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According to David Tong’s notes on statistical physics on pages 139 to 140, a superheated liquid lies on an isotherm between the spinodial curve and the coexistence curve. Hence, does this mean that the superheated liquid has the same temperature as an otherwise ordinarily boiling liquid in equilibrium with its gas phase?

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The superheated liquid has a temperature above an ordinary boiling liquid at the same pressure. That's the "super" part of "superheated".

If you take a cup of water, and boil it thoroughly in a microwave (ten seconds or more), then you'll boil out most (or possibly all) of the dissolved air. This removes nucleation sites for boiling. Now let it calm down, and hit it again for 10-60 seconds, depending on your microwave.

It should be superheated. Dropping a tea bag into it will be exciting (and can lead to scalding -- be careful, and do not hold the cup in your hand when you do this). If it's really superheated, then dropping in the tea bag will make it boil violently. It's hard to describe, but if you look at the surface just right it'll look oddly rippled -- you can tell by looking that it's superheated, but it's definitely a look, not something to be described by words.

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  • $\begingroup$ If the surface of the water looks "oddly rippled," then I'd wager that the water is in motion (i.e., there's convection currents circulating in the cup). I don't know what you could add to the water to reveal the motion without triggering sudden, violent boiling though; and I'm not eager to try the experiment without a full-face shield, a heavy waterproof apron and heavy, waterproof, elbow-length gloves; none of which I currently own. $\endgroup$ – Solomon Slow Jun 6 at 21:06
  • $\begingroup$ Oh, you don't need that stuff unless you've forgotten that you haven't made tea yet, and just put creamer into the superheated water. Then the creamer gets to the bottom of the cup and initiates boiling down there, making your tea cup into a boiling water cannon. (It turns out that many office coffee makers dispense de-aerated water, which is just a bit cold for perfect tea -- which is how I found out about the microwave superheated water experiment in the first place). $\endgroup$ – TimWescott Jun 6 at 21:10
  • $\begingroup$ I know somebody who was scalded in a kitchen accident. I will not intentionally re-create the superheated tea cup trick with any less safety equipment than what I mentioned above. $\endgroup$ – Solomon Slow Jun 6 at 21:31
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    $\begingroup$ By the way, @alfred, I recommend that you do NOT try the microwave oven experiment where you try to superheat water. Superheated water is extremely unstable, and it is very easy to get scalded. $\endgroup$ – David White Jun 6 at 21:45
  • $\begingroup$ @david thanks! That sounds fairly dangerous indeed. My main confusion about superheated water is that somehow it is on an isotherm, which means it has the same temperature on the pv plane as any other point on the isotherm. This seems to suggest that the liquid actually has the same temperature regardless of whether it is superheated. Is my reasoning wrong? $\endgroup$ – alfred Jun 7 at 17:49
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Superheated liquid is a liquid which is not undergoing phase transition into gas in bulk (=boiling) even if its temperature is higher (that's why it is called superheated) than the minimum temperature necessary for this phase transition to occur (high enough temperature means high enough vapor pressure which can overcome external pressure). Superheated liquid is metastable state, a strong enough mechanical excitation or introduction of nucleation facilitators (dirt, radiation, charged particles) can destroy it and the phase transition then will occur.

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  • $\begingroup$ If the superheated water is on the same isotherm as boiling water? Why should the superheated water have a higher temperature? $\endgroup$ – alfred Jun 7 at 17:51
  • $\begingroup$ It is not "on the same isotherm as boiling water". $\endgroup$ – Ján Lalinský Jun 7 at 20:20
  • $\begingroup$ It seems to me that boiling or internal evaporation is a phenomenon accompanying a first-order liquid-vapor phase transition. The first-order phase transition is the transfer of the interphase boundary. If the threshold for the formation of vapor bubbles in a liquid is sufficiently small, then even at low satiety (overheating), a boiling will be observed as a way to increase evaporation (due to the increase in the interface area). $\endgroup$ – Aleksey Druggist Jun 7 at 22:25

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