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January 8

Hot peppers

I know why peppers hurt going in, but why do they hurt coming out? Mingmingla (talk) 04:29, 8 January 2012 (UTC)[reply]

Same reason. The capsaicin is still there and active. It doesn't hurt in between as much due to the mucus lining of the digestive tract. StuRat (talk) 04:39, 8 January 2012 (UTC)[reply]

(after ec) Agreed, capsaicin is a physical irritant. From the article Trinidad Scorpion Butch T pepper: "The Trinidad Scorpion Butch T pepper is so strong that those who handle it must wear protective gloves." That should give you the general idea. If this is correct, I found it out just by reading a couple of articles, although I admit the article Chili pepper doesn't seem to make the skin irritant factor explicit. IBE (talk) 04:45, 8 January 2012 (UTC)[reply]

Hence the saying Ring of fire--92.25.100.174 (talk) 11:30, 8 January 2012 (UTC)[reply]

Shortest life-span

Which animal has the shortest life-span? --Amoeba159 (talk) 05:30, 8 January 2012 (UTC)[reply]

The mayfly has brief adult life, but can live a year as a nymph. StuRat (talk) 05:33, 8 January 2012 (UTC)[reply]

A fruit fly lives around a month. StuRat (talk) 05:38, 8 January 2012 (UTC)[reply]

Thanks, and which bird and mammal has shortest life-span? --Amoeba159 (talk) 05:43, 8 January 2012 (UTC)[reply]

According to this article by WonderQuest, the gastrotrich has the shortest lifespan, at 3 days. --184.218.118.2 (talk) 07:21, 8 January 2012 (UTC)[reply]

Does an amoeba live for long? (Just noticed the OP's username, and couldn't find the answer in the article.) HiLo48 (talk) 07:26, 8 January 2012 (UTC)[reply]

Amoebas are protists. DRosenbach ^{(Talk | Contribs)} 02:48, 11 January 2012 (UTC)[reply]

The average life span of an amoeba is little more than two days. Because they reproduce by dividing they are more or less immortal. Clones of amoebae are normally immortal". The normally immortal Amoeba proteus could be switched into a state of limited lifespan by restricting their food intake for several weeks. Von Restorff (talk) 07:44, 8 January 2012 (UTC)[reply]

Since amoebas reproduce by cell division, it's hard to define "life span" for a single individual. Is it the longest branch starting from a given cell? The shortest? The average? However, since in the big picture, the number of amoebas has to remain more-or-less constant, the average life time has to be the same as the average time between reproductions. --Stephan Schulz (talk) 10:47, 8 January 2012 (UTC)[reply]

Why would the number of amoebas have to remain more-or-less constant? Von Restorff (talk) 13:55, 8 January 2012 (UTC)[reply]

If the number is declining then they will eventually all die out. Since they haven't, we know that isn't happening (or is happening too slowly to be of interest). If the number is increasing, then they'll eventually run out of resources and then they'll have to stop increasing. It doesn't take long for amoeba introduced to a new environment to reach maximum numbers, since they reproduce so often and show exponential growth. --Tango (talk) 14:16, 8 January 2012 (UTC)[reply]

Even if the total number has to remain constant that does not necessarily mean that the average life time has to be the same as the average time between reproductions; for instance if we imagine that the 50% of the current population that was cloned before the other 50% has a huge advantage. Von Restorff (talk) 15:27, 8 January 2012 (UTC) p.s. A long time ago the amoeba population must've been smaller than it is now. Except if you assume there was a god who created the earth and created the same number of amoebas that are alive now; but that obviously never happened.[reply]

Do the math. If the average lifetime is longer than the time between reproductive cycles, then there will be some amoebas left over after each cycle. This results in exponential growth. Unchecked exponential growth drowns the planet in grey goo very quickly. If the older generation has a huge advantage, then the young generation has to die immediately to avoid population increase. Of course there are some simplifications in the model (the number of amoebas will go up and down, and only be approximately constant in the long run and an eco-system in equilibrium). But that does not change the basic result - as long as the number of amoebas stays roughly the same, the same number of amoebas that is born needs to die, and hence the average lifetime is the time between two successive cell divisions. --Stephan Schulz (talk) 22:31, 8 January 2012 (UTC)[reply]

Imagine 10% of the first generation is still alive. Then do the math. Von Restorff (talk) 15:23, 12 January 2012 (UTC)[reply]

How do the total number of Amoebas in a specific ecosystem remain constant? --Amoeba159 (talk) 15:16, 8 January 2012 (UTC)[reply]

Statements from old sources about the "immortality" of unicellular organisms should be interpreted carefully. See [1]. Even E. coli experiences aging, despite its superficially symmetric replication, because one cell, which ages, retains the old cell pole, and the daughter cell creates a new cell pole. In some cases, as with budding yeast, you can see which is the daughter cell; sometimes you can't. Saying that a culture of amoebas is immortal because some still survive is like saying that the human race, or a branch thereof, is immortal because it hasn't been wiped out to the last ~~man~~ woman. Wnt (talk) 15:27, 8 January 2012 (UTC)[reply]

Of course they are not really immortal, I just liked the quotes, but on Wikipedia we would probably consider all three sources reliable enough for someone to use them as a source without being reverted within a couple of minutes.

Von Restorff (talk) 15:46, 8 January 2012 (UTC)[reply]

combustion - volume of products

An experiment demostrated in most schools: You invert a beaker over a trough of water with a candle burning in it. The water moves up the beaker ‘proving’ that air contains 20% oxygen. My question: oxygen has been used up but carbon dioxide and water vapour have been produced – therefore there are MORE atoms now in the air around the candle – so the water should move down. Why does it move up? — Preceding unsigned comment added by YakovKorer (talk • contribs) 12:55, 8 January 2012 (UTC)[reply]

The "explanation" neglects the more important factor of temperature. The candle heats the air which expands. When the flame is extinguished, the remaining gas cools and contracts, allowing normal air pressure to force water into the beaker. Dbfirs 13:00, 8 January 2012 (UTC)[reply]

Using the ideal gas law approximation (PV = NkT), N, the number of particles in the gas, is a count of gas molecules, not atoms, but you are exactly right YakovKorer, ~~for every O₂ molecule consumed, you would expect two CO₂ or H₂O molecules produced (with about twice as many of the latter than the former, depending on the composition of the fuel~~ (oops, see below) -- (CH₂)_nH₂ with 20 ≤ n ≤ 40 for paraffin wax). Depending on the humidity, some of the water vapor may condense, complicating matters a bit, but, as Dbfirs says, T is the driving force. If elementary science texts explain this via the consumption of oxygen, then it is an example of a lie told to children gone wrong, as the behavior is the opposite of what you would expect when taking the combustion gases into account. It would be like using a radiometer to demonstrate radiation pressure. -- ToE 15:40, 8 January 2012 (UTC)[reply]

Oop - careful, ToE, your molecule count is off. Combustion of one one atom of carbon to form one molecule of CO₂ consumes one molecule of O₂, not two. (All other things being equal, the net volume change would be zero if the experiment involved complete combustion of pure carbon.) Given a paraffin wax made from saturated long-chain hydrocarbons – molecular formula (CH₂)_nH₂ – combustion will involve:

(CH₂)_nH₂ + (1.5n + 0.5)O₂ → nCO₂ + (n+1)H₂O

For long chains (larger n), that means that about 1.5 moles of oxygen gas will be consumed to produce one mole each of carbon dioxide and water vapor—leading to a presumed volume increase, because the number of molecules in the gas phase has increased. However, we're not quite done. The experiment will be mostly at room temperature; the water below the gas and the beaker around it are all big heat sinks sitting at room temperature. At that temperature, the vapour pressure of water is about 0.02 atmospheres. Once the gas phase gets above 2% water vapor, water will start to condense out—so the combustion reaction really converts 1.5 moles of oxygen into 1 mole of gaseous carbon dioxide and 1 mole of liquid water (with effectively negligible volume compared to the gases). This will mean that the volume of gas in the beaker really is reduced — though not by the full, advertised 20%.

Incomplete combustion, incidentally, will also skew this distribution. Candle soot is almost pure elemental carbon ([2]); to make it, the reaction is stripping most of the hydrogen from the hydrocarbon and forming much more water than carbon dioxide. All that said, I am inclined to agree with the other posters here that volume contraction caused by the cooling of trapped hot gas is a contributing factor to the observed effect in most such demonstrations. It is possible to design the demo in such a way that this problem is avoided (or at least minimized), but I doubt that most demonstrators will go to the trouble. TenOfAllTrades(talk) 16:39, 8 January 2012 (UTC)[reply]

Thanks for catching my faulty stoichiometry. I picture your refined experiment involving an unlit water-surrounded candle covered with a beaker with the water levels equalized (perhaps by temporarily slipping a small tube up into the inverted beaker to equalize pressures). The candle is then remotely lit (probably using an electric resistive heater) and the water level is observed during and after combustion. If the water level initially drops low enough for air to bubble out, then it wasn't deep enough to begin with. If we can imagine it, shouldn't there already be a YouTube video if it? -- ToE 17:15, 8 January 2012 (UTC)[reply]

Athlete's foot fungi

How long can the fungus that causes athlete's foot live outside the human body? How can you kill the athlete's foot fungi on carpeting?11virginia (talk) 15:45, 8 January 2012 (UTC)[reply]

The Wikipedia articles Trichophyton rubrum and Athlete's foot don't contain any information on that. However, if you use Google or Google scholar, and search for Trichophyton rubrum, you may find the information you seek. --Jayron32 18:30, 8 January 2012 (UTC)[reply]

Achhhh, that's not an answer! We want to get the data up and front so we can discuss and evaluate it, and leave it in an archive for some day when we get our act together and have all these answers ready on cue if you ask an artificial semi-intelligence a question. Now looking I do quickly find [3], a 1979 study that found that Trichophyton mentagrophytes remain viable on guinea pig scales for 182 days at 4-30 C but not 37 C. (Working up this study further sounds like an excellent high school class experiment, which might even be publishable) According to [4] (speaking very generally of many species), "Dermatophyte spores are susceptible to common disinfectants such as benzalkonium chloride, dilute (1:10) chlorine bleach, or strong detergents. Chlorhexidine is no longer considered to be a good environmental decontaminant for these fungi. The mechanical removal of any material containing keratin, such as shed skin and hairs, facilitates disinfection. Vacuuming is considered to be the best method in many cases." Wnt (talk) 23:00, 8 January 2012 (UTC)[reply]

The real secret, is avoid keeping ones feet in an environment that encourages the fungi that causes athlete's foot to take hold – the fungi itself, is unavoidable as it is every where. I discovered by accident, when I was a student (before man landed on the moon and all that), that the reason it is called athletes foot, is that whilst sock got washed frequent (well almost) my running shoes/ plimsolls etc. didn't. Being constructed differently from regular (leather) foot wear, some sports-ware was obviously harboured the this fungi because the problem magically disappeared, when thy too got a regular trip in through the washer. Therefore, I suggest that using medications/carpet treatments to mitigate this problem is not the right approach.--Aspro (talk) 23:28, 8 January 2012 (UTC)[reply]

Also, I might add, that the skin has a pretty good defense again fungi, bacteria and viruses, providing it is not over whelmed by numbers. In other words, the natural contact with the odd fungi spore here and there is no going to bother it. Just vacuuming the carpet should be enough to keep the natural fungal population benign.--Aspro (talk) 23:46, 8 January 2012 (UTC)[reply]

My feeling is that susceptibility to fungal infections varies wildly. I don't have that exceptional an immune system where bacteria are concerned, but my idea of treating athlete's foot has been to absent-mindedly scratch away the dead skin between two toes and (correctly) expect not to see it again. Whereas other people have constant trouble with it, onychomycosis, even strange diseases like blastomycosis. An example of a susceptibility factor is a gene CARD9.^[5]

what is the typical expected ultimate recovery (EUR) of traditional gas wells?

Google searches keep reporting breakeven prices and initial production rates, but I am interested in ultimate recovery. I know for example, that according to Chesapeake Energy, a 7.5 million horizontal fracturing shale gas well (that will typically replace 24-32 vertical wells) will have an average EUR of 3.75 mmcf (or MMBTU). What is the typical EUR of a vertical well, and what is the typical EUR of a conventional gas well? Can someone help me find some figures? I realise they are different whether they are onshore, shallow offshore, or deep offshore. Thanks. elle _{vécut heureuse} ^{à jamais} (be free) 17:41, 8 January 2012 (UTC)[reply]

Maybe you want to read UK Gas Reserves and Estimated Ultimate Recovery 2011. Von Restorff (talk) 23:40, 8 January 2012 (UTC)[reply]

Those don't cite per-well statistics =( elle _{vécut heureuse} ^{à jamais} (be free) 01:23, 9 January 2012 (UTC)[reply]

Sorry I am unable to find the data you want. I would recommend contacting a company that owns a gas well. Von Restorff (talk) 13:40, 9 January 2012 (UTC)[reply]

IC Engine braking

When changing down gears to slow down in a car with an IC engine, some of the kinetic energy loss of the car must go into speeding up the engine, and the rest into heating up the clutch. How do I calclulate the distribution of this energy? Also, assuming my gear change was instantaneous, does the speed of the engine also change instantaneously? BTW not homework, just wondering.--92.25.99.162 (talk) 20:09, 8 January 2012 (UTC)[reply]

I don't think there would be a way to easily calculate the distribution, it would depend on too many other factors: The efficiency of the clutch, the weight of the engine and drive train components, the speed difference. You'd probably have to measure it. For the second bit, the engine could only change speed instantaneously if the clutch didn't dissipate any energy at all, so if you assume the clutch is doing some of the work, the engine won't change speed instantaneously. Vespine (talk) 21:22, 8 January 2012 (UTC)[reply]

What's an IC engine? HiLo48 (talk) 22:05, 8 January 2012 (UTC)[reply]

internal combustion Von Restorff (talk) 22:11, 8 January 2012 (UTC)[reply]

You are also absorbing energy by compressing the gas in the cylinders before the valves open again - that is what is meant by engine braking!.... Sorry, got that back to font. Expanding the gases in the cylinder.--Aspro (talk) 22:19, 8 January 2012 (UTC)[reply]

Our article engine braking does a good job describing this, though it is in dear need of references (and so tagged since June 2008). -- ToE 00:44, 9 January 2012 (UTC)[reply]