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Imagine you put a tea bag in a cup and half fill it with boiling water. Then after one minute you take the tea bag out and then fill the cup up to fill with hot water. Will the tea be weaker than if you had filled the water up to the normal level with boiling water in the first place and then left it for one minute before taking the tea bag out?

Intuitively I can make an argument that it should be but is there a mathematical description for how much weaker it would be?

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    $\begingroup$ Why don't you do an experiment? $\endgroup$
    – jinawee
    Jun 8, 2014 at 16:06
  • $\begingroup$ @jinawee I did and that is what inspires this question! $\endgroup$
    – Simd
    Jun 8, 2014 at 16:44
  • $\begingroup$ So if you did do the experiment you must've got some result... What was it? $\endgroup$
    – user43470
    Jun 8, 2014 at 18:11
  • $\begingroup$ @user43470 I mean I made tea and tasted it :) It tastes weaker when you make it in a half cup of water. $\endgroup$
    – Simd
    Jun 8, 2014 at 18:17
  • $\begingroup$ Ohh then maybe it's got something to do with the amount of tea the water can absorb. More water around the bag, more tea absorbed. But I have no clue about the mathematics! $\endgroup$
    – user43470
    Jun 8, 2014 at 18:19

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In answer to your question - the math you seek is somewhere in the Noyes-Whitney equation as pointed out in the following link analysis of teabag dynamics unfortunately the author doesn't expressly calculate the rate of dissolution but the ammount of detail will either A/ give you a good start on figuring it out - or B/ help you decide you don't need to know the exact answer that bad.

My take on it is that the rate of dissolution starts the same in both cases and the value for the bulk mass concentration [Cb] will increase faster for the half cup slowing its dissolution rate faster.

To determin a definative answer you would need to integrate the rate of dissolution for both cups to find out the total ammount of tea compounds delivered into the water on your calculator.

My gut says that brewing in the full cup will be stronger.

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Each substance in the tea leaves will have a certain maximum soluble concentration in water.

If there are enough tea leaves in the bag to reach saturation of a full cup of water for a given compound, then for half a cup only half as much of that compound will dissolve.

For a compound that doesn't reach the saturation point even for half a cup of water, there may be no difference.

So it is not simply that the tea will be weaker. The ratios of the compounds in the tea solution will also be different.

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  • $\begingroup$ Another minor factor: tea leaves, and the tea bag itself, are porous, and the bag will retain some of the solution after it is removed from the cup. So if you brew a half-cup, remove the tea-laden tea-bag, and then dilute to full volume, twice as much tea solute is lost due to the tea-bag removal as is lost when you simply brew a full cup of tea, and then remove the tea bag. $\endgroup$ Jun 9, 2014 at 1:34
  • $\begingroup$ I had assumed there would be a formula which tells you that the rate of transferable of tea will be proportional to the difference in strength of tea between the water and the tea leaves (or something like that). This would then imply that having less water would slow down this transferral as it would be more concentrated with the same absolute amount of tea in it. Is there nothing like that? $\endgroup$
    – Simd
    Jun 9, 2014 at 20:56
  • $\begingroup$ felix, tea has multiple individual compounds. This reference: monographs.iarc.fr/ENG/Monographs/vol51/mono51-8B.pdf lists hundreds of compounds. Each compound will have its own solubility in water, for a given temperature. The Noyes-Whitney equation mentioned by user263399 says that rate of dissolving will be proportional to the difference between solubility and the concentration in the solvent at a given time. en.wikipedia.org/wiki/Noyes-Whitney_Equation $\endgroup$
    – DavePhD
    Jun 9, 2014 at 21:21

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