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Suppose, that in a particular area of the ocean on an earth-like planet, all water molecules are transmuted into an exotic variety of $H_2O$ that is identical to normal water in every way except with a boiling point 95 degrees C lower than ordinary water at any given level of pressure (i.e. a graph like this would be shifted down 95 on the Y axis). Any normal water entering the area is also transmuted, and any exotic water that leaves reverts to normal

Let's assume that the affected area is a cylinder 100km in diameter, extending from the ocean floor 3.5km below sea level (the average ocean depth on earth) to the edge of space. Assume further that it is located in a stretch of open ocean at a latitude similar to, say, San Diego (where average surface temps are above 15C or so year-round).

What would area of ocean look like once it reached equilibrium?

For certain living creatures at or near the surface would die instantly as their blood boils.

Water on the surface would immediately boil away, but it quickly gets complicated deeper down. Atmospheric pressure increases by about 1 atmosphere for every 10 meters of ocean depth, so initially the exotic water below 10-20 meters would remain liquid at normal ocean temps, but then the pressure would decrease as the top layers boil off. At the same time, normal water would be flowing in from outside, but would also boil rapidly. There are probably other effects still that I'm not considering. Probably as fairly-cold water vapor drifts out into the normal region you'd also have a considerable amount of rain or fog along the perimeter.

I've described above the general factors at play, what I can't figure out is how these balance out. What would this area look like from the perspective of a ship crossing into the area?

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    $\begingroup$ If the heat of water evaporation and ice melting point are not handwaved away, water would quickly freeze. $\endgroup$ – Alexander Mar 20 at 21:52
  • $\begingroup$ @JBH Fair point, tried to specify $\endgroup$ – rsandler Mar 20 at 22:08
  • $\begingroup$ To explain @Alexander's observation: water has a massive latent heat of vaporization, about 2.3 kJ per gram, compared to the specific heat of liquid water of about 4.2 J per gram kelvin. Since boiling water can only extract heat from neighboring water, and water starts at 15° C, each little boiling droplet will find itself surrounded by a shell of freezing water... $\endgroup$ – AlexP Mar 20 at 23:13
  • $\begingroup$ @Alexander, between you and AlexP, it sounds like you have an answer... shipping would run into an icy reef... that's actually kinda cool. $\endgroup$ – JBH Mar 20 at 23:33
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    $\begingroup$ I do not understand why the blood of the sea creatures would boil. The temperature of the water would be locked at 5 degrees Celsius, and the steam produced would e at 5 degrees Celsius. That is, the temperature of the water would not exceed five degrees Celsius, unless under pressure. This is definitely not hot enough to boil blood, in fact it would super-cool it in any warm-blooded animal. $\endgroup$ – Justin Thyme Mar 21 at 2:09
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What you need to realize is, just how much energy is needed to turn liquid water into steam: You need 2257kJ to turn a single kg of water into steam. So, with the same amount of heat that can turn 1l of boiling hot water into steam, you can heat more than 6l of water from 20°C to 100°C.

As such, the water at the very surface of the ocean will indeed start flash-boiling immediately. However, assuming a 20°C ocean, this boiling will instantly cool the remaining water to 5°C. For every liter of water that evaporates, over 35 liters of 5°C water remain.

The remaining 5°C water is significantly heavier than the hot water around it, both because of temperature, and because the salt remains in the liquid phase, so the cooled water will also have a much higher salinity. As such, it will try to sink below the surface immediately. Convection will start to constantly supply the surface with fresh hot water that boils off to cool, and then start sinking into the depths of the sea.

After some time, there will be so much liquid water lacking from the area, that the surrounding water starts flowing in from the top. In the end, I guess you will get a giant downward convection of water in your region, and a ring of evaporation around it.

For the climate, the amount of water vapor that's added to the atmosphere means, that you'll get a stationary hurricane above your region. And the strength of this hurricane will simply dwarf all the hurricanes that we know. If normal hurricanes can go up to the fifth category, this hurricane will likely be in category 10.

So, the area that's affected will immediately turn into a no-go area for (robotic) ships and low-flying drones (humans entering the region would die instantly due to our body temperature). The wind and the waves will rip any vessel apart. You may get some nice big and constant waves for surfing some place a few thousands of kilometers away, though.

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  • $\begingroup$ The salt remains in the liquid phase? As in the salt has 'melted'? Or the salt remains in suspension? $\endgroup$ – Justin Thyme Mar 21 at 2:23
  • $\begingroup$ I am not sure you have the 'sinking water' part correct. The sinking water would have to be at 5 degrees, until the pressure increases, in which case the temperature of the water would increase with increasing depth. That is because the water around it would be at a higher temperature, and thus would supply the heat. Eventually, the pressure effects would mean that the surrounding water would be at equilibrium temperature with the 'special' water. $\endgroup$ – Justin Thyme Mar 21 at 2:29
  • $\begingroup$ Another complication with the depth thing is that, as soon as the special water moves out of the cylinder, it reverts back to the normal boiling point, in which case the water that WAS special, at the 'special' temperature, would revert to regular water but still be at a lower temperature then the surrounding regular water. I suspect there would also be interesting effects at the point where the regular water, at lower depths, is just as cold as the 'special' water at boiling-point-under-pressure. The special water would act just like the regular water. $\endgroup$ – Justin Thyme Mar 21 at 2:35
  • $\begingroup$ "it will try to sink below the surface immediately" Cooling and salinity would increase density a little. But steam would decrease it several dosen times. So it would act like airlift and boiled water would go upward. $\endgroup$ – Vashu Mar 21 at 4:36
  • $\begingroup$ @JustinThyme The salt remains in solution, to be correct. Apart from that, what happens will be pretty much what happens in your cooking pot when you boil water: Hot water rises to the surface, forming vapor bubbles along the way, which expedite its upwards progress, and cold water moves down. Since there is no pot bottom in the ocean, the cold water will continue to move down until it reaches a layer of similar density. The renormalization of the water that moves out of the region doesn't change much: The heat has been lost at the surface, the water will remain cold as it passes the border. $\endgroup$ – cmaster Mar 21 at 7:22
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The water will freeze over.

This may not be very intuitive to understand, but the water has very high heat of evaporation, much higher than its heat capacity. And since there is no external source of energy, this heat must come from surrounding water - which would inevitably would make it cooler. So the water temperature would drop below the boiling point, thanks to relatively small temperature differential (5 C vs 15-20 C).

"But why freezing?" - may you ask - "Why the water just don't stay liquid at 5 C?" The key to understand it is that while boiling happens at specific temperature, evaporation happens at any temperature (with different speed). And whenever evaporation happens, temperature of the liquid drops, sometimes considerably lower than the temperature of the environment.

Water will freeze in a vacuum. Why? "Because it's cold!" Wrong. Suppose, we have not a deep vacuum, but rather a very low-pressure gas at room temperature. Water will still freeze, because it will boil at very low-pressure, and boiling will cool it off below the ambient temperature.

In our example, difference between the boiling point (5 C) and freezing point of saltwater (-2 C) is very low. Boiling water will begin to freeze almost instantaneously. Moreover, because this difference is so low, the ice will continue to sublimate at -2 C.

So, in less than an hour, our affected area will become one large ice field. Hot water under the ice would cool off and come into thermal equilibrium with the surface. This ice field would still be producing more vapor that the surrounding ocean, so this area would be enveloped in a fog that is escaping it.

Next, because of ocean currents this ice field would not stay in one place. On the side of oncoming current, new water will be continuously entering the affected area. We will see active boiling (and a lot of fog) at that edge. On the opposite side, ice will drift away from the area. Its active cooling will stop, and it will be slowly melting and breaking.

Overall, this event would not produce any significant local drop in the ocean level.

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  • $\begingroup$ If the area became one large ice field, how could water flow into it? Ice does not 'flow'. $\endgroup$ – Justin Thyme Mar 21 at 2:16
  • $\begingroup$ @Justin Thyme ice floats on water. It's not going to freeze all the way to the bottom. $\endgroup$ – Alexander Mar 21 at 2:54
  • $\begingroup$ @JustinThyme tell that to glaciers... Glaciers are effectively rivers of ice, they flow down hill at a glacial (yes...) pace. Icebergs calving at the shore are the result of those glaciers flowing over the sea and sinking, causing stresses that at some point become too great. $\endgroup$ – jwenting Mar 21 at 8:35
  • $\begingroup$ So more appropriately, regular water would flow in UNDER the ice, and then turn into the 'special water'?. Would this super-sized ice burg of special water float outside of the area, thus becoming regular ice water, or would the resultant currents ensure the special water ice burg is held captive in the area, and remains special water? $\endgroup$ – Justin Thyme Mar 21 at 13:41
  • $\begingroup$ @Justin Thyme this ice field will be pushed by the current, and there is no way assume that there would be an opposite current to keep the ice field in place. Once out, it will be just regular ice. And yes, incoming current will be getting under the ice, causing continuous boiling at that end of the field. $\endgroup$ – Alexander Mar 21 at 16:15
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The outcome would be sensitive to the exact nature of the magic causing the effect.

Water boils at 100°C at standard pressure because that reflects how much kinetic energy the molecules need to overcome the formation enthalpy of intermolecular bonds.

If the molecules are made “less sticky”, then conservation of energy dictates that (a lot of) intermolecular bond energy is released as water flows into the magic zone; by default we’d assume this energy appears as heat. Also, the above arguments relating to the vaporisation enthalpy of water will not apply because the substance in the zone is not water; in terms of molar mass and boiling point, it’s somewhere between water and methane. So when water enters the zone, I think it becomes hot methane and boils explosively, creating a high-pressure fountain of gas. As that gas escapes the zone, it turns into water vapor, at maybe 20°C or so, which condenses and saturates the air with moisture. This lowers the local pressure, while the boiling on the inside of the zone raises it.

I think the net result is a permanent, violent convection cell. Depending on the exact details (which I couldn’t begin to model), it may well be that this becomes a giant stationary hurricane, continually dumping moisture into the air, possibly until the whole planet is covered in storms. A lot of energy is released when water enters the zone, and not all of it is recovered when the water vapor leaves the zone, because (if the surrounding air was not saturated to begin with), some of the water vapor doesn’t condense right away.

I suspect this doesn’t entirely add up thermodynamically – it seems like it continually sucks heat from the ocean and turns it into storm energy – but that’s as far as I get.

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  • $\begingroup$ The question states "identical to normal water in every way", so I assume vaporisation enthalpy argument should stay. $\endgroup$ – Alexander Mar 21 at 3:09
  • $\begingroup$ @Alexander It does say that, but it also says water boils at 5°, so something’s got to give! I just chose to partly ignore that bit of the question. $\endgroup$ – bobtato Mar 21 at 10:32
  • $\begingroup$ +1 for the methane bit. I do think it helps tremendously to envision the water replaced by something with a similar boiling point to the 'special water' that we do understand. When you imagine it turning into another substance and then behaving like that substance, it does help to model the simulation a bit better. However, I posit methyl bromide or phosgene as more appropriate. Even butane. Butane we understand, 'special water' not so much. So the question sort of becomes 'What happens if the water takes on the boiling properties of butane in this confined area?' $\endgroup$ – Justin Thyme Mar 21 at 13:34
  • $\begingroup$ @bobtato normally this is how "handwavium" works - laws of physics are discarded. Unless there are some logical contradictions in the question itself, all should be good. $\endgroup$ – Alexander Mar 21 at 16:20
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The world's oceans will converge on the singularity point, until the sea level has dropped sufficiently so that new water cannot reach it (imagine pulling the plug out of a basin full of water).

"For certain living creatures at or near the surface would die instantly as their blood boils."

Why? The "water" might be boiling, but keep in mind that that doesn't make it hot. Air hasn't killed us yet, has it? ;-)

cmaster's answer is just plain wrong. The proto-water has a lower boiling temperature: If the temperature of the surrounding environment is above 5C, then it has already attained the energy to boil and evaporate. Also, the proto-water isn't going to freeze. That makes even less sense.

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  • $\begingroup$ Welcome to the site Ouroboros, please take the tour and read up in our help centre about how we work: How to Answer The question has a science-based tag, read up to find out the constraints that places on answerers. When you have achieved a little reputation, you'll be able to directly comment on questions and other answers without distracting from your own answwer by doing so. $\endgroup$ – Don Qualm Mar 21 at 0:12
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    $\begingroup$ "energy to evaporate" should come on top of thermal energy contained in a liquid at given temperature. Without external source of heat, any liquid stops boiling and its temperature quickly comes below the boiling point. $\endgroup$ – Alexander Mar 21 at 0:21
  • $\begingroup$ Water (or any liquid) does not boil just because its temperature has reached the boiling point; to actually convert water into vapor you need to put in the latent heat of vaporization. In the case of water, the energy required to vaporize one gram of water which is already at boiling point is about 500 times larger than the energy required to warm the same gram of water by one degree centigrade. This energy has nowhere to come from but from the surrounding water. $\endgroup$ – AlexP Mar 21 at 0:53

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