Apologies if this is a chemistry question

I've read that drinking water contains dissolved oxygen to the tune of $10\:\rm{ppm}$.

I've also read that raising the temperature of water will remove some dissolved gases.

Given standard temperature and pressure, what percentage of disolved oxygen will be removed from the water by boiling?

Specifically there's a running arguement in the office that when making tea one should never reboil the kettle, the reason being that the tea will taste bad because the dissolved oxygen has been removed.

Thanks for your help, apologies if this question doesn't fit.

  • $\begingroup$ I never knew that dissolved oxygen can make stuff taste different. Then again, I don't drink tea/coffee often (I value my sleep), so I may not know. $\endgroup$ Mar 23, 2012 at 16:38
  • $\begingroup$ I personally think this is on-topic -- physical chemistry is just physics which is used more by chemists. It's still physics. $\endgroup$ Mar 23, 2012 at 16:53

2 Answers 2


enter link description here
(source: engineeringtoolbox.com)

The graph(blue lines are what you want) only goes till $50^o C$, but we can extrapolate and say that the oxygen level will become a fifth or so by the time the water reaches its boiling point.

The solubility comes from Henry's law, but I don't know the temperature dependance of the proportionality constant $k_H$--I'll check it out tomorrow and edit it in..

  • 1
    $\begingroup$ Can you address the last paragraph of the question? Is re-boiling the kettle likely to make much difference to the amount of dissolved oxygen? Or, to put it another way, when you raise the temperature of the water, how long does it take for the oxygen to reach its new level? $\endgroup$ Sep 2, 2015 at 19:48

I was just googling that (so not saying I know a lot about it) but colder water holds more gases and so the water would probably take the gases back after it has cooled.

When you have water (or any liquid) in a container there is a constant process of some of the liquid evaporating into vapor and some vapor condensing into liquid. In equilibrium, these two processes exactly cancel and you have liquid with some vapor over it. The equilibrium pressure of that vapor depends on temperature(as well as the particular liquid).

As you heat up the liquid, the vapor pressure rises. At 100 degrees Celsius (212 F) the vapor pressure of water is about equal to the atmospheric pressure at sea level. At that point, as water evaporates inside the container, the vapor pressure inside the bubbles is high enough to keep the bubbles from collapsing again from the pressure of the water around it. Then the bubbles rise (why?) and break the surface.

Hence boiling.


  • $\begingroup$ Hi user150868, we typically prefer our material to be self-contained (i.e., not links only but a description of the relevant part and link as a reference), so I copied the relevant portion of the link into the body for you. $\endgroup$
    – Kyle Kanos
    Apr 4, 2017 at 10:23
  • $\begingroup$ Your linked paragraph isn't relevant. Water vapour is very different from dissolved gasses. $\endgroup$
    – JMac
    Apr 4, 2017 at 10:59

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