I always make tea with boiling water - not water that is almost boiling, not water that has boiled and stopped boiling a few seconds earlier - in my opinion, for the best tea flavour the water must be boiling as it comes out of the kettle into the pot (or cup).

My wife is not so obsessive, and will often leave the kettle to switch itself off, then, after as long as a minute, pour the water onto the tea (she does admit that tea made by me is far superior to the tea she makes). Anyway, I've found that tea made by be (boiling water) cools to a drinkable temperature faster than tea made by her (boiled, then very slightly cooled water). But that seems scientifically impossible - or could it be explained in the same way that warm water freezes (when put in a domestic freezer) quicker than cold water? Just clutching at physics straws here!

Maybe the whole thing is my imagination.

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    $\begingroup$ I feel like you should just get a couple thermometers, identical cups, and test it. It sounds like it's your imagination. $\endgroup$
    – Vectorjohn
    Commented Apr 26, 2019 at 22:22
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    $\begingroup$ Related, but probably not directly applicable: the Mpemba effect (which is somewhat disputed). $\endgroup$
    – R.M.
    Commented Apr 26, 2019 at 23:30
  • $\begingroup$ you should control your variables and give a precise statement,maybe you can make a short film or take pictures to tell us what really is happening $\endgroup$ Commented May 6, 2019 at 14:16

3 Answers 3


There is a reasonable chance that your tea is cooling faster.

This is somewhat counter-intuitive, but it's because the only variable here isn't the temperature.

Evaporation is also very important in the cooling of the tea. The kettle probably has much less available area to release the steam, so the air above is saturated. This will lead to a much slower rate of evaporation, compared to when the boiling water is exposed to open dry air. This would mean if you poured the same amount of water into each tea, yours would lose more mass through evaporation which could be the deciding factor.

The rate of evaporation depends on the surface area and vapour saturation in the air. A kettle has a small opening to exchange it's saturated air with the surroundings, while the open cup can easily move away the saturated air through natural convective currents and new fresh air can come in to take it's place.

If you put some sort of loose covering on your cup that emulated a kettle while it cools for the first few seconds (be careful to make sure steam can still get out, just not as easily) you should see it remain quite a bit warmer.

Edit: Adjusted the answer to reflect brought up in Vectorjohn's answer. I was getting too invested into evaporation and lost track of why I brought it up in the first place. It changes the mass of the tea you have to cool if you pour both to the same level.

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    $\begingroup$ So to be specific, you're saying the extra variable is the mass of water: if more is lost to evaporation because of starting hotter, there's less thermal mass to cool off. (And the ratio of area to volume changes). So the smaller amount of non-evaporated water that started hotter can overtake the water that started not quite boiling. $\endgroup$ Commented Apr 26, 2019 at 21:51
  • $\begingroup$ @PeterCordes Ignore my other comment if you saw it, this would be one of the major factors here. $\endgroup$
    – JMac
    Commented Apr 26, 2019 at 22:38
  • $\begingroup$ The kettle/evaporation thing is not relevant - maybe I should clarify the different methods: My method: $\endgroup$
    – Paul
    Commented May 7, 2019 at 20:39
  • $\begingroup$ The kettle/evaporation thing isn't relevant - maybe I should clarify the different methods: My method: Boil water in kettle, pour water into both cups as the kettle is boiling. Start a timer. Allow tea to brew for 4 minutes, add milk, remove teabag (I'm not being posh and using a teapot), allow tea to cool until "drinkable" temp is reached. Stop timer. Wife's method: Boil water in kettle. Allow kettle to switch itself off. Faff around doing something else for one minute. Pour this hot water into both cups. Start timer. Brew 4 minutes, milk in, bag out, wait until "drinkable", stop timer. $\endgroup$
    – Paul
    Commented May 7, 2019 at 20:44
  • $\begingroup$ @Paul It's still relevant. Your wife's tea has more resistance to evaporation while in the kettle, and the time when it is the hottest is the time that it would have the most evaporation. The amount of heat that the evaporating mass carries away can be significant. You would have to control for variables such as the amount of tea lost to evaporation, which I doubt you are doing in your tests. If you measure out the exact same mass of tea for both scenarios, making sure the masses are the same when you start the timers, you could determine if this is the reason or not. $\endgroup$
    – JMac
    Commented May 7, 2019 at 22:24

The answer is no. It does not cool faster to a drinkable temperature.

The key thing other answers seem to be ignoring is that you're asking about the time between pouring the water and the tea becoming a drinkable temperature (emphasized because someone else interpreted your words as "drinkability"). It doesn't matter what's happening in the kettle, weather the air is saturated in it or not. Because you are asking about tea that is in a cup not a kettle.

Your tea certainly cools faster at first, because it's still evaporating quickly and heats up the cup more. The temperature differences and steam all lead to faster cooling. But at some point in time T+X, the temperature of your tea is the same temperature as your wife's was when she poured it. At that point, there is no magical process that will keep it cooling faster than hers, and in the mean time hers has already had X seconds of cooling. It will always be cooler than your tea.

If you really want to answer the question, it's a trivial experiment to do and might be fun. Make sure you use identical room temperature cups.

  • $\begingroup$ This is a good point, and my answer was inaccurate. I lost track of why I brought up evaporation in the first place. The extra heat has the benefit of increasing the evaporation potential, which reduces the overall mass (if poured to the same level); which has the potential to overtake the temperature difference. $\endgroup$
    – JMac
    Commented Apr 26, 2019 at 22:44
  • $\begingroup$ I think it's still incorrect to say the answer is "yes" his cup cools faster. He specifically is asking about how long it takes to go from pouring the tea to the same temperature. His wife's tea has to have a delta-T of X, and his tea needs a delta-T of (X+Y). There is no way, once the tea is poured, that it could do that faster. Consider, that once his tea has cooled slightly, it stops evaporating as much and the rate of cooling slows down. $\endgroup$
    – Vectorjohn
    Commented Apr 26, 2019 at 22:50
  • $\begingroup$ There's also a delta M though, that his warmer cup gets extra of because it is allowed to evaporate more during the cooling. It is possible that the effects of the reduced mass could lead to a bigger heat difference than the heat difference caused by his wife's water cooling $\endgroup$
    – JMac
    Commented Apr 26, 2019 at 23:44
  • $\begingroup$ I wonder if I interpret "my" tea as more "drinkable" sooner just because it's nicer than tea made by my wife's method! If I find a cheap thermometer that goes to 100˚C I might try to do an experiment. $\endgroup$
    – Paul
    Commented May 7, 2019 at 20:48

Does tea made with boiling water cool faster than tea made with boiled (but still hot) water?

In short, yes. The rate of cooling (temperature drop) will be higher when the liquid is boiling than when it has already reached a temperature closer to ambient temperature. However, as it cools, its rate of cooling will slow down, and it will reach the ambient temperature of the room more slowly than another cup in the same environment with a starting temperature closer to ambient.

I've found that tea made by be (boiling water) cools to a drinkable temperature faster than tea made by her (boiled, then very slightly cooled water)

Your classification for drinkability probably includes how well steeped the tea is and how much of the detectable compounds have transferred into the water. That process happens faster with the hotter starting temperature. There may also be some psychological bias in wanting to be right.

could it be explained in the same way that warm water freezes (when put in a domestic freezer) quicker than cold water?

The notion about warm water freezing more quickly than cold water is probably because water passing through the hot water heater has picked up more minerals/sediment/seeds for crystallization than water which has not.

Think about it this way: You bring a pot of water to a boil. Once it's well boiling, you pour equal amounts into identical mugs, one with tea and the other without. You start the timer for tea made your style. You let the other cool down for a minute, then add tea to the second and start the timer for tea made your wife's way. You stop the timers when the cups reach the temperature you classify as "drinkable." You are going to stop both of those timers at pretty much the same time, but the timer for tea made your way is going to display more time on it, because you started the other one after the cooling process was already underway.

Note: If you leave the second cup in a closed, insulated tea kettle not exposed to open air above, and put the first cup in a poorly insulated mug with an open top where steam and heat can easily leave, that first cup might indeed reach "drinkable" temperature more quickly than the second.

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    $\begingroup$ @Paul In other words, it all depends on multiple factors as well as your bias. :) $\endgroup$
    – Bill N
    Commented Apr 26, 2019 at 16:59

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