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January 7

Hand warmer heat

I'm trying to find out how much energy is produced by an iron-oxidation style of chemical hand warmer, and not having much luck with wikipedia or web search. Any help? The reaction itself is described here but I don't know how to calculate the heat released by the reaction. Let's say the typical unit is about 100g which includes both the "fuel" (iron powder) and the oxidizer. Any help would be appreciated. Thanks. 2601:648:8200:990:0:0:0:B9C2 (talk) 07:37, 7 January 2023 (UTC)[reply]

According to various sources, 4Fe(s) + 3O₂(g) → 2Fe₂O₃(s) releases 1652kJ/mol heat, and one mole of iron weighs 55.845 g, so for every 55.845 g of Fe we get 1/4 × 1652 kJ = 413 kJ. To proceed, we need to know how much of the content by weight is iron. The reaction as described uses pure O₂ as the oxidizer. We may assume the oxygen is not included in the unit but supplied by the ambient air, but according to the info in Hand warmer § Air activated (iron), such units contain, next to iron, also cellulose, activated carbon, vermiculite and salt. --Lambiam 09:05, 7 January 2023 (UTC)[reply]

I remember seeing a packet a few days ago, and reading the ingredients. I don't fully remember it though, I do remember seeing cellulose and vermiculite, but I also saw "a polymer" and not activated carbon, which does not appear to be a polymer. 67.165.185.178 (talk) 13:40, 7 January 2023 (UTC).[reply]

Thanks. I guess I'll assume the packet is 55.845% iron, so if the energy of a 100g packet is released over 8 hours, that's a little over 14 watts. 2601:648:8200:990:0:0:0:B9C2 (talk) 20:23, 7 January 2023 (UTC)[reply]

Are the following calculations related to the distance between gold and alpha in Rutherford's gold foil experiment properly referenced?

When an object traveling speed v reverses direction, speed v will become zero at the time the direction reverses. The object’s KE also becomes zero. In Wikipedia’s article on the Geiger-Marsden experiments the alpha particle’s mass $=6.645e^{-27}$ kg, and initial velocity $=1.53e^{7}$ m/s. Therefore, the alpha particle’s initial KE is:

KE={\frac {1}{2}}mv^{2}={\frac {1}{2}}6.645e^{-27}*(5.53e^{7})^{2}=1.016e^{-11}Joule

The electric potential energy due to distance (r) between charges is:

U_{E}(r)=K_{E}{\frac {qQ}{r}}{}

An alpha particle’s velocity becomes zero, then the electric potential energy between the alpha and the gold can equal the alpha’s initial KE:

{\frac {K_{E}qQ}{r}}=KE

The distance between charges (r) when electric potential energy equals kinetic energy is:

r={\frac {K_{E}qQ}{KE}}

In the Geiger-Marsden experiments the Coulomb constant = $8.998e^{9}N.m^{2}/C^{2},q=1.266e^{-17},Q=3.204e^{-19}$ . Therefore, the distance between the alpha and the gold when alpha velocity is zero is:

$r={\frac {kqQ}{KE_{\alpha }}}={\frac {8.99e^{9}*1.266e^{-17}*3.204e^{-19}}{1.016e^{-11}}}=3.59e^{-15}$

The radius of gold’s atomic nucleus is $7e^{-15}$ meters. Therefore, when alpha velocity becomes zero, the alpha particle will be within the gold nucleus. Vze2wgsm1 (talk) 13:12, 7 January 2023 (UTC)[reply]

In your computation of KE you used 5.53e⁷ (sic!) instead of 1.53e7 (also written as

1.53\cdot 10^{7}

but never the way you wrote it). The correct result is

r_{\mathrm {min} }=4.69\cdot 10^{-14}\,\mathrm {m}

, which keeps the alpha particle outside the gold nucleus. --Wrongfilter (talk) 20:33, 7 January 2023 (UTC)[reply]

If the question truthfully is a search for proper references, I seem to recall an entire chapter in Modern Physics - Chapter 4, in fact - that introduces the conceptual ideas, explores them in great depth, and then works the math. It's got diagrams, it's got numbers, it's got explanations, it dispels the common "gotchas" and experimental subtleties, ... why, it's almost as though this is so standard that someone chose to write it down in one place, where thousands of educated people look at and study from it during a typical class in elementary physics.

Not only was this in the book - my edition even had a special multimedia enhancement, "More of Chapter 4": Rutherford’s Prediction and Geiger’s and Marsden’s Results, which has been mirrored at this University of California at San Diego's physics website.

If the query is a weakly-veiled request for someone to spend the time and effort to vet your personal effort to work-out-your-own-math, I think we can safely refer you to our standard reference on working-out-your-own-math. It's just not something we do, frankly.

Nimur (talk) 22:49, 7 January 2023 (UTC)[reply]

My apologies. I was not replicating Rutherford. I was expanding. To be more complete, Rutherford should have also used alpha particles accelerated to a higher energy. Then the alpha could have either collided with or penetrated gold's nucleus. Vze2wgsm1 (talk) 04:17, 8 January 2023 (UTC)[reply]

He could not? For one thing, he didn't himself perform the experiments, they were conducted by two colleagues and he theorised about the results. From the details (possibly incomplete) mentioned in the article (which you linked above), the alpha particles were not accelerated at all, they merely radiated from radioactive sources which were all that was available at the time. The earliest particle accelerators seem to have been introduced around 1929, some 20 years later. {The poster formerly known as 87.81.230.195} 51.194.245.235 (talk) 06:02, 8 January 2023 (UTC)[reply]

Does not matter who performs the high-energy experiment or when. Without a collision or penetration of the gold nucleus, Rutherford could have concluded (from infinite force when distance between charged particles becomes zero) that a nucleus is a single point (not necessarily volume) that contains electrostatic force from multiple protons and gravitational force from mass. Vze2wgsm1 (talk) 12:46, 8 January 2023 (UTC)[reply]

This doesn't make any sense. Any alpha particles with sufficient energy (and sufficiently accurate trajectory) to overcome the Coulomb barrier and enter the nucleus wouldn't have been reflected in ANY direction, and wouldn't be observed directly.

You'd have one of the reactions 197Au(alpha, n)200Tl, 197Au(alpha, 2n)199Tl, or 197Au(alpha, gamma)201Tl, all of which have minimum energies WELL in the range that requires a particle accelerator to produce on earth.[1]. PianoDan (talk) 19:16, 8 January 2023 (UTC)[reply]

You could define a nucleus as a zero-volume center of an atom's mass and charge. The Geiger-Marsden experiment and Rutherford's calculations are consistent with this definition. The experiment and the calculations are not proof that a gold nucleus has volume. Vze2wgsm1 (talk) 03:23, 9 January 2023 (UTC)[reply]

You could APPROXIMATE a nucleus that way, and that's what the calculations are - an approximation making that assumption. The approximation is invalid as soon as you get into regimes where you have to take nuclear forces into account, i.e., higher alpha particle energies. This is easily proven by the fact that gold irradiated by sufficiently high energy alpha particles becomes activated, due to the creation of the aforementioned Thallium isotopes. PianoDan (talk) 15:31, 9 January 2023 (UTC)[reply]

Furthermore, while you say "The experiment and the calculations are not proof that a gold nucleus has volume.", that is strictly true, but it wasn't what the experiment and calculations are used to prove. The experiment and calculations are used to show an upper bound for the size of the nucleus, which is considerably smaller than the atomic radius of the atoms in question. The nucleus could be any volume from zero up to the size implied by the experiment, and that size is so much smaller than the size of the atom it shows that the concentration of positive charge in the atom is considerably, on the order of at least several magnitudes, smaller than the size of the atom as a whole. The experiment proves the nuclear model of the atom; that the atom has a small, dense, positively charged nucleus and a diffuse cloud of electrons around said nucleus. It gave experimental evidence of such a model. The main controversy about the structure of the atom was that, in 1904, two different physicists proposed different models: J. J. Thomson proposed the so-called "Plum pudding model" and Hantaro Nagaoka who proposed the first workable nuclear model of the atom. The GM Experiment and Rutherford et al. showed that the atoms did not conform to the plum pudding concept, as the results of the gold-foil experiment would have been very different, and the nuclear model does work with the results. Also also, things get very tricky with defining what volume and radius even mean at these scales. Even the small, dense nucleus doesn't have a "skin" or a "wall" or any kind of physical real boundary defining where it ends and "not the nucleus" begins. We define volumes and radii and the like at these scales by things inductively, by (say) how close two particular particles get to each other in a specific experiment. To say that a nucleus (or even a whole atom) has a specific size merely means "under these conditions, this is the closest we can get to it's center". One needs to disabuse oneself of any kind of analogy between the nucleus as a particle and physical objects we handle on a daily basis. Such facile analogies always break down. --Jayron 32 16:22, 11 January 2023 (UTC)[reply]

Thanks PianoDan. Yes, as alpha KE increases, total energy can eventually equal the energy required for a quantized molecular energy change. That is not necessarily an approximation of nuclear size.

For example, the energy increasing could be thermal energy, instead of KE. The condensation of a steam H2O unit by a N2 unit in air. A

100^{o}C

steam H2O unit that deposits its enthalpy of evaporation into a single nitrogen (N2) molecule without changing H2O temperature would require an N2 molecule

1962^{o}C

colder than the H2O unit. Air’s molar heat capacity is

20.7J/mol^{o}C

. H2O enthalpy of evaporation is

40613J/mol^{o}C

.

\delta T={\frac {40613{\frac {J}{mol}}}{20.7{\frac {j}{mol^{o}C}}}}=1962^{o}C

For both the H2O and the N2 to have the same temperature before and after the reaction could require chemically combining the adjacent N2 units to the H2O unit until enthalpy of evaporation can be deposited in the adjacent N2 units. Millions of N2 units could join the H2O during the attosecond duration of a Quantized Molecular Energy Change. Vze2wgsm1 (talk) 15:31, 11 January 2023 (UTC)[reply]

These sort of thermodynamics calculations you are (for some utterly inexplicable reason) showing here are based on the statistics that arise from very large numbers of atoms interacting. They have absolutely nothing to do with the Rutherford scattering experiment.

And "Quantized Molecular Energy Change" is not a useful term for the activation of a single atomic nucleus, since the structure of whatever molecule the atom may or may not be a part of us completely irrelevant. PianoDan (talk) 16:55, 11 January 2023 (UTC)[reply]

“They have absolutely nothing to do with the Rutherford scattering experiment.”

Your linked article detailed alpha collisions with gold, causing nucleosynthesis, with the production of X-rays and gamma rays. Classical physics does not have a mechanism for producing electromagnetic radiation. Therefore, emission of X-rays and Gamma rays only occur during Quantized Molecular Energy Change (QMEC).

Likely, for an alpha collision with gold that results in a QMEC, the alpha need only be as near the gold as necessary for a chemical reaction. Vze2wgsm1 (talk) 15:47, 13 January 2023 (UTC)[reply]

There's no chemical reaction. At best there's a nuclear reaction. --Wrongfilter (talk) 15:54, 13 January 2023 (UTC)[reply]

? Nobody said that nucleogenesis is a chemical reaction. I said nucleogenesis is a QMEC. Vze2wgsm1 (talk) 17:27, 13 January 2023 (UTC)[reply]

Are you claiming that nucleogenesis does not occur during a QMEC? Vze2wgsm1 (talk) 17:41, 13 January 2023 (UTC)[reply]

"Likely, for an alpha collision with gold that results in a QMEC, the alpha need only be as near the gold as necessary for a chemical reaction." is what you posted above and user Wrongfilter responded to. Perhaps you meant to say "... necessary for a nuclear reaction". Perhaps you know by who and when was "QMEC" coined? Modocc (talk) 17:52, 13 January 2023 (UTC)[reply]

Quantized Molecular Energy Change is not a standard term, and I suspect is just something Vze2wgsm1 made up. PianoDan (talk) 18:25, 13 January 2023 (UTC)[reply]

If an +2 charged alpha left a nucleus full of plus charges, alpha KE would be based on distance between the alpha and the rest of the nucleus. Too much energy.

Instead, a QMEC occurred. This is the time total energy redistributes. Change of quantized energy (mass) equals the change of non-quantized energy. Vze2wgsm1 (talk) 14:42, 14 January 2023 (UTC)[reply]

Nope, you don't create an entirely new terminology and process in your head based on a premise that you haven't even demonstrated to exist. Why would a scattered alpha particle leave a nucleus "full of plus charges" when those charges are being scattered away from the nucleus? --OuroborosCobra (talk) 15:06, 14 January 2023 (UTC)[reply]

Was not referring to modern terminology or processes. Rutherford could calculate the energy due to force between charged particles. Vze2wgsm1 (talk) 22:07, 14 January 2023 (UTC)[reply]

Thanks Nimur. I plan to get the book. Vze2wgsm1 (talk) 13:30, 8 January 2023 (UTC)[reply]

Is the length unit in this thermal lance graph correct?

The graph on https://pubs.acs.org/doi/full/10.1021/acs.iecr.1c00532 shows the length unit as cm. I know almost nothing about thermal lances but comparing it with the graphic and looking at how quickly the temperature drops, I think metres is more reasonable. Is this correct? Thanks, cmɢʟee⎆τaʟκ 14:38, 7 January 2023 (UTC)[reply]

Idon't know either but looking at a picture of a thermal lance I don't think meters would be correct as they don't look like glow all that much any distance from the tip. They burn down fairly rapidly and the oxygen would cool as it expands near the end. NadVolum (talk) 16:46, 7 January 2023 (UTC)[reply]

The abstract alone at [2] does not justify changing the graph scales (by 100x) or adding a 300 kelvin asymptote unless you can find verification in the full article or other WP:RS. Philvoids (talk) 19:52, 7 January 2023 (UTC)[reply]

In the publication, Figure 6, the asymptote is drawn as a fat straight line, just as fat as the graphed curve. Due to its fatness, given the y-scale, it covers the range from 258.4K to 337.4K. Its middle value, 297.9K, is close enough to 300K. --Lambiam 21:47, 7 January 2023 (UTC)[reply]

My assumption of the assumption of the pre-ignition temperature is 293 K (20 °C) that fits conventional STP while resisting the lure of False precision. Philvoids (talk) 19:31, 8 January 2023 (UTC)[reply]

The mathematical model has a constant term

T_{\infty }

that is not further specified, but I assume the authors mean this to be just whatever the ambient temperature happens to be in any situation where the lance is used, and not some standard temperature. Also due to fatness, the tail in the image above covers the range from 260 K to 341 K; changing the displayed value to 293 can IMO also, and perhaps even more so, be considered false precision. (Otherwise I must insist on using NIST's true standard temperature, 293.15 K.) --Lambiam 00:11, 9 January 2023 (UTC)[reply]

"The Lure of False Precision" from a fairy opera.

FAIRY QUEEN: Oh, this is weakness! Subdue it!

FAIRY CHOIR: We know it’s weakness, but the weakness is so strong. (With apologies to W. S. Gilbert and any offended fairies.) Philvoids (talk) 15:47, 9 January 2023 (UTC)[reply]

Thermal lances can be several metres long, but the temperature drops quickly as a function of the distance from the tip, flattening out well within a distance of 1 cm, at least in the theoretical model on which the authors of the publication linked to base their graph. In the abstract they write their mathematical model is "validated", which I assume means it conforms to experimental data. --Lambiam 20:57, 7 January 2023 (UTC)[reply]

Thanks very much, everyone. I've updated the image and note on thermal lance. cmɢʟee⎆τaʟκ 05:17, 8 January 2023 (UTC)[reply]