# How can I calculate the speed at which water evaporates when held at 100 degrees Celsius?

If I have a mixture of some higher bp solvents and water, and I heat it to 100 degrees celcius so that the water in the mixture begins to boil, how can I calculate the rate at which water leaves the solution, assuming the temperature is held relatively constant.

I did some reading into this and found out the humidity plays a role, so I suppose we can assume 10% humidity, though, I haven't been able to find any relevant formulas that are clear cut about this.

The device is 500w, but it has a digital thermometer control, so the hotplate is always at a particular temperature. Not sure how this would interfer with calculating heat transfer based off wattage soley.

$n.b:$ I just realised it's a mixture of water and solvents. I don't think this answer would be correct, then, because heat will be taken up by the solvents too. Also, the boiling point of water may increase in presence of impurities, which means that at a $100^\circ C$, the equation may not be valid!

The mass of water $m$ evaporated depends on the amount of heat $Q$ given to it:

$$Q = mL$$

where L is the enthalpy of vaporisation. For water, $L = 2.26 \times 10^6 J/kg$.

Differentiating the equation, we get: $$\frac{dQ}{dt} = L\frac{dm}{dt}$$.

You're using a water heater: the power of the device is the rate of heat transfered. $500W$ is $500J/s$, so the value for your $\frac{dQ}{dt}$ is 500. Substituting it into the equation will give you the value for $\frac{dm}{dt}$, which will be in $kg/s$!

Of course, the factor of efficiency of the heater plays a role too, but I can't think of a way to calculate that. Anyway the value you get from the equation should be pretty much accurate.

Another way of doing this is using a weighing scale, and calculating how fast the weight decreases.