What will exactly happen if a small amount of iron from the sun's core were teleported to the Earth's surface?

Question

It is obvious that the surrounding area will become extremely hot, but I want to know other consequences if enough iron to make a sword were to be teleported to earth. Would it be possible to make a sword of that iron?

Answer 1

Other answers have already covered why the iron would be dangerous to harvest because of its gaseous state... but the other important thing to consider is that even if you could teleport pure iron in from another source, you still have a lot of work to do to make it into a useful metal for making a sword out of.

Pure, un-worked, un-tempered iron is so soft, that its material properties are very similar to annealed copper.

SOURCE: https://nvlpubs.nist.gov/nistpubs/jres/28/jresv28n5p643_A1b.pdf

To put this in perspective, a very low quality sword like you would find in the bronze age or early iron age would generally have BHN of ~160 with a UTS of ~50,000lb/sqin. If you want to make something more like a late medieval tempered, medium carbon steel sword, you'd expect a BHN in the 200-400 range with a UTS that could exceed 100,000 lb/sqin.

So, to make a useful iron blade, you'd still need to alloy it with carbon, magnesium, and/or phosphorous to harden it, you'd need to hammer it out to improve its crystalline structure, and you need to alternate its temperature between various highs and lows to properly temper it. So, if your goal is to extract hot iron from another source and turn it directly in a sword, you will get a very poor quality blade. Most of the stuff that ancient smiths did to smelt and form thier steel can't just be skipped over by having a source of really hot iron to start with.

Answer 2

Iron becomes a gas at 5182F, about half the temperature of the Sun. There is probably plenty of iron in the sun, from asteroids and comets falling in. Certainly enough to make a sword. But any iron in the sun is in a very diffuse gaseous state, teleporting enough to make a sword would likely take a miles wide chunk of the sun, and would explode so quickly that the atomic iron would be spread over most of the Earth.

Unless you have some magical way to extract just the iron and cool it down enough to condense into at least a liquid (2800F), I think this idea is a non-starter.

Answer 3

Move your teleporter a bit, to target Mercury's core instead of the sun

Consider teleporting iron from Mercury's core. It has huge amounts of pure iron, and it is solid.

Be careful !

Mercury's core iron is probably much cooler than Earth's 5200 degrees celcius.. Mercury has a lower mass.. but on reception you'll have to cool it down and contain it ! The pressure will be lower, it could explode. To make sure you can harvest a useful amount, teleport it into a cave or under water, and await the teleportation on a safe distance.

https://agupubs.onlinelibrary.wiley.com/doi/10.1029/2018GL081135

Forge the sword

Then, the solid iron can be forged into a sword. You can also melt it and use graphite powder, to harden the iron. You can also use the ashes.

The magic of a Mercury sword

According to the usual magic symbol system for planets, a Mercury sword will make a warrior agile and smart.

Answer 4

Summary

My concern here would be the pressure difference. The Sun's core has a pressure of about 265 gigabar. That's about 265 billion times higher than sea level on Earth. Simultaneously, the temperature is about 15 million Kelvin, compared to Earth's 300 K, or about 50 thousand times higher than Earth. (Wikipedia, Solar core)

My original answer suggested the energy release would be quite small. I made the mistake of trusting mathematician definitions of the Boltzmann constant, leading to incorrect dimensional analysis.¹. I restarted the calculations from scratch and got a value that's much closer to what I expected. That doesn't make it correct, but I'm more confident in this answer.

This would detonate like a very large bomb. Just the 1kg of iron would explode like 20 tons of TNT. If you had to get the 500kg or so of raw star stuff required to get that much iron, it would be roughly like the Little Boy nuke.

Ideal Gas Law

The ideal gas law tells us:

$PV=nRT$

n is the number of molecules, which is constant for a given mass, and R is the gas constant for that composition. Since P (pressure) and T (temperature) are given by the problem, that only leaves V (volume) to normalize to Earth standard.

$V=\frac{nRT}{P}$

$V_{new}=nR\frac{\frac{T}{2.65\cdot 10^{11}}}{\frac{P}{5\cdot 10^4}}$

$=nR\frac{T}{P}\frac{2.65\cdot 10^{11}}{5\cdot 10^4}$

$=(5.3\cdot 10^6)\frac{nRT}{P}$

$=(5.3\cdot 10^6)V$

Assuming R stays the same during cooldown (probably an invalid assumption since we're converting from plasma to gas, but I think the premise is still valid) this means the mass you teleported is going to expand in volume by over 5 million times, with the radius increasing by about 170 times.

The speed of sound in the core as you teleport it is about 50 km/s (Stanford, Results of solar model calculations, very last graph on the page). This is somewhere between the blast wave speed of C4 (1.5 km/s, PubMed abstract of The disguised face of blast injuries: shock waves) and a nuclear blast (882 km/s @ 2m, Worldbuilding.se answer to How fast is the shockwave of a nuclear bomb from 2-5m away?).

None of this directly helps us, but is essentially the basis of my gut feeling.

Plasma Physics

I'm not at all familiar with plasma physics, so any errors are probably in this section. I've done the best I can to figure out what's correct based on other reading.

A typical sword weighs 2-5 lbs, meaning it masses about 0.9-2.3 kg. Let's call it 1 kg for simplicity. The heat capacity of a plasma is given by $\varepsilon=\frac{3}{2}n k_B T$ (Physics.se, answer to Heat capacity of plasma?. $\varepsilon$ is energy per unit volume, so we can get energy as:

$E=\varepsilon V$

$=\frac{3}{2} n k_B T V$

$n$ is the plasma density. Star plasma has a density of about $10^{26}\frac{e}{cm^3}$ (Cern, Introduction to Plasma Physics, page 2).

$k_B$ is the Boltzman constant, $1.380649\cdot10^{−23}\frac{J}{K\cdot e}$, where $e$ is the number of electrons.¹

$T$ is temperature. Plasma physicists tend to use units of eV (electron-volts). 1 eV corresponds to 11600 K. The Cern paper gives star plasma a temperature of $2\cdot 10^3$ eV, which corresponds to 23 million K, which is pretty close to the 15 million K we're using.

$E=\frac{3}{2} n k_B T V$

$=\frac{3}{2} 10^{26}\frac{e}{cm^3} \cdot 1.38\cdot 10^{-23}\frac{J}{K\cdot e}\cdot 1.5\cdot 10^{7} K\cdot V$

Originally at this point, I kept calculating, then tried to calculate the volume and substitute that back in:

$=3.1\cdot 10^{10}\cdot V\cdot \frac{J}{cm^3}$

But volume doesn't matter. $\frac{e}{cm^3}\cdot V$ must equal the number of electrons. And we can calculate the number of electrons directly.

Iron is about $56 \frac{g}{mol}$, or $18 \frac{mol}{kg}$. Since we have 1 kg of iron, that means n=18 moles in conventional terms. However plasma physics counts in terms of electrons, which is about the number of protons per atom. Iron atoms have 26 electrons per atom, so our plasma particle count is $18\cdot 26$ $=468\frac{mol}{kg}$. 1 mole is $6.02\cdot 10^{23}$ electrons, giving $2.82\cdot 10^{26} \frac{e}{kg}$. Since we have a 1kg sword, that's $2.82\cdot 10^{26} e$.

So we can replace particle density times volume with number of particles. Then instead of energy per volume, we'll have energy, which is what we actually want.

$E=\frac{3}{2} 2.82\cdot 10^{26} e \cdot 1.38\cdot 10^{-23}\frac{J}{K\cdot e}\cdot 1.5\cdot 10^{7} K$

$=8.8\cdot 10^{10} J$

A ton of TNT is about $4.2\cdot 10^9 J$ (WolframAlpha), so this is the equivalent of about 21 tons of TNT.

I can't find a good source, but various forums suggest the Sun consists of about 0.2% iron. In order to get enough iron to make a sword, you'd then need about 500 kg of Sunstuff, increasing the energy substantially.

$E=4.4\cdot 10^{13} J$, which is about a 10.5 kT TNT, or about 0.7 times the energy of Hiroshima nuke (Wikipedia, Little Boy).

Notes

Star Trek did a similar concept in an episode (S5E13, The Masterpiece Society, where they had a chunk of the core of a neutron star for some hand-wavy purpose. There might (or might not) be something useful you can get inspiration from there.

¹ If you look up the Boltzman constant in most cases, you'll typically find it listed with units of $\frac{J}{K}$, but that's because mathematicians tend to ignore ad-hoc units like "cycles" and "particles". The Boltzman constant is just the ideal gas constant when particle number is in units of "individual particle" instead of "moles" (which is just $\frac{\text{number of particles}}{\text{Avagadro's number}}$) (Wikipedia, Boltzmann constant). Since the ideal gas constant has the number of molecules and we're dividing by a dimensionless quantity, the Boltzmann constant also has to have the number of molecules.

In my original equations, this caused me to try cancelling out the number of molecules when calculating volume, so I was inadvertently dividing by very large numbers when I shouldn't have. This is why engineers tend to prefer including all the units, not just the neatly-packaged units with exact definitions. It's also why it helps to actually understand the material in question.