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I'll give an alternate view of how second order phase transitions may look like. Let us study a parameter $F$. If there is a second order phase transition in $F$, then the second derivative would be discontinuous but won't diverge, as for the heat capacity below: and first derivative should look like: Interestingly, the plot of $F$ itself may be ...

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You need to define boiling more accurately. For nucleate boiling it will actually slow the process down as agitation mixes the system. For a full boil less rocking does decrease your energy loss due to sound but thats likely minimal. In reality the agitation will likely decrease the time to a full boil by both dispersing the nucleated regions increasing ...

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The FRG can be thought of as a modern version of Wilson RG, although the technical details are of course very different. But all in all, if one could do all calculations exactly, these different versions would all be the same. Now, about these technical differences. In Wilson RG (and in Polchinski's functional version) one work with a low energy action for ...

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In single phase system, you can have $$dQ =m C_p dT$$ or $$C_p = \frac {dQ}{m dT}$$ When two phases co-exist, the heat added will not increase system's temperature. The heat is consumed by latent heat for phase changing from for example water to vapor. So if you still use, $$C_p = \frac {dQ}{m dT}$$, you have finite value for dQ but you have zero for dT. ...

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It is indeed possible to change between these phases adiabatically. Since, as you noted, the ground state changes between being a superfluid and a Mott insulator, starting in the ground state and making an adiabatic change means that you track that change in state by definition. Note that this diagram is only formally true for the Grand Canonical ensemble, ...

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While the answer of wbeaty is very interesting in showing points relevant in practice, I think all the answers are still missing an important and simple theoretical point, which you should consider to understand the process. vapour pressure does mean two different things as used above. First, the pressure, the existing water vapour would have (if it were ...

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When the vapor pressure is equal to the external pressure, there will form a bubble. Not true. Instead, when the vapor pressure is equal to the external pressure, then any existing bubbles will begin growing continuously. And, if no bubbles are already present, then the water will superheat far above the boiling temperature, yet no bubbles will ...

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