thermodynamics - 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?

Answer

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:

1. 
   

   to heat the water

   
   


2. 
   

   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.