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September 30

biochemistry

what colour could be observed if glucose solution is added to hydrochloric acid + iodine solution — Preceding unsigned comment added by 41.78.77.246 (talk) 07:27, 30 September 2011 (UTC)[reply]

I don't know. When you added glucose to a mixture of hydrochloric acid and iodine in your chemistry lab, what happened for you? --Jayron32 19:09, 30 September 2011 (UTC)[reply]

You may wish to read Iodine test and Benedict's reagent. Graeme Bartlett (talk) 08:21, 2 October 2011 (UTC)[reply]

second moment of area

hi, how can we visualise second moment of area ? i men, its easy to visualise normal moment of inertia, but i am having difficult in physical visualisation of second moment of area — Preceding unsigned comment added by 117.192.193.0 (talk) 13:09, 30 September 2011 (UTC)[reply]

There is a COMSOL Multiphysics demo which shows this, in a beam strength and stress simulation. They will give you a free 30 day demo CD and key if you ask. 69.171.160.77 (talk) 18:09, 30 September 2011 (UTC)[reply]

Radio frequency intervals

Warning: the topic below will require some basic knowledge of music theory, physics of sound harmonics and radio astronomy.

Hi. Last night, I was experimenting with my radio. It was on the AM radio dial, and at approximately 600 AM (probably a bit lower), an ambient humming sound filled the room. I found that the beat frequency was almost exactly 0.5 Hertz, and that the pitch was at an E natural, more than two octaves higher than middle C. At frequencies of both 640 and 680 AM, I detected a near-monotome, static buzzing hovering at B flat, just below middle C. Experimenting further on my electric keyboard, I found that the distance between these two notes was two octaves plus six semitones. Thus, I found the note between these two, as well as another note of a lower pitch at the same frequency. The chord formed by these four notes is a G-diminished 7th.

My question is: what causes these specific channels to produce these ambient buzzing and humming sounds, even in the lack of clear channel reception? What is the Hertz interval between the two main notes mentioned (B flat below middle C and E natural 2 octaves above)? Moreover, what actually caused the 0.5 Hertz beat? Thanks. ~AH1 ^(discuss!) 14:15, 30 September 2011 (UTC)[reply]

You are probably hearing intermodulation distortion (tone injection) due to complicated interference between the transmitter's carrier frequency and some other frequency (possibly your local oscillator, or some other frequency in your radio). Do you know what type of radio you have? Do you know what type of signal you're tuning? (If it's commercial AM radio in the United States, it's probably DSB-SC; but you can still get intermodulation distortion even without a tonal carrier, it's just harder to describe in simple terms).

Also, to convert musical-note to numeric frequency, you only need to know this equation:

f=f_{0}2^{n/12}

...where n is the number of half-tones above f₀. For example, Middle C is three half-tones above A₄₄₀ - so f = 440 Hz × 2^0.25 or about 523 Hz.

Finally: to diagnose the root-cause of a 0.5 Hz beat, we'll... need some more technical information. How much do you know about your radio? Does your receiver have a built-in frequency auto-tuner? AGC? ... Can you hook a spectrum analyzer up and see what else is going on? If you can't answer those more precisely, we probably can't say any more than "it's probably interference from something." These kinds of questions are better probed with the sort of equipment you find in a HAM base station, rather than an off-the-shelf AM radio. Nimur (talk) 18:25, 30 September 2011 (UTC)[reply]

If the sounds stays around the same pitch, but the buzzing changes every few seconds, then that can be interferences from a television. Buzzing that drifts in frequency can be switchmode power supply. A very clean n x 1kHz whistle could be interferences from different radio station carriers, including those from overseas using different band plans. Computers are also great at interfering with AM radio making a a bunch of whistles and "sssh" blocks of white noise. Have a look at Note to work out your frequencies in Hz. Graeme Bartlett (talk) 08:41, 2 October 2011 (UTC)[reply]

What is the average size of human tonsils?

I've read the WP article on tonsils including the passage "Tonsils tend to reach their largest size near puberty, and they gradually undergo atrophy thereafter. However, they are largest relative to the diameter of the throat in young children." So what is that largest size, on average? I don't have a medical encyclopedia handy but an initial G-search proved fruitless. §everal⇒|Times 19:03, 30 September 2011 (UTC)[reply]

What measure are you looking for? This paper PMID 20304270 reports tonsil size (in terms of height, width, thickness, and weight) in overweight and normal-weight children who have problems with breathing during sleep. If that does not answer your question, perhaps you could provide a more specific question. -- Scray (talk) 12:16, 1 October 2011 (UTC)[reply]

I was looking for a statistic of diameter or of volume. I'm aware that there's often no true "average" size of any particular human body part and that any such measurement is subject to differences in age, weight, diet, etc. When medical WP articles discuss the size of certain parts relative to others, however, it's useful to have some specific numbers involved in the discussion. The paper you provided is helpful (though I suspect that tonsilar volume may be related to incidence of sleep disorders and/or breathing difficulties, so it's not quite what I was looking for). Thanks! §everal⇒|Times 17:06, 3 October 2011 (UTC)[reply]

Why exactly does the enthalpy of vaporization fall to zero at the critical point?

Is it because the heat capacity of the gas ends up increasing much faster than the liquid? If so, this qualitative explanation doesn't appear to be included in our critical point (thermodynamics) article, though it delves into derivatives. And if so, why does this overtaking trend accelerate ? I note that 10 degrees Celsius below the critical point, the heat of vaporization is still considerable, but at the critical point, this disappears completely. elle _{vécut heureuse} ^{à jamais} (be free) 19:43, 30 September 2011 (UTC)[reply]

The reason is actually tautological. At the critical point, the substance becomes a Supercritical fluid, for which there is no distinction between a gas and a liquid. The fact that the enthalpy of vaporization at this point drops to zero is because vaporization has no meaning above the critical point. The two statements (enthalpy of vaporization = zero, distinction between liquid phase and vapor phase doesn't exist) are simply two statements of the same fact; which is the critical point of the substance. That is what the critical point is. --Jayron32 19:48, 30 September 2011 (UTC)[reply]

Well yes, but I was looking for a more quantitative explanation. This doesn't really solve problems of the heap: what's the difference in enthalpy of vaporization for a substance 0.000001 K below the critical point versus at the critical point? Surely we cannot resolve problems of analysing a quantitative trend by merely using hand-waving definitions. AFAIK, the two distinct phases don't exist because Hvap is zero, not the other way round. The two phases disappearing is a result of this fact. So I want to know -- what causes this fact? elle _{vécut heureuse} ^{à jamais} (be free) 20:00, 30 September 2011 (UTC)[reply]

What causes the critical point? The article Critical point (thermodynamics) covers that rather well, I thought. The reason that the critical point exists is that the kinetic energy of the molecules overcomes the energy holding the molecules together in the condensed phase. After you exceed that temperature, no amount of compression will cause the molecules to condense. In some ways, you can think of the enthalphy of vaporization as related to this energy difference. In one perspective, you can think of the enthalpy of vaporization as the extra energy needed for that fraction of molecules not yet in the vapor phase to get into the vapor phase. The closer you get to the critical point, the greater proportion of molecules have that necessary energy, so the less extra energy you'd have to supply to get there. The article Enthalpy of vaporization has a very nice graphy which shows that the enthalpy of vaporization decreases until it reaches zero. The difference between the critical temperature and an arbitrary temperature below that is that, at the lower temperature enough molecules have a low enough kinetic energy that they cannot overcome the intermolecular forces holding them together, if there is enough pressure. At the critical temperature, this number of molecules functionally drops to zero. --Jayron32 20:56, 30 September 2011 (UTC)[reply]

Yes, but that's all nice and qualitative. I guess I want a semi-quantitative explanation (e.g. a relation like dG/dT = -S) for why heat of vaporization should accelerate to zero near the critical point. What exactly is going on? Why does it accelerate? elle _{vécut heureuse} ^{à jamais} (be free) 21:51, 30 September 2011 (UTC)[reply]

Oh. In that case I think you want to look at the Clausius–Clapeyron relation for a general mathematical treatment of phase changes. This page discusses some of the mathematics involved at the critical point. This page here represents some early research on the topic, and discusses the derivation of the relationship between heat of vaporization and critical temperature. --Jayron32 23:34, 30 September 2011 (UTC)[reply]

Unidentified bird in France

Does anyone know what kind of bird this is? I thought it was some sort of finch, and then I realized, hey, I don't know anything about birds (and even if it is a finch, I have no idea what kind). If it helps, this was taken in Nantes, France. Apologies for the rather inadequate photo. Adam Bishop (talk) 20:44, 30 September 2011 (UTC)[reply]

Looks like a wagtail, perhaps an immature white wagtail. Mikenorton (talk) 20:55, 30 September 2011 (UTC)[reply]

Thanks! Adam Bishop (talk) 18:39, 1 October 2011 (UTC)[reply]

How would you calculate the change in freezing point (with increasing pressure) without the Clausius-Clapeyron equation?

At first if am only given density difference, difference in pressures, and the freezing point at one temperature. If I assume the change in entropy of vaporization is constant -- +22.014 J/(K*mol) and I know the change in volume per mol is +217.05 cc/mol (yup this is water). Then the difference in atmospheric work between 10 MPa and 1 atm is 2.0785 kJ/mol.

ΔS = ΔH/T T = ΔH / ΔS

then our new temperature point = ΔH/ΔS = (6.01 kJ/mol - 2.0785 kJ/mol) / +22.014 J/K = 178.6 K = -94.6 C

This is a very large change in the melting point of the ice.

But if I use Clausius-Clapeyron engineering equation (which I don't think I'm supposed to do):

T2 = 0C

ln(1 atm / 10 MPa ) = 6.01 kJ/mol / R * (1/T1- 1/(273.15 K) )

R * ln(1 atm / 10 MPa ) / 6.01 kJ/mol = (1/T1 - 1/273.15 K)

R * -4.592 / 6.01 kJ/mol + 1/273.15 K = 1/T1

Which brings me below absolute zero (if I flip this -- I get 98.35 C)

If I use the slope equation mentioned in the Clausius-Clapeyron article I get ΔT = ΔP * T * ΔV / L = (10 MPa - 1 atm) * 273 K * 0.875 cc/mol / 8.68 kJ/mol = -0.272 Kelvin

Which is wayyy less than the ~100K drop predicted before.

What is the cause of the discrepancy? Is it just me? elle _{vécut heureuse} ^{à jamais} (be free) 22:25, 30 September 2011 (UTC)[reply]

[The heading “Freezing point change calculation” is adequately brief and adequately informative.

—Wavelength (talk) 23:18, 30 September 2011 (UTC)][reply]

Earnshaw's Theorem

Just trying to comprehend something the teacher said in class. So if I have 8 identical point charges (protons let's say) positioned at the corners of a cube (like a physical octopole except all charges have the same polarity), then if I place a positive test charge at the center of the cube, the test charge will be in equilibrium, right? From what I understand of Earnshaw's theorem, the test charge will be in equilibrium but it won't be in a stable equilibrium so if I perturb the test charge slightly (just push it in any one direction), the test charge will fly away as fast it possibly can. My teacher sounded like as if there is no equilibrium and no matter what (even theoretically) the charge just won't stay at the center. I was like how can this be. If I put a charge dead center it will stay there. All of the components of the force vectors should cancel. In the real world of course, I understand its impossible to confine a single charge like this let alone plasma or something. Am I right?128.138.138.122 (talk) 22:38, 30 September 2011 (UTC)[reply]

That isn't quite an accurate description. Instability means that if you push the charge away from the center, it will start to move farther away. It does not necessarily mean that it will move quickly -- and in fact it won't. If you only move it a small distance, it will move very slowly. The point is which direction it moves: away from the center. Looie496 (talk) 22:52, 30 September 2011 (UTC)[reply]

A stable equilibrium will return to the equilibrium point when perturbed (in chemistry, this is called Le Chatelier's principle, but the concept applies to any stable equilibrium system). An unstable equilibrium (usually termed metastable) will drift away from the equilibrium point when perturbed. --Jayron32 23:09, 30 September 2011 (UTC)[reply]

Note that Earnshaw's theorem implies that there are no completely unstable equilibria either, because any such point would be a stable equilibrium for a test particle of the opposite charge. So the center point in your example must be stable against perturbations in some directions and unstable against perturbations in other directions. -- BenRG (talk) 23:39, 30 September 2011 (UTC)[reply]

As expected, we have an article for Earnshaw's Theorem. I wouldn't say it's an accessible article or that helps understand the topic well:( DMacks (talk) 19:36, 1 October 2011 (UTC)[reply]