This post is about the mathematics of making a good cup of tea. I drink a lot of tea and for a long time I would ignore the direction to use water below boiling, say 75° or 85°. Because – if you don’t have a boiler with a temperature setting how do you get water at less than 100°? Wait for it to cool? Come on!

Of course you can do it easily by adding cold water, you just have to figure out how much. While I was waiting for a delivery this morning I thought I’d look into that and it turned out to be a fun little practical application of the scientific method.

Theory 1: the average temperature

Let’s assume tap water is 22° and you add 500ml of it to the same amount of boiling water. My starting assumption was that the result will be 1l of water at the average of the two temperatures, 61°:

\[t_r = \frac{100 \cdot 500 + 22 \cdot 500}{500 + 500}=61\]

Put more generally, my theory was that if you mix water of two different temperatures then the temperature of the result will be the average over the two temperatures, where each temperature is weighted by the volume of water

\[t_r = \frac{100 v_b + t_t v_t }{v_b + v_t}\]

(Here \(t_r\) is the resulting temperature of the water, \(v_b\) is the amount of boiling water which will be 100°, \(t_t\) is the temperature of tap water, and \(v_t\) is the amount of tap water.)

If you measure the tap water temperature and know how much water your tea pot will hold you can solve for the amount of boiling and tap water to use to make any given water temperature. For instance, my tap water is 22° and my tea pot holds 700ml so the formula for how much to boil for a given temperature is,

\[t_r = \frac{100 v_b + 22 (700 - v_b)}{700}\] \[t_r = \frac{100 v_b + 22 \cdot 700 - 22 v_b}{700}\] \[t_r = \frac{78 v_b + 22 \cdot 700}{700}\] \[700 (t_r - 22) = 78 v_b\] \[\frac{700}{78} (t_r - 22) = v_b\]

So to get a pot full of 85° water, the amount of boiling water should be

\[\frac{700}{78} (85 - 22) = 8.97 \cdot 63 = 565\]

Boil 565ml, add the rest up to 700ml from the tap (that would be 135ml) and you’ve got 700ml of 85° water. That’s the theory.

Ok, so having solved the problem I went to make tea, my delivery having still not arrived (spoiler alert: it arrived while I was at the dentist, obviously). It turns out though that the theory doesn’t work. When I did it in practice it turned out that the water wasn’t 85°, it was 77°. Bummer. The theory is too simplistic.

Theory 2: the tea pot constant

The problem was fortunately obvious: I was pouring the water into a cold tea pot. The formula only says what happens to the water itself, it doesn’t take into account that it all ends up in a pot that’s initially at room temperature.

What I thought I’d do then was to model the effect of the cold pot as if it were extra tap water. How much exactly I don’t know, but it seems reasonable to expect that the cooling effect of a particular pot works basically the same way as if you’d added a bit of extra cold water. So we replace \(v_t\) in the formula above, the amount of tap water, with \(v_t + v_p\), the amount of tap water and some additional amount, \(v_p\), to account for the pot,

\[t_r = \frac{100 v_b + t_t (v_t + v_p) }{v_b + v_t + v_p}\]

You have to determine \(v_p\) experimentally, it is a property of a particular pot just like the volume. I determined it by pouring boiling water into the cool pot and then measuring the temperature; it had dropped to 90°. Using this I could find \(v_p\),

\[t_r = \frac{100 v_b + t_t (v_t + v_p) }{v_b + v_t + v_p}\] \[90 = \frac{100 \cdot 700 + 22 (0 + v_p) }{700 + v_p}\] \[90 \cdot 700 + 90 v_p = 100 \cdot 700 + 22 v_p\] \[-10 \cdot 700 = -68 v_p\] \[\frac{10 \cdot 700}{68} = v_p = 103\]

In other words: the pot cools boiling water as much as adding 103ml of extra tap water.

Again, it’s just a theory that this makes sense, but it’s a theory we can test. Earlier, I mixed 565ml boiling and 135ml tap water and got 77° instead of the 85° I expected. What does the new theory predict will happen?

\[t_r = \frac{100 v_b + t_t (v_t + v_p) }{v_b + v_t + v_p}\] \[t_r = \frac{100 \cdot 565 + 22 \left( 135 + 103 \right) }{700 + 103} = 76.9\]

That’s so close to the 77° that you’d almost think I fudged the numbers, but it’s really just a coincidence.

With this new and more plausible general formula we can plug in the values for my pot and water and get a specific formula for how much boiling and tap water to use to produce a particular temperature in my pot,

\[t_r = \frac{100 v_b + 22 (700 - v_b + 103)}{700 + 103}\] \[t_r = \frac{100 v_b + 22 \cdot 803 - 22 v_b}{803}\] \[t_r = \frac{78 v_b + 22 \cdot 803}{803}\] \[803 (t_r - 22) = 78 v_b\] \[\frac{803}{78} (t_r - 22) = v_b\]

It turns out to be the same as the one before except with 803 instead of 700. Using this formula, then, to get 85° water I need

\[\frac{803}{78} (85 - 22) = 10.29 \cdot 63 = 648\]

ml of boiling and the rest, 52ml, of tap. I tried this out and the result was… drumroll… 86°. So the formula seems to work. Yay!

Summary

So, to sum all this up, here’s how you apply the formulas in practice.

Measure how much water it takes to fill your pot, call that \(v\). Boil that much water and pour it into the pot. Measure the temperature and call that \(t_p\). Measure the temperature of your tap water, call that \(t_t\). Plug them into this formula to get the tea pot constant for your tea pot, \(v_p\)

\[v_p = v \frac{t_p - 100}{t_t - t_p}\]

Then, given a temperature \(t_r\), the amount of water to boil to get that temperature can be calculated by,

\[v_b = \frac{(v+v_p) (t_r-t_t)}{100-t_t}\]

And that’s it. You only have to calculate these once for a given pot so even though it’s a hassle, it’s a one time hassle. And it’s possible that most tea pots have approximately the same constant, in which case you can just assume that yours has the same one. But that’s for future research into the fertile field that is the science of tea.

Ironically, I got so caught up calculating the ideal amount of water that I never actually made tea, just water of different temperatures. I think I’ll go make some actual tea now.