The YouTube video How Hot Can it Get? contains, at the 2:33 mark, the following claim:
A pin head heated to 15 million degrees will kill everyone in a 1000 miles radius.
On what basis can this claim be true? Some of the things I can think of:
Radiation of the metal as it cools down
Energy released in fusion (not sure if this works for an iron pin)
Would the damage be only to organic matter or will it destroy other structures within that radius?
In this occasion Vsauce rather dropped the ball, I should think. As the other answers show, the claim as stated doesn't make much sense when you put in the numbers, and if you chase the source to its origin there's some crucial context that got dropped along the chain.
The video description attributes the quote to the book The Universe and the Teacup: The Mathematics of Truth and Beauty, by KC Cole, which contains the quote, attributed to James Jeans (but without an actual reference), in the second page of chapter 2,
A pinhead heated to the temperature of the center of the Sun, writes Jeans, "would emit enough heat to kill anyone who ventured within a thousand miles of it."
The quote itself comes from The universe around us (Cambridge University Press, 1930), p. 289, and it reads
The calculated central temperature of 30 to 60 million degrees so far transcends our experience that it is difficult to realise what it means. Let us, in imagination, keep a cubic millimetre of ordinary matter ─ a piece the size of an ordinary pin-head ─ at a temperature of 50,000,000 degrees, the approximate temperature at the centre of the sun. Incredible though it may seem, merely to maintain this pin-head of matter at such a temperature ─ i.e. to replenish the energy it loses by radiation from its six faces ─ will need all the energy generated by an engine of three thousand million million horse-power; the pin-head of matter would emit enough heat to kill anyone who ventured within a thousand miles of it.
OK, so having filled in the references, let's pick apart the calculation and see what the claim actually is. What Cole and Vsauce missed in the quoting is a crucial qualifier:
merely to maintain this pin-head of matter at such a temperature ...
The claim is therefore that an object at that high a temperature, were it to radiate away as a blackbody, while also having a magical energy pump to keep it at that temperature, would be as deadly as claimed.
To see whether this is true, let's put in some numbers. A blackbody at temperature $T$ radiates away a power determined by the Stefan-Boltzmann law, which reads $P=\sigma AT^4$, where $A=6\:\mathrm{mm}^3$ is the area of the cubic pinhead of the paragraph and $\sigma = 5.67\times 10^{-8}\:\mathrm{W\:m^{-2}\:K^{-4}}$ is the Stefan-Boltzmann constant, and this power then gets distributed equally over a sphere of radius $R=1000\:\mathrm{mi}$, so this gives a power density at that 1000-mile radius of $$ j = \frac{\sigma \, A \, T^4}{ 4\pi R^2} \approx 0.0104 \left(T/\mathrm{MK}\right)^4 \:\mathrm{W/m^2}. $$ Now, here we get to one of the sticky points: the claim in Jeans' book has over-estimated the temperature of the core of the sun by about $50/15\approx 3.33$, but Vsauce has missed that discrepancy and he's repeated the claim for the more modern value of $15\:\mathrm{MK}$. Normally, this would not be a problem, because factors of $3$ are pretty ignorable in Fermi analyses, but the Stefan-Boltzmann law has a quartic dependence in $T^4$ and this can mount up quickly, giving a discrepancy of $(50/15)^4\approx 120$ between the Vsauce claim and its Jeans source.
In this case, the difference does matter, probably because Jeans has chosen his numbers so they're roughly at the edges of what they can give. If we put in the numbers for the modern value of the core temperature, we get $$ j_\mathrm{Vsauce} \approx 0.0104 \times 15^4 \:\mathrm{W/m^2} \approx 530\:\mathrm{W/m^2}, $$ which curiously enough is just under half of the solar constant, i.e. the energy flux density from the actual Sun at the surface of the Earth. Thus, from a pure heat-flow perspective, if you were exposed to this for an extended period of time, you might get slightly sunburned, but it's very far from deadly.
The Jeans claim, on the other hand, is somewhat different because of that factor of a hundred, giving $$ j_\mathrm{Jeans} \approx 0.0104 \times 50^4 \:\mathrm{W/m^2} \approx 65\:\mathrm{kW/m^2} = 6.5 \:\mathrm{W/cm^2}, $$ and that's a lot closer to the damage thresholds. Going by Safety with Lasers and Other Optical Sources: A Comprehensive Handbook (Sliney and Mellerio, Springer, 1980, p. 162), the threshold for flash burns is at around $12\:\mathrm{W/cm^2}$, while second-degree burns start at $24\:\mathrm{W/cm^2}$ - for a flash exposure under half a second in duration. Stick around for more than a minute and it sounds about right that you'll very quickly develop some very severe burns, and succumb to them not long after that.
However, as pointed out in the comments, the bulk of the radiation that carries this energy will be in the form of high-energy photons, peaking at around $1.3\:\mathrm{keV}$ (for $T=15\:\mathrm{MK}$; it's $4.3\:\mathrm{keV}$ at $T=50\:\mathrm{MK}$), and that's at the beginning of the ionizing-radiation regime (more specifically grenz rays), which means that the effects are somewhat harder to model, and the detailed radiometry of what would happen is maybe an interesting exercise for an xkcd What if? episode.
As a rough estimate, if you assume that all of the radiation is absorbed (reasonable given this plot of attenuation lengths in water), and taking a surface area of $1\:\mathrm{m}^2$ and a body mass of $75\:\mathrm{kg}$, the Jeans energy flux, when seen as ionizing radiation, is equivalent to an absorbed dose of about $870\:\mathrm{Gy/s}$, which immediately gets out of hand. The lower-temperature source at $15\:\mathrm{MK}$ delivers an equivalent dose of about $7\:\mathrm{Gy/s}$, which I guess is a bit over the ballpark of a Chernobyl liquidator after a couple of seconds. It seems, then, that at both temperatures you're likely to die through radiation sickness, though the details will be messy to work out - but then again it's not really what either of the original sources imply.
To emphasize, though ─ this calculation assumes that you have a magical source of energy that can supply the ${\sim}2\times 10^{18}\:\mathrm{W}$ (!) required to keep that pinhead at $50\:\mathrm{MK}$. It's a reasonable thing to assume if you're already in hypothetics land, but it's a completely different question to the energy that's actually stored in that tiny bit of highly ionized iron plasma, and it's important to state that up front.
I'll keep this around for the next time I need to scare someone into checking their sources ─ it's a monument to academic carelessness, if you will ─ because it's such a good example of how things fall apart if you don't look carefully enough. The claim, in its original context, is roughly reasonable - but the Vsauce claim falls flat under even mild scrutiny.
Thus, if you were exposed to this for an extended period of time, you might get slightly sunburned, but it's very far from deadly.
– N. Virgo
May 17 '17 at 06:38
530W/m2 would do to you.
– Seth
May 17 '17 at 08:33
A pin head is maybe equivalent to a spherical piece of iron with a diameter of 2 mm. That gives it a volume of about 4 mm$^3$ and a mass of $3.2 \times 10^{-6}~\rm{kg}$; computing the heat capacity of matter at these kinds of temperatures is hard, but whatever method you use, the energy content of the pin that you calculate would be insufficient to kill all living things at that distance.
But what if you went to the extreme limit - the matter somehow converted entirely to energy? In that case, the energy would be
$$E = mc^2 = 3\cdot 10^{11} \:\mathrm{J}\, .$$
That's a lot more energy - but if you spread that over a sphere of 1000 mile radius, you would have 9 mJ (milli Joules) of energy per square meter: this is clearly not enough to kill "everything" at that distance.
On the other hand, if you could heat a pin head to that temperature, and keep it that hot, that would require (and release) a very significant amount of energy.
Assuming a perfect black body radiator with a 1 mm radius at 15 MK, the power emitted per unit time would be
$$P = A\sigma T^4 = 4\pi 10^{-6} \cdot 5.67\cdot 10^{-8} (15\cdot10^6)^4 = 4\times 10^{16} \:\mathrm{W} \, .$$
That is some serious power, but when we distribute that power over a sphere with 1000 miles radius, the power density is about 1.2 kW/m$^2$, which is roughly the intensity of sunlight.
Now it's worth noting (as pointed out by David Hammen) that the wavelength distribution of this power is "far beyond the visible". In fact, Wien's displacement law tells us that the peak is at $\lambda = \frac{b}{T}$ where $b=2.9\cdot 10^{-3} m K$. At a temperature of 15 MK, that puts the peak wavelength at 0.2 nm - the realm of X rays. In fact, the handy conversion from wavelength to eV is E = (1240 eV nm) / $\lambda$, so 0.2 nm has an energy of about 6 keV. Lucky for you, that is an energy that is well absorbed by air - according to this table the attenuation coefficient at 6 keV is about 23 cm$^2$ / g. With the density of air at about 1.2 kg / m$^3$ or 1.2 mg/cm$^3$, the attenuation length in air is 0.027 cm$^{-1}$. That means none of that radiation would reach very far. The local air would be massively ionized, but would then re-emit energy at progressively longer wavelengths; at 1000 miles you would be quite well shielded.
Clearly, getting very close to such a hot body would kill you, but at 1000 miles you get "just" the power of the sun - which you should be able to survive. Just don't look directly at the pin - you will probably go blind.
And given the power needed - no, you can't make (and keep) a small blob of matter that hot.
A small atomic bomb converts about 1g of its mass into energy.
So even in the best (worst?) case of a pinhead being perfectly turned into energy this is only about the amount of energy needed to severely redevelop a small city's downtown. It certainly didn't kill everyone within 1000mi
It depends how quickly you heated it. If you did not do it instantly like within a fraction of a second, but lets say took an hour to do it. the real small mass of steel that you speak of would gradually go through the following phase changes as it was heated up.
solid to liquid
if somehow confined to the pinhead volume:
liquid to gas
finally gas to plasma
Since according to Boyle's law, the volume of a quantity of gas increases with temperature, you would have to confine the original pinhead mass somehow to stay in that pinhead volume, and even more so as the mass changes form gas to plasma. In any sense, a gradual heating will not cause the thermal shock wave that a rapid instantaneous heating will cause.
Pinhead is made out of steel which is around 80% iron and 20%carbon. While you may get some carbon nuclei to fuse through quantum tunneling at 15 million degrees, this temperature is still insufficient for iron to iron fusion or carbon to iron fusion.
Even if you were able to cause fusion of all the carbon nuclei in this minute sample, you would not have enough energy to accomplish said devastation; not even a fraction of it.
