# Tag Info

98

Energy is needed to convert water to steam. This is called the latent heat of vapourisation and for water it is 2.26MJ/kg. So to boil away 1kg (about a litre) of water at 100ºC the kettle would need to supply 2.26MJ. Assuming the kettle has a power of 1kW this would take 2260 seconds. Given the unexpected interest in this question let me expand a bit on ...

26

There are three phenomena that occur before vigorous boiling of water that produce sound. 1) Air dissolved in water on heating forms small air bubbles at the bottom of the container. These air bubbles get released from the bottom of the container on reaching a sufficient size. The process of release produces a sound of frequency ~ 100Hz. 2) On boiling, ...

21

I have read that true steam is clear (transparent) water vapor. According to this theory, the white "steam" you see is really a small cloud of condensed water vapor droplets, a fine mist in effect. So what you are seeing is not more steam, but more condensation and more mist. The speed with which the steam/vapor/mist rises and disperses may also change.

11

Yes, of course, the freezing point will decrease by the pressure developed, while part of the water freezes. But do not underestimate the pressures! In such an experiment easily some thousand bares may be developed. (Depends on the rigidity of the vessel and the volume of water) Here is a video showing how freezing water cracks a cast iron sphere. ...

11

Since neither of the answers given so far really answers the question, here's my $0.02: between convection (the flow of water of various temperatures around the kettle), and the fact that the heating element is at the bottom, the water is at various temperatures at various parts of the kettle at any time. Usually, the hottest is at the bottom, if the kettle ... 10 I'll give a very qualitative answer / overview. The classification 'first-order phase transition vs. second-order phase transition' is an old one, now replaced by the classification 'first-order phase transition vs. continuous phase transition'. The difference is that the latter includes divergences in 2nd derivatives of$F$and above - so to answer your ... 10 If the metal pan was cool then you would expect to see water droplets staying in the same place once any original movement had dissipated. You would have a combination of cohesive forces within each water droplet and adhesive forces between the water and metal surfaces. With the metal having a temperature well above the boiling point of water, the water ... 9 Generalities on Conformal Invariance In two dimensions, a lot is known / conjectured about statistical models at criticality. For instance, at$T_c$, the spin configuration that you see will not only be self-similar (what others here have been calling "fractal") but actually fully conformally invariant (in the continum limit); that is, the probability ... 9 It's certainly possible for ice to sink in water under the right conditions. The diagram this section of Wikipedia's ice page will show you the conditions under which the various types of ice can form. Most of the "exotic" ones such as XII will form only at pressures greater than around 200MPa. These high-pressure forms are all denser than water, so they ... 9 Temperature is a measure of average kinetic energy. When you have a kettle of water at 100˚C, some of the water molecules will have more-than-average energy, and some will have less. The more-than-average molecules are the ones that will turn to steam, carrying off their energy and lowering the average (and thus the temperature) for the remaining water. ... 9 Boiling is clearly not a surface phenomenon. But vaporising is. Boiling happens at all the points inside the liquid whereas when vaporising only the molecules at the surface escape into the space above. And it is true that a liquid boils when its saturated vapour pressure equals external (room) pressure. But it is not to be confused with vaporising. ... 8 As mentioned in the comments, this is an instance of supercooling. When you cool a liquid below its freezing point, the molecules are still moving around quite a lot and any two that stick together are likely to be broken up by a subsequent impact. Liquids freeze better when the molecules have something to latch onto -- either a block of the same ice they ... 8 Let's define temperature to be a measure the kinetic energy of the atom. A single atom has limited numbers of ways it can store energy. It can translate in X, Y or Z. It can't really rotate (well it does rotate, but it takes so little energy to make it rotate that we can ignore it). It can't vibrate. It does have electronic modes where adding energy can ... 8 In vacuum and with only the particles we know about the answer is no. Let's look at the symmetries we know exist in nature:$SU(3)$colour: confined, only colourless states exist below the QCD phase transition$SU(2)\times U(1)_Y$electroweak: Higgsed to$U(1)_{EM}$electromagnetism$U(1)_{EM}$: Here we have opportunity. See below...$U(1)_{B-L}$: Global ... 7 A simple material will not undergo a liquid to solid transition as the temperature is raised. When you see this it means somthing more complicated than a simple phase transition is going on. In the example of egg white, what you are seeing is denaturation of the protein albumin. The heat causes the protein to lose its tertiary structure then form cross ... 7 In physics, critical behavior means the behavior in which there are no localized boundaries between phases. More quantitatively, the correlation length diverges (is infinite). For example, at the critical point of water, one sees clouds of vapor at all possible length scales. This is only possible because the relevant laws of physics around this point ... 7 No, the boundary doesn't suddenly "end" or "fade away", as the liquid-gas boundary fades away near the critical point. Instead, the sudden end indicates that many other things may happen in the region of these extremely high pressures and the diagram doesn't want to discuss those because they're outside the limits of interest of the author of the diagram. ... 7 Neutron degenerate matter can undergo a phase transition to a superfluid state. The process is thought to be analogous to Cooper-pairing, but the coupling interaction due to the long-range nuclear force is of order 1 MeV, so can occur at temperatures below about$10^{9}$K in neutron star interiors. The neutrons (fermions) form bosonic pairs in an analogous ... 6 The amount of heat added to the system is the integral of the specific heat wrt temperature: $$Q = \int C(T)dT$$ So in the link you give it's just the area under this graph: Although it's true that the specific heat tends to infinity at the lambda point it does so sufficiently suddenly that the area under the graph remains finite. That means the ... 6 If you want to see all water in a container immediately turn to steam, you need a transparent container that you can seal. Fill the container 50% with water and tightly seal it. Place the container on an open flame and let it heat up. While it is heating, walk far away and watch the container through binoculars from some distance (e.g., 50-100 m should do ... 6 The situation is well represented in the following very pictorial picture but this is a very active field of study. It is interesting to note that a real proof of existence for the critical endpoint (CEP, indicated as a critical point in the figure), both from a theoretical and numerical point of view, does not exist yet. The reason, at least for the ... 6 Option 3. An equilibrium phase transition is a non-analytic point of the thermodynamic free-energy. For a finite number of particles, the free-energy is always analytic. So you cant get a phase transition. Kardar discusses this point. 6 Yes mean-field theory is wrong for the one-dimensional case (and wrong for the two and three dimensional cases as well, where the transition exists but the mean-field approximation gets the wrong critical temperature and exponents). In fact it's a typical first year exercise to solve the 1D Ising model exactly using transfer matrices, and I suggest you look ... 6 I do not think Mainwood makes any argument against what he calls the "theoreticians case", much less a compelling one. The "theoretician's case" is that phase transitions do not exist in finite size systems but only as features which become infinitely sharp in the infinite size limit (also user10001's comment). In fact Mainwood briefly dismisses the case and ... 6 I am curious to know under what conditions of the air pressure(atm), temperature, solute density in the water would cause the Niagara fall frozen? In general, the answer is "a bit lower than 32 Fahrenheit". Here's two things which one might think would come into play, but actually do not to an appreciable extend. Solute concentation The major ... 5 Unlike the exponential, the logarithm function can actually be defined on physical quantities with a dimension, it was discussed here. Not by power series, but that's not "the function" but just one often-useful way of calculating it! Mathematically, the preferred definition is via differential equations or integral identities. Even if you avoid the ... 5 It seems pretty clear that if you take a very diluted subset of, say, the horizontal line through$0$, then you'll be able to make a Peierls argument. For example, put$h=+\infty$(worst possible case, amounting to fixing the corresponding spins to$+1$) at all vertices with coordinates (10^k,0), with$k\geq 1$. Then, when removing a contour surrounding a ... 5 Assuming you have a strong enough container to resist the force you will create a whole range of unusual forms of ices 5 The size of these bubbles is growing nearly by the speed of light. So if the boundaries of two such bubbles are close enough to each other, the exponential expansion of the parent space in between them, even if this expansion exists, is negligible relatively to the shrinking distance between the bubbles due to their growth. That's why the bubbles collide ... 5 Let me state first that one should not speak about energies but about free energies, as soon as temperature is not absolute zero. Discussion of energies has in this case no sense. Second, a very simple answer to the original question is pointing out that according to its definition the heat,$Q = T \Delta S$, where$S$is entropy,$\Delta S\$ is its ...

Only top voted, non community-wiki answers of a minimum length are eligible