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Wikipedia:Reference desk/Archives/Science/2021 March 6

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March 6

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Equation regarding amount of heat in a certain amount of matter.

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Hi, I was looking for an equation which tells me how much a certain amount of heat in a certain amount of matter will translate into.

As a reference, it has been said that the head of a pin as hot as the surface of the sun (10,000 Fahrenheit) could kill a person from 90 miles away.

I was looking for a specific equation though - for example, assuming no other variables - If I was standing 90 miles away from a pinhead (let's say 1.5 mm diameter) as hot as the sun's surface (10,000 F) - how hot would it be where I'm standing? Thanks.--IBBishops (talk) 00:04, 6 March 2021 (UTC)[reply]

Simple answer: The Sun subtends an angle of approximately 0.52° as observed from here on Earth. Since a pinhead at 90 miles will appear much, much smaller, you would feel much, much less radiant heat from it than you feel from the Sun.
If your pinhead has a diameter of 1.5 mm, then at 90 miles it subtends only 5.9×10-7 °. Since you are interested in the solid angle of the heat source, you want the square of the ratio of the subtended angles: Thus the received radiant heat from the pinhead will be (5.9×10-7 ° / 0.52°)2 = 1.3×10-12 that of the sun. Inversely, the sun will feel 770 billion (7.7×1011) times warmer to you. -- ToE 03:50, 6 March 2021 (UTC)[reply]
Nearly as simple answer: Per Solar radius: 1 R☉ = 695,700 km, so your pinhead is 0.75 mm / 695,700 (km/R☉) = 1.1×10-12 R☉ in radius.
Per Astronomical unit: 1 AU = 92955807 miles, so your pinhead is 90 mi / 92955807 (mi/AU) = 9.7×10-7 AU away.
Given its temperature, the amount of heat radiated by the sun (or your pinhead-sun) is a function of its surface area, which varies with its radius squared. The portion of that heat your body intercepts is inversely proportional to your distance squared from the heat source. So the heat you receive from your pinhead will be (1.1×10-12 / 9.7×10-7)2 = 1.3×10-12 that of the sun. Inversely, the sun will feel 770 billion (7.7×1011) times warmer to you. -- ToE 04:43, 6 March 2021 (UTC)[reply]
These two back-of-the-envelope calculations match because they are in essence the same geometrical argument.
But they both ignore the practical issue that if you were to be in 90 mi line-of-site of your pinhead, its radiant heat would be further attenuated by passing through much, much more atmosphere than that of the mid-day sun. -- ToE 04:43, 6 March 2021 (UTC)[reply]
But you want a black-body radiation calculation? Stefan–Boltzmann will oblige.
The total power emitted by your pinhead is 4πr2σT4 = 4π(0.00075 m)2(5.67×10−8 W⋅m−2⋅K−4)(5800 K)4 = 454 watts.
(That's slightly less than 1/3 that of a typical 1500 W electric space heater!)
Divide that be the area of a 90 mi radius sphere = 4π(90 mi)2 = 2.63×1011 m2 to determine the irradiance of 1.72×10-9 W/m2.
The solar constant is 1361 W/m2, which is 791 billion times greater than the calculated irradiance of the pin, which is a close enough match given the precision of these calculations.
(Both the solar constant and the calculated irradiance of the pin do not consider any atmospheric attenuation.) -- ToE 05:48, 6 March 2021 (UTC)[reply]
But you asked "how hot would it be where I'm standing?" and I instead answered what the radiant flux density would be.
We could instead calculate your planetary equilibrium temperature as if you were a rotating (or heat conductive) black body in a universe empty of everything else except your 5800 K pinhead 90 miles away. How warm would it keep you?
 
Where:
Aabs/Arad is the ratio of your absorption area to your radiation area. In this case it is 1/4, which is the ratio of the area of the disk, πr2, you present perpendicular to the pinheadlight to that of your entire surface area, 4πr2, with which you reradiate that energy via your own black body radiation. (We are determining your equilibrium temperature where those energy fluxes balance.)
L/(4πD2) is the irradiance we calculated above, with L being the total luminosity of your pinhead and D the 90 mi between you two.
The 1/σ and fourth root are from us solving the Stefan–Boltzmann law for temperature.
And for a black body albedo a = 0 & emissivity ε = 1 so those terms don't appear.
Thus Tyou = ( (1/4)(1/(5.67×10−8 W⋅m−2⋅K−4))(1.72×10-9 W/m2) )1/4 = 0.295 K.
So as hot as that pinhead is, it does almost nothing to keep you warm. (Though there are colder spots in Florida.)
If you wish to recalculate for objects of different sizes and distance, it's much better to substitute in our luminance formula and simplify.
 
Redoing our calculation:
Tyou = ( 0.00075 m / (2 ⋅ 144841 m) )1/2 ⋅ 5800 K = 0.295 K.
Más sabe el diablo por viejo que por diablo, and the parallel moral here is that the power of the sun comes more from its enormous size than simply from its temperature. -- ToE 14:53, 6 March 2021 (UTC)[reply]
None of the above calculations consider how your pinhead manages to hold together at such a temperature.
Nor do they consider the 454 watt power source which maintains it temperature.
Without power input it will quickly dim due to its tiny surface-to-volume ratio.
V = (4/3)π(0.075 cm)3 = 1.77×10-3 cm3.
If we assign it the density and specific heat capacity of steel, roughly ρ = 7.87 g/cm3 and c = 0.466 J/(g⋅K),
and we assume that specific heat capacity remains constant across a wide range of temperatures and there are no phase changes to deal with,
then your pinhead's heat capacity will be 1.77×10-3 cm3 ⋅ 7.87 g/cm3 ⋅ 0.466 J/(g⋅K) = 6.48×10-3 J/K, and even multiplying that by the full 5800 K gives us only 37.6 J. So it clearly can't maintain it's 454 watt luminous output for but a fraction of a second, and it will rapidly dim.
This is no different than a pinhead of steel being blasted by the 40,000° F jet from a plasma cutter. It will be bright but ephemeral. If you are in the workshop, then you should be wearing welding goggles, but 90 miles away, you won't even be aware of its existence.
Compare this to the sun where, other than the sudden drop in neutrino flux, we wouldn't notice anything for 10,000 years were its fusion to somehow suddenly stop. -- ToE 15:58, 6 March 2021 (UTC)[reply]
As pointed out by ToE, the solid angle has to be similar to that of the Sun. This means that at a large distance the object needs to be huge to lead to enough radiant flux. Take e.g. a fireball from a large impact. Or large gas fires: "The fires from the wells and the oil and gas lines (all of which ruptured, one by one) had produced flames with a height of about 200 metres and a peak rate of energy consumption of ~100 gigawatts, three times the rate of UK total energy consumption." In this case the radiant heat was so large that it was felt inside rescue helicopters at a distance of 1 km away. Count Iblis (talk) 02:59, 7 March 2021 (UTC)[reply]

RBMK part 2

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So I've read that paper about the Chernobyl disaster which was linked in response to my previous questions about the RBMK reactor (where the hell did the article go?!), and while its main hypothesis is not plausible (it claims there was an actual nuclear explosion during the disaster, among other implausible claims), it does bring up at least one salient point: it says that the RBMK main circulation pumps have a built-in low-voltage/low-frequency trip (which would explain why they wound down so fast all 3 times after the turbine was tripped), and that this, rather than the graphite tips of the control rods, was the reason for the thermal runaway at Chernobyl. So my question is: (1) is it true that the pumps trip automatically due to low voltage and/or frequency, and (2) if so, is it plausible that at least part of the role which the INSAG-7 report attributes to the graphite tips in causing the reactivity spike was actually played by the abrupt stoppage of the circulation pumps (as the paper alleges)? 2601:646:8A01:B180:7446:DB1C:3033:BF8E (talk) 03:22, 6 March 2021 (UTC)[reply]

To answer your first question, see RBMK. Since the "R" part of the acronym stands for "Reaktor", I guess calling it the "RBMK reactor" or "High-Power Channel-type Reactor reactor" is considered tautological. {The poster formerly known as 87.81.230.195} 2.125.75.168 (talk) 05:32, 6 March 2021 (UTC)[reply]
We have an article on that: RAS syndrome. -- ToE 06:08, 6 March 2021 (UTC)[reply]
Sorry, the article does not answer my first question -- it says that the coolant pumps are prone to cavitation (which I already knew from reading said article, and which was specifically mentioned in both INSAG-1 and INSAG-7 as a contributing factor to the disaster), but it says nothing about the power being cut to said pumps! 2601:646:8A01:B180:7446:DB1C:3033:BF8E (talk) 07:57, 6 March 2021 (UTC)[reply]
The first question you asked in this thread seems to be "where the hell did the article go?!". It may not have been numbered, but since it was the first question you asked, it seems reasonable to call it "your first question". The article is called RBMK not RBMK reactor for the reasons stated which is an answer to that question. (It doesn't look like it was ever called anything else [1] although I don't know if any redirects were deleted. It looks like it was only ever wikilinked under RMBK in Wikipedia:Reference desk/Archives/Science/2021 February 22#RBMK reactor.) No one said the article answers any of your other questions AFAICT. To avoid confusion, it might be better to ensure you number all your questions, or don't number any. Nil Einne (talk) 14:28, 6 March 2021 (UTC)[reply]
I did number all my questions: (1) is it true that the pumps trip automatically due to low voltage and/or frequency, and (2) if so, is it plausible that at least part of the role which the INSAG-7 report attributes to the graphite tips in causing the reactivity spike was actually played by the abrupt stoppage of the circulation pumps (as the paper alleges)? 2601:646:8A01:B180:21E9:FB1D:6AAA:EDD3 (talk) 08:12, 7 March 2021 (UTC)[reply]
Where I'm from, "where the hell did the article go?" is a question, numbered or not. Should we refer to it as your zeroth question?  --Lambiam 11:23, 7 March 2021 (UTC)[reply]
It's worth asking the question why is the claim that there was "an actual nuclear explosion" implausible? It's not a great practice in science to just think something is implausible because "it seems patently absurd," or some such. Why is it actually implausible? --OuroborosCobra (talk) 18:44, 7 March 2021 (UTC)[reply]
Because for one to occur, it is not enough for the reactor merely to become prompt-critical -- it has to be prompt-critical by a very wide margin so that there is an energy spike of trillions of joules within a small fraction of a second, and the power output at 1:23:46 on that morning was nowhere even close to that (not even by the highest estimates!) Anyway, the question was about the pumps, not about what exactly exploded, right? 2601:646:8A01:B180:DDEB:56CE:4137:67EA (talk) 08:53, 8 March 2021 (UTC)[reply]
What if it fizzled? I'm not saying it was a full on, successful nuclear bomb. --OuroborosCobra (talk) 19:07, 8 March 2021 (UTC)[reply]
Nuclear fuel for power plants, of any type, is not in the same form as it is for nuclear bombs. There's not really anything you could do to it, short of reprocessing it into a completely different form factor, to make it explode. As noted at criticality accident, "Though dangerous and frequently lethal to humans within the immediate area, the critical mass formed would not be capable of producing a massive nuclear explosion of the type that fission bombs are designed to produce. This is because all the design features needed to make a nuclear warhead cannot arise by chance." Nuclear bombs require seriously complex engineering to make them go boom, they simply don't happen by chance. --Jayron32 13:22, 11 March 2021 (UTC)[reply]
My point exactly -- you're just better at putting it into words than I am! 2601:646:8A01:B180:D1FF:AC7:8199:8F42 (talk) 12:49, 12 March 2021 (UTC)[reply]
[un-indent] Anyone happen to know the answers to my original questions about the pumps??? 2601:646:8A01:B180:D1FF:AC7:8199:8F42 (talk) 12:50, 12 March 2021 (UTC)[reply]
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