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What would be the effect of placing an object that cold in an environment that warm? Would the room just get a little colder? Would it kill everyone in the room like some kind of cold bomb? What would happen?

Don't think about how the cube got there, or the air which it would displace.

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    $\begingroup$ Did you try to do a back-of-an-envelope-calculation? $\endgroup$
    – Qmechanic
    Dec 29 '20 at 0:33
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    $\begingroup$ Also, don't think about pink elephants. $\endgroup$
    – Richard
    Dec 30 '20 at 22:54
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Nothing overly dramatic, though it would be cool to look at. The cube would very quickly become covered by a layer of nitrogen/oxygen ice as the air which came into contact with it froze. Further away, you'd see condensation of water vapor into wispy clouds, which would swirl around the block due to the air currents generated by the sudden pressure drop.

Other than that, as long as you aren't in immediate thermal contact with the block, you wouldn't notice much other than that the room cools down. Here's a video I took of a vacuum can that was just removed from a dewar of liquid helium at 4 kelvin. It's maybe 5 kg of copper, not 10 kg of lead, but I'd say that's close enough to get the idea.

You can see one of my coworkers climbing down into a pit below it; he had to be careful not to bump his head on it, which would have really ruined his day, but there was no fatal cold bomb :)

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    $\begingroup$ @NorbertSchuch Yes, but the latter is proportional to the former, and the specific heat capacities of iron and copper (out of which the vacuum can is mostly made) only differ by about 10%. $\endgroup$
    – J. Murray
    Dec 29 '20 at 0:49
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    $\begingroup$ You should at least specify that the vacuum can is made entirely of copper. $\endgroup$ Dec 29 '20 at 1:05
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    $\begingroup$ @NorbertSchuch Done. I would add, though, that my answer was not at all intended to be quantitative. If the can had been made of aluminum or stainless steel or whatever other material you’d like, the answer would have been the same. $\endgroup$
    – J. Murray
    Dec 29 '20 at 1:15
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    $\begingroup$ "it would be cool to look at it": not sure if clever pun or ambiguous term $\endgroup$
    – YSC
    Dec 29 '20 at 8:12
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    $\begingroup$ @stackzebra Remember that scene from A Christmas Story? Your fingers would briefly feel stuck to it, and when you pulled them away you'd likely be leaving some of yourself behind. Even if that doesn't happen, you'll experience frostbite on your skin. You wouldn't notice for a minute or so, but then it would become .... very unpleasant. $\endgroup$
    – J. Murray
    Dec 29 '20 at 13:32
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The heat capacity of iron at room temperature is 0.444 J/K per gram (it changes with temperature, but let us leave that aside). That means it will want to absorb about 1,332,000 J of heat. That is a lot, but a bathtub with 300 litres of 40 °C water (about 10 degrees above a 300 K) will have about 12,900,000 J of internal energy - if you dumped the iron into the warm water it would not become cold (ignoring losses to the environment).

The real issue is how fast it would cool. Convective heat transfer from air to the metal gives a heat flow $\approx hA(T_{hot}-T_{cool})$ where $h\approx 10$ to 100 W/(m$^2$K) and $A$ is the area (about 0.0726 square meters for the iron). So the heat flow would in theory be on the order of 217.8-2178 J/s initially: sounds fairly impressive, but you would get the same flow in the other direction from a 600 °C cube. That flow will also soon slow, since the cube would be surrounded by cold air and whatever it is sitting on.

So the sad news is that the cube would not do anything super impressive. It would sit there, making air condense like around liquid nitrogen and soon be covered with rime frost. Probably some interesting crackling and perhaps cracking as it changed volume while heating. But no explosions, just a room with cold air along the floor.

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    $\begingroup$ I would assume that frozen air (and water) would decrease heat flow relatively quickly. $\endgroup$
    – Michael
    Dec 29 '20 at 14:37
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    $\begingroup$ 300 liters would be an unusually large bathtub. $\endgroup$ Dec 30 '20 at 8:06
  • $\begingroup$ @pericynthion these people think otherwise badeloftusa.com/ideas/how-much-does-bathtub-hold $\endgroup$ Dec 31 '20 at 0:12
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    $\begingroup$ @pericynthion Your bath tubs must be tiny then. Even I underestimate my bathtub at 4' x 2' x 1.5' that still works out to be 324 litres. $\endgroup$
    – DKNguyen
    Dec 31 '20 at 5:08
  • $\begingroup$ I wonder why this wasn't accepted as the answer. $\endgroup$
    – Helen
    Jan 21 at 20:15
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There are about 30 m3 of air in a room, and the density of air is about 1.2 kg/m3. So the mass of air in the room would be about 36 kg. The heat capacity of air is about 1 kJ/kg-K. So the mass times heat capacity of the air is about 36 kJ/K. The mass times heat capacity of the 10 kg of iron is about 4.5 kJ/K. So, if the room started out at 300 K, the final equilibrium temperature (neglecting the walls) of the room would be about $$\frac{(36\ kJ/K)(300\ K)}{36\ kJ/K + 4.5\ kJ/K}=267\ K$$

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    $\begingroup$ This ignores the heath capacity of the walls etc. $\endgroup$
    – fqq
    Dec 29 '20 at 0:45
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    $\begingroup$ @fqq Isn't that what I said? $\endgroup$ Dec 29 '20 at 0:50
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    $\begingroup$ That is quite chilly-below freezing-but the air would pull heat from the solids around it and also pull more warm air in from outside the room. $\endgroup$ Dec 29 '20 at 6:07
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    $\begingroup$ I like those back-of-the-envelope calculations. $\endgroup$
    – Jens
    Dec 29 '20 at 8:47
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    $\begingroup$ In other words : not a big deal. If you ventilate 15 minutes during a cold winter day, with all windows open, the room air temperature sinks very fast and can reach close to 0°C. Once you close the windows, the room temperature increases fast again, even without any heating, simply because the building has a much higher thermal mass than the air inside. $\endgroup$ Dec 29 '20 at 10:58
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Let's crunch a few numbers.

As Anders observes, "The heat capacity of iron at room temperature is 0.444 J/K per gram (it changes with temperature, but let us leave that aside)."

So your entire mass needs to absorb 4440 joules per degree K it rises. Just getting that to 273K (freezing) will call for (coarsely) 1212120 joules.

Water has the impressive 4.18 joules per gram per degree C of heat capacity. Its enthalpy of vaporization is 2265 J/g which means it must give up that much energy simply to liquefy. Its enthalpy of fusion (freezing) is 334 J/g. So that's 2599 J/g water vapor must give up to freeze, plus another 26.85x4.18 = 112 J/g to cool from 300K to freezing, totalling 2711 J/g.

Why am I talking about water?

Because water will be doing all the heavy lifting here. After all, water and its potent enthalpies moderate the temperature of this planet!

How many grams of water vapor must freeze to bring the block to freezing? 1212120 J / (2711 J/g) = 447 grams.

Almost exactly 1 pound of water. I should've used British Thermal Units!

Is that even in the room? Well, if the air was 27C and at 100% relative humidity, it would contain 27 g per cubic meter. 16.5 cubic meters would hold that much water. Say we're at 50% RH, 33 cubic meters, that's a rather modest bedroom! So the room surely has enough water to do the entire job.

The sequence of events

Yes, instantaneously, it will freeze the nitrogen and oxygen. But very quickly, that will re-vaporize as it exchanges heat energy with water vapor, which eagerly freezes in its place.

This ice will act as an insulating layer to slow heat transmission further. It will expand and expand. Though it won't be terribly thick: after all, we're dealing with less than a pint/half litre of water. 1 water bottle.

Eventually, this will slow down. Even before the iron core has reached 273K, water on the outside will start being melted by the heat in the room, either convecting from objects in the room, or the thermostat has switched on and made a call for heat.

In room air conditioned to 27C, it is easier for water to gather the 334 J/g needed to melt than the 2265 J/g needed to boil, so once the water has melted, it infuses into the (now dry) air much more slowly.

TLDR: You get a puddle.

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    $\begingroup$ "(coarsely) 1212120 joules." If 5-6 significant figures is "coarse" to you, what's precise? $\endgroup$ Dec 31 '20 at 21:34
  • $\begingroup$ @RussellBorogove Oh, it's the source data that's coarse, the 0.444 J/K/g number as mentioned in that quote. To mutilate Mark Twain, "I used a long number because I didn't have time to write a short one" lol. $\endgroup$ Jan 1 at 0:42
  • $\begingroup$ "the heat in the room". Please don't. $\endgroup$ Jan 1 at 4:17
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@PeterMortensen suggests it will result in frozen oxygen and nitrogen, but misses the obvious: Water vapor.

This cup is full of frozen CO2: https://www.youtube.com/watch?v=9OII401xcPI

I think it will be hard to tell if the cube is 0 K or 194 K just by looking at it.

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A lot of conjecture in this answer, but I thought I'd highlight a couple of other risks.

If the room is small and air tight, the problem is likely to be that the block freezes the air, creating a partial vacuum and killing the occupants with a slow vacuum exposure.

With a bigger room, or one with a way for a little air to get in it could potentially still be pretty bad. Air will be sucked in and solidify on the block to begin with. Then as an insulating layer of air ice forms, it will liquefy into a pool on the floor. As it warms, nitrogen will boil off first, leaving the air seemingly breathable, but with no (or very little) oxygen. You feel the need to breathe when the CO2 level in the blood is high, not when the oxygen is low. Therefore you would be likely to suffocate without even realising it.

If you survive that, beware. You now have a puddle of liquid oxygen on the floor with all the explosive fun that can lead to.

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    $\begingroup$ Hi Geoff, welcome to Physics.SE and thanks for contributing. Generally speaking, answers that contain demonstrable arguments are preferred over answers that rely mostly on conjecture or speculation. $\endgroup$
    – kaylimekay
    Dec 29 '20 at 11:51
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    $\begingroup$ I think other answers with back of the envelope calculations are closer to the truth. To raise the temperature of all of the iron to 90k so oxygen won't liquify, the iron needs to absorb about 400,000J. At the high end heat flow would be about 2,000 J/s, so you have 3 minutes at least. Phase changes use a lot of energy, on the order of 200,000J/kg. With 36kg of air in the room, at most 2kg (1/18th) would liquify. $\endgroup$ Dec 29 '20 at 12:14
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    $\begingroup$ A vacuum sufficient to kill someone would be impossible to generate in most rooms without enormous pumping capacity - normal rooms are much too leaky. If you could, they would not be able to support the vacuum structurally and would likely collapse/implode. $\endgroup$
    – J...
    Dec 31 '20 at 0:51
  • $\begingroup$ Counter-conjectures do not count as rebuttal of primary conjecture; especially when the primary conjecture is extensible as in - 'observer suffocated calmly for lack of oxygen' but was not able to collect all the proof on account of preempted death ... as per conjecture. hence conjecture remains positively true. $\endgroup$
    – MKhomo
    Jan 20 at 18:57

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