When I boil a kettle, what stops all the water from turning (exploding!) in to steam in one go once it reaches 100°C? While making a cup of tea in the office kitchen, a colleague asked me this question and neither of us could answer with any certainty. We're guessing it has something to do with the pressure of the column of water, or temperature differences between the top and bottom, but does anyone know the real reason?
 A: 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 it).  Assuming that the container is a strong one, the water will heat up to a temperature much higher than 100 C before rupturing, meaning that once it does rupture, the water that is exposed to the atmosphere will be substantially superheated.  At that point, very much of the water will immediately flash to steam.  The vigorous expansion that results will hurl pieces of the container in all directions at high velocity, which is why you want to be far away.
In other words, do not try this in your kitchen.
A: Summary


*

*Water's temperature corresponds to the concept of average kinetic energy. The actual molecules exhibit a distribution of various kinetic energies.

*It takes a quantity of energy to boil a quantity of water.

*For some quantity of water to boil approximately instantly, all its molecules would have to be at approximately the same kinetic energy, which is thermodynamically improbable.

*Therefore, a quantity of water does not "hoard up" energy as it is heated, and then evaporate suddenly. Rather, there is gradual attrition of the fast-moving molecules.

*Thus, we need the concepts of latent heat of vaporization, together with the concept of thermodynamic temperature to understand why water doesn't boil all at once.


Discussion
What stops the water from exploding all at once when it reaches the boiling temperature is that the notion of temperature corresponds to the average kinetic energy of the liquid.
Water, at its boiling point, is a mixture of molecules which are moving about at various velocities and thus kinetic energies (according to a distribution depicted here). Only a small number of molecules have exactly that energy which corresponds to 100 degrees Celsius. Most have less, or more.
The reason that water heats until the boiling point and then stays there is that those molecules which have a large amount of kinetic energy break free of the attraction which holds them together and evaporate.  This evaporation is a form of negative feedback which keeps the temperature stable: the molecules which leave are invariably ones with high kinetic energy, which leaves behind the slow ones, and that keeps the average kinetic energy (temperature) down.  If the incoming flow of heat is increased, that increases the boiling rate, while the temperature remains stable.
A quantity of water will "explode" if a large amount of heat is applied instantaneously, and/or if there is some other event such as a sudden depressurization. (Of course, for instance, a nuclear explosion can suddenly vaporize a lake.)
That brings us to another reason why boiled water typically will not just explode all at once into steam. Vaporization requires energy: the "latent heat of vaporization". Instant vaporization of a quantity of water requires all the energy to be available at the same time. Whereas the typical heating mechanism used for boiling water supplies heat slowly over time. The time required to boil the water can be calculated as the total energy required to boil it divided by the intensity of the heat being transferred into the water. (The wattage of the heater, modified by the efficiency of heat transfer.)
If some quantity of water requires 10,000J of energy to boil off, and the heater is 1000W, of which 150W is delivered into the water (let us assume that this 15% efficiency is steady over the entire vaporization period, which is certainly unrealistic), then the boiling takes about 66 seconds: $10000J / 150W \approx 66s$ (a Watt of power/intensity is a Joule of energy per second).
However, if we follow this latent heat explanation only, we have no reason to believe that the water could not accumulate heat for 66 seconds and then suddenly vaporize. We need to take into account the varying distribution of kinetic energies of the molecules of water, which makes it vanishingly improbable that they all reach the required energy to escape nearly simultaneously.
A: Since neither of the answers given so far really answers the question, here's my 2 cents' worth : 
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 is on top of a heating element.  Indeed, you can see with a glass pot of water that the bubbles in fact form at (or very close to) the bottom.
Also, there's a factor that's not so important for kettles, but otherwise can be relevant: nucleation points.  When a phase change (like liquid to gas) occurs, it normally begins at a location that has some sort of disturbance, maybe a speck of impurity in the water, a locally significant temperature fluctuation, a slight imperfection in the surface of the (inside of the) bottom of the kettle, that sort of thing.  That's why, even though lowering the temperature increases the solubility of CO2 in H2O, dropping an ice cube into soda releases a lot of gas: the abundance of nucleation points on the surface of the ice allows the dissolved CO2 to un-dissolve and form bubbles.  
So, heating a mug of water in a microwave where the volume of water is small enough to be uniformly heated, and it's heated from the inside rather than the outside, it is possible for the whole mug to get to 100°C at the same time.  If the inside of the mug is really smooth and the water's really pure, it could even super-heat to a bit above 100°C.  At this point, a slight disturbance (like putting in a spoon or some sugar, or moving it and making waves) can create a nucleation point and lead to much of the mug boiling at the same time.  This is both very cool and quite dangerous: if you go to a big city ER you will find they treat people with badly burned faces due to this effect, once in a while.
A: 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.
This is why you start seeing steam before the kettle reaches 100˚C, why you need to keep adding heat after the kettle reaches that temperature, and why the water in the kettle doesn't all boil at once.
A: Energy is needed to convert water to steam. This is called the latent heat of vapourisation and for water it is 2.26 MJ/kg. So to boil away 1 kg (about a litre) of water at 100 °C the kettle would need to supply 2.26 MJ. Assuming the kettle has a power of 1 kW this would take 2260 seconds.
Given the unexpected interest in this question let me expand a bit on what happens to the water. Suppose we start with water at room temperature and we turn the kettle on. We'll take the power of the element to be $W$ (units of joules per second) so we have $W$ J/s going into the water. This power can be used for two purposes:


*

*to heat the water

*to evaporate (boil away) the water
Let the rate of temperature increase per second be $\Delta T$, then the power used for this increase is $C\,\Delta T$, where $C$ is the specific heat of the water. Let the rate of evaporation be $\Delta M$ kg/s, then the power used to evaporate the water is $L\,\Delta M$, where $L$ is the latent heat of vapourisation. These two must add up to the power being supplied so:
$$ W = C\,\Delta T + L\,\Delta M $$
When we start heating, and the water is cool, the rate of evaporation is very low so we can ignore it and say $\Delta M \approx 0$. In that case we find the water heats up at a rate of:
$$ \Delta T = \frac{W}{C} $$
When the water is boiling the rate of temperature increase is zero because the water can't get (much) hotter than 100 °C so $\Delta T = 0$. In that case we find the water evaporates at a rate of:
$$ \Delta M = \frac{W}{L} $$
So at the start the water is mainly getting hotter at a rate of $W/C$ degrees per second, and when boiling the water is turning to steam at a rate of $W/L$ kilograms per second. In between the water will be both getting hotter and evaporating at some rate lower than these two limits.
A: What should be mentioned in addition to what has been said in the answers already given, is that at  100°C, the vapor pressure of water equals the standard atmospheric pressure. Now, a glass of water at room temperature is not in thermal equilibrium as long as the relative humidity in the room is not 100%. This leads to water evaporating into the room. But the water can only evaporate from the surface. As air molecules cannot easily penetrate the water, this means that the water is held at atmospheric pressure. In the bulk of the water it is not possible for water vapor to form, because the equilibrium vapor pressure is smaller than the ambient pressure.
If we raise the temperature to the boiling point, then  we reach the point where bubbles consisting of water vapor can form and grow larger. So, the water which was metastable in liquid form (it could only slowly evaporate from the surface), will now become unstable to rapidly changing phase. Now, the nucleation of bubbles can be suppressed or enhanced. As mentioned in Jeffiekins' answer, by heating water in a microwave without a turntable, you can heat water in excess of its boiling point. When you take the water out of the microwave, the small shaking will cause it to boil explosively.
A safer experiment is to let a large amount of water (say 4 liters) boil for about 5 minutes and then you wait a few minutes and then you let it boil again. The second time it will boil with less air bubbles in the water, you can see far larger bubbles appearing. Then you switch off the heating and wait a few seconds until the water looks perfectly still. You then poor some macaroni or rice in the water. This will cause the water to boil explosively due to introducing air bubbles.
A: A kettle contains a heating element - either protruding into the water or in the base of the kettle. The heating element heats the water that is in immediately contact to the element through heat conduction.
Since the water is free to move within the kettle and that hotter water is less dense than cooler water the water that is heated by the element rises to the top allowing cooler water to come into contact with the element.
As this continues the water, on average, gets close to boiling point, but the water that comes in contact with the element begins to turn to steam. This happens slowly at first but becomes more violent as the water is heated.
At the point when the kettle's thermostat turns off the kettle the water that hits the heating element is instantly turning into steam, but the rest of the water is still yet to receive enough energy to turn to steam.
So the answer really boils down to, pardon the pun, that the water in the kettle does not uniformly heat up and so each molecule of water can have different energies to the other molecules.
If you could uniformly heat the water then it would, theoretically, all instantly turn to steam. But a kettle does not evenly heat the water, so it does not.
