look at - yet transparent to IR light. Clouds are somewhat like that - but they are tricky to manage. Water vapor is a greenhouse gas, like CO2 - but when the water droplets are just the right size and temperature to form clouds, they become opaque and reflect away sunlight. But it's tough to control water vapor - to have it form clouds and reflect light without becoming a greenhouse gas.

You'd also need something that was very cheap and non-damaging to manufacture (if, for example, it took a lot of energy to make - it might cause more problems than it solved). You need an ungodly amount of 'stuff' to fill all of that atmosphere densely enough. It would also have to not damage ozone, not be poisonous, etc. Ideally, you'd want something that would break down naturally - at just the right rate so that we could control it's effects - you wouldn't want something where some small misunderstanding of the Earth's environment caused us to use too much of the stuff and plunge the planet into a massive cooling spell. Overall, this kind of approach is very dangerous...it's hard to imagine that doing this would be sufficiently non-risky to be acceptable. The politics of doing this would be very tough to negotiate too...suppose one country decided to do this kind of risky planetary engineering without the agreement of all of the other countries on Earth?

SteveBaker (talk) 11:56, 27 July 2010 (UTC)[reply]

The earth has natural cycles of heat and cold, and our industry only makes a small difference to the temperature of the earth. Yes, the risks and expenses of making such a system would make it completely impractical. --Chemicalinterest (talk) 12:02, 27 July 2010 (UTC)[reply]

Let's not derail this into a discussion about whether climate change is anthropogenic, please. --Mr.98 (talk) 12:35, 27 July 2010 (UTC)[reply]

In any case, if the earth is only 7000? years old it's questionable how we can know whether the earth has natural cycles or not and these natural cycles must be so short that we better be bloody scared. Nil Einne (talk) 21:31, 27 July 2010 (UTC)[reply]

No. There was a warmth period (global warming) in the time of the Vikings, which is why they moved to North America. They left when it started cooling, making it too cold for them to survive profitably. It was still cold throughout history until recently, when it started warming again. During that cold spell, the early immigrants to the US (the Pilgrims) had a hard time being established here because of the cold. --Chemicalinterest (talk) 13:18, 28 July 2010 (UTC)[reply]

See Medieval warm period and little ice age, both of which are fairly insignificant compared to the observed and projected warming of the 20th and 21st century, especially the MWP which was cooler than year 1900 depending on the proxy source. ~A H 1^(TCU) 01:52, 2 August 2010 (UTC)[reply]

Clouds reflect a lot of visible light, but don't they also reflect IR? That's why it's generally warmer at night if there is cloud cover. --Tango (talk) 14:19, 27 July 2010 (UTC)[reply]

If you have an infrared thermometer you can actually measure this. If on a day with some clouds in the sky you point your thermometer in the direction of clear sky you get a very low reading (like -50 °C), if you point it toward a cloud you get a much higher reading (e.g. 0 °C). Count Iblis (talk) 15:38, 27 July 2010 (UTC)[reply]

Isn't that detecting the thermal radiation from the cloud, rather than reflected thermal radiation from the ground? --Tango (talk) 21:59, 27 July 2010 (UTC)[reply]

Why not paint large areas of the Earth's surface white? Count Iblis (talk) 15:16, 27 July 2010 (UTC)[reply]

Direct modification of the Earth's albedo is discussed in the article on solar radiation management, linked in the first answer above. TenOfAllTrades(talk) 15:19, 27 July 2010 (UTC)[reply]

Thanks! Count Iblis (talk) 15:38, 27 July 2010 (UTC)[reply]

There is a big misconception circling around here - that gases "reflect" visible light or IR radiation. They do not - gases can scatter and absorb, but there is not mirror-like reflection unless there is a flat surface of some sort (which gases don't have). Gas molecules with polar bonds (e.g. C=O, C-H, O-H, C-F, etc.) tend to be good at absorption of long-wave IR radiation (The same gases also emit a lot when warm) which makes them greenhouse gases. An "anti-greenhouse" gas would need to be poor at absorbing long-wave IR (so it must not contain such polar bonds) and good at scattering or absorbing visible light. However, the common gases with nonpolar bonds such as N₂ and O₂ are transparent. Gases without polar bonds can be colored/non-transparent such as Cl₂, but it is toxic, harmful and unstable so clearly we shouldn't be pumping it in massive quantities into our atmosphere. I am not aware of any gases which i) have no polar bonds, ii) are not transparent to visible or short-wave-IR, and iii) Could be released in massive quantities necessary to achieve a substantial cooling effect without disastrous side effects. However, if you remove the restriction to gases, sulfate aerosols might work (see Geoengineering.) 129.2.46.178 (talk) 01:02, 28 July 2010 (UTC)Nightvid[reply]

Some people think that there may be a way to protect the earth via a gas (See this article). Basically, a relatively small amountof SO2 in the stratosphere could 'reflect' some of the light that would have entered the atmosphere. Not sure if it really reflects (see above regarding gases absorbing or scattering light), but it is possible that it would exist as a fine particulate, maybe sulfates as mentioned above, that would reflect. As for cheap sources, there is quite literally tons of sulfur sitting in the Alberta tar sands that some people would love to find a use for...24.150.18.30 (talk) 02:50, 28 July 2010 (UTC)[reply]

A recent article in New Scientist argued that reduction of the UK's power generation and industrial emissions of Sulpher dioxide (by using 'cleaner' fuels) to lessen the acid rain impact on Scandinavia had, ironically, had little impact on the acid rain (because the main sources of Scandinavian SO₂ were actually Continental), but had increased the UK's local warming by decreasing SO₂ reflection of sunlight. 87.81.230.195 (talk) 10:16, 28 July 2010 (UTC)[reply]

Direct injection of sulphur dioxide into the stratosphere is one of the possibilities discussed in geoengineering. However it would be a problem to remove all those trillions of microscopic particles should something go wrong, and when there are removed the temperature on Earth could skyrocket by 7°C. Also take a look at carbon dioxide air capture, water vapor feedback and Iris hypothesis. ~A H 1^(TCU) 01:52, 2 August 2010 (UTC)[reply]

There is not enough darkness in all the world to put out the light of even one small candle

Is this true?

Namely, if you had a small unobtrusive clear container of STP air in a remote and (perfectly?) dark region of space with a small wax candle burning inside, would black-body radiation drop the temperature low enough to extinguish the flame? -Craig Pemberton 18:16, 27 July 2010 (UTC)[reply]

I assume you're thinking of something like the fact that water left outside on a clear night can freeze even if the air temperature is above freezing, because it loses heat to space, and air (not being very efficient at radiating heat) does not radiate enough heat back to the water to compensate. Whereas if the night is cloudy, it won't happen, because the clouds will radiate enough heat to keep the water liquid.

I am quite sure that effect cannot put out a candle flame. The limit of the effect is the heat that would otherwise be transferred by radiation from the air to the candle wick, if air followed the black-body law. But that's a pretty trivial amount of heat compared to what a candle generates. --Trovatore (talk) 18:23, 27 July 2010 (UTC)[reply]

Let's try some back-of-the-envelope calculations and see what they reveal. Let's assume that the container is a cube 10cm on each side, that the space is at absolute zero (actually, cosmic microwave background radiation makes it about 3K, but that's close enough to zero), that the container is a perfect black body and that the candle continues to have enough oxygen to burn by magic. The relevant law is the Stefan–Boltzmann law, which states that the power lost is

A\sigma T^{4}

, where A is surface area, T is temperature and

\sigma =5.7\times 10^{-8}{\textrm {J\,s}}^{-1}{\textrm {m}}^{-2}{\textrm {K}}^{-4}

. According to candle, a typical candle emits about 40W. We can now rearrange the Stefan-Boltzmann law and find the equilibrium temperature.

T={\sqrt[{4}]{\frac {40}{0.06\cdot 5.7\times 10^{-8}}}}=330\mathrm {K} =56^{\circ }\mathrm {C}

. The candle will have no problem burning with the container at 56C (the air right next to the candle will be hotter). If we increase the size of the container, then the equilibrium temperature will be lower, for example a cube 1m on each side would have a temperature of -170C, which is probably too low for the candle to keep going (the air in the centre would be much hotter, but the outside would be very close to the point where oxygen condenses, which would cause problems). --Tango (talk) 18:47, 27 July 2010 (UTC)[reply]

As a minor aside, I will note that bringing liquid oxygen together with a combustible material in the presence of a spark or other ignition source is actually a very effective way of making a very impressive fire. While I have not tested this personally – and I would discourage any but the most qualified from making the attempt – I strongly suspect that dropping a lit candle into liquid oxygen would be very...exothermic...indeed. Making interesting reading (and viewing) are the numerous accounts of people lighting charcoal barbecues using a lit cigarette, a pile of charcoal, and a bucket of liquid oxygen — attached to a very long pole. TenOfAllTrades(talk) 22:03, 27 July 2010 (UTC)[reply]

(ec)A candle can burn as hot as 1930K - i.e. a temperature rise of 1637K, even if you froze the whole thing to absolute zero the candle would still release enough energy to burn (at least once you managed to ignite it). Ariel. (talk) 18:50, 27 July 2010 (UTC)[reply]

I quite like Tango's analysis above, but I fear that once the box gets large (and the corresponding mass and volume of air around the candle become significant) we will no longer be able to approximate the system as reaching a uniform air temperature. Instead, we will have a gradient of air temperatures ranging from 'quite hot' adjacent to the candle flame down to 'very cold' at the walls of the box. Assuming the box is under gravity then convection currents will also play a role. (If the box is not under gravitational or other acceleration, then we might have problems with the candle depleting the oxygen around itself and going out for that reason.) Conductive and convective heat transfer will become increasingly important in larger boxes. TenOfAllTrades(talk) 22:03, 27 July 2010 (UTC)[reply]

Indeed, my calculations are for the container, rather than the air. The air will have a temperature gradient, from the temperatures I gave at the edge to thousands of degrees at the candle itself. The exact details of that gradient are beyond my ability to calculate. --Tango (talk) 22:13, 27 July 2010 (UTC)[reply]

Per Fourier's law, the gradient will tend towards a linear isotropic temperature falloff after a long period of time. In other words, dT/dr = constant, where r is the vector from a test point to the candle heat source. In reality, two factors confound this calculation; there is not spherical symmetry (if the box is a "cube"), so depending on whether we have "imaginary nonconductive walls" or some walls with real thermal characteristics, that will break the linear isotropic assumption. We could call the walls "very conductive" and therefore at constant temperature at all locations along the wall, in which case they would serve as a boundary condition for the heat flux equation. Or, we could call them "very non-conductive" (much less conductive than air), in which case they would simply truncate an otherwise spherically symmetric solution. The realistic case, where the conductivity is "comparable" but non-equal to that of air, would be a complicated boundary value problem. The second problem is that air will convect and turbulence will exist; if the candle burns for an "indefinite period of time", we can assume a steady-state will be reached eventually, but it may be a very long time before that is the case. Tiny fluctuations in initial conditions of the momentum and angular momentum of each air particle will persist in a very unpredictable way. I agree with Tango, any realistic solution for the air temperature that does not rely on trivial-solution assumptions are extremely difficult. (Not surprisingly, these are the same calculations used to determine planetary energy-balance and surface-temperature equations in planetary science. The results can provide bounds on material composition, internal seismicity or radiogenic heat, and so on. There is good coverage in de Pater and Lissauer's Planetary Science text). The scenario described above is a lot like a miniature "hot gas planet" with an internal heat source. Nimur (talk) 02:34, 28 July 2010 (UTC)[reply]

Point of correction: Fourier's Law implies that

{\vec {T}}(r,\theta ,\phi )=T(r){\hat {r}}

and

r^{2}{dT \over dr}

is constant in a spherical geometry. Dragons flight (talk) 03:01, 28 July 2010 (UTC)[reply]

Ah, yes, I forgot about the conductivity of the container. I made an implicit assumption that it was highly conductive, which I should have included in my list of assumptions at the beginning. Thanks! --Tango (talk) 13:05, 28 July 2010 (UTC)[reply]

Volume of blood in the human body

On average, how many cm³ of blood are there in the human body? --138.110.206.99 (talk) 19:28, 27 July 2010 (UTC)[reply]

Humans typically have between 5 and 7 liters of blood, so 6,000cc would be a good estimate. Googlemeister (talk) 19:36, 27 July 2010 (UTC)[reply]

Blood gives a typical volume of 5 liters, which is in sync with the 4.7–5.7 L range given in circulatory system. Of course none of this is specifically cited. Anyone with a medical text or similar ref handy? DMacks (talk) 19:41, 27 July 2010 (UTC)[reply]

10-12 pints. DRosenbach ^{(Talk | Contribs)} 22:46, 27 July 2010 (UTC)[reply]

(Which is, roughly speaking, the same as 5-6 liters). TenOfAllTrades(talk) 23:07, 27 July 2010 (UTC)[reply]

Yes, that would be correct, Ten. DRosenbach ^{(Talk | Contribs)} 23:36, 27 July 2010 (UTC)[reply]

"Ten" in this case does not refer to the question, but the username of the user whom the above user responded to, TenOfAllTrades. ~A H 1^(TCU) 01:44, 2 August 2010 (UTC)[reply]

Bermuda Triangle

Did we find out what really happened in this triangle?75.73.152.238 (talk) 22:40, 27 July 2010 (UTC)[reply]

Yes. The answer is: Nothing special. Our Bermuda Triangle article has details. Comet Tuttle (talk) 22:41, 27 July 2010 (UTC)[reply]

How much paneer cheese would I get per litre of milk?

Having read the paneer article I'm intrigued by the idea of making some myself, rather than buying salt-laden fetta cheese from the supermarket. What weight of paneer cheese would I get from a litre of milk please? Is it possible to make low-fat paneer cheese by using skimmed milk? Thanks 92.29.116.34 (talk) 23:19, 27 July 2010 (UTC)[reply]

It is not specific to paneer, but this website [1] describes turning 5 gallons (roughly 40 pounds) of milk into 6 pounds of cheese. So, using that ratio, 1 liter of milk should make about 150 grams of cheese. Dragons flight (talk) 23:42, 27 July 2010 (UTC)[reply]

I think it can vary a lot though by type of cheese, softer cheeses requiring less milk per final product. Rckrone (talk) 04:31, 28 July 2010 (UTC)[reply]

Softer cheeses contain more whey, so they have greater yields. Cheddared cheeses are the opposite. Regardless, the total cheese and whey combined that is produced should be roughly invariant. Whey can be used in the place of water in a lot of cooking and is nutritious. -Craig Pemberton 07:13, 28 July 2010 (UTC)[reply]

I've looked into and made paneer before. Skimmed milk usually isn't recommended because the paneer is said to be grainy and rubbery (haven't tried it myself) but I've seen it suggested half skim and half full works okay. I've used reconsituted (full cream) milk powder myself because it's cheaper and I don't think it will make that much of a difference to the taste or texture but I haven't actually tested it to see. I can't remember the yields but the above sounds roughly write, in any case if you use milkpowder the amount isn't going to be that different from the solids you put in, 1 kg of full cream milk powder yields 8 litre IIRC which means 125g for 1 litre which is about what Craig Pemberton found in the ref above. Nil Einne (talk) 09:43, 28 July 2010 (UTC)[reply]

Do you need non-homogenized milk to make paneer? Googlemeister (talk) 15:23, 28 July 2010 (UTC)[reply]

Thanks. What about filtering it - where could I get some muslin from in the High Street? Or is something else suitable? The cost of full-fat milk means it would be more expensive than shop-bought cheese - is there anything I could add to it to stretch or bulk it? I do not want to use dried milk. 92.29.121.86 (talk) 12:29, 29 July 2010 (UTC)[reply]

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