16
$\begingroup$

A little thought experiment, similar to this one: Imagine you are making a cup of tea when the door bell rings. You've poured the boiling water into a cup with a teabag in it. As you're just about to pour milk in to the mix, the door bell rings.

My question is: what will keep the resulting drink hottest after my chat with the person at the door: quickly pouring the milk in, or leaving it in the carton until after the interruption ends?

Being a bit of a mathematician I'm happy to have the questions answered with a formula in terms of $t$ the length of the interruption, $m$, the temperature of the cold milk, $w$ the temperature of the hot water, $a$ the ambient temperature etc. I guess you'll have to use a model of how an insulated body loses/gains heat... feel free to use simplifying assumptions like the environment stays the same temperature etc.

My starting point from a physics point of view, is that heat is lost (or gained) proportional to the temperature difference between the object and the environment. So I imagine that would create some sort of exponential curve?

Improve this question
$\endgroup$
5
  • 3
    $\begingroup$ Your premise is faulty: clever people put the milk in first so that it does not scald. $\endgroup$
    – dmckee --- ex-moderator kitten
    Commented Aug 2, 2011 at 16:49
  • $\begingroup$ @dmckee What means scald in this context? My dictionary gives too many possiblilities. And "first" means first in the cup, before hot water is poured? I remember faintly some British resarcher in "Science" in the 70ties to discuss this problem :=) $\endgroup$
    – Georg
    Commented Aug 2, 2011 at 17:58
  • $\begingroup$ Georg: Scald in this instance is to overcook the milk giving it a strong taste and sometime leaving curdled strings floating in your tea. First mean you make the tea in a pot, put the milk in the cup and pour tea over that. It doesn't work if you want to make the tea in the cup, alas. $\endgroup$
    – dmckee --- ex-moderator kitten
    Commented Aug 2, 2011 at 19:13
  • 1
    $\begingroup$ Aha, thats what we call "gerinnen". That British researcher from the 70ties discussed this phenomenon, and blamed social ("underprivileged") reasons for it, namely not having a fridge to store milk. The scalding only occures for milk when lactic acid fermentation was initiated before, he wrote. You see, in early 70ties everything got a turn to privileges :=) $\endgroup$
    – Georg
    Commented Aug 3, 2011 at 10:15
  • 1
    $\begingroup$ See also the lady tasting tea story in the history of statistics. $\endgroup$
    – Retarded Potential
    Commented Mar 27, 2013 at 18:07

2 Answers 2

Reset to default
18
$\begingroup$

Pouring the milk in early on will leave the tea warmer than keeping it hot, then later pouring the milk in.

Your assumptions (in the final paragraph) would be correct for most kinds of conductive heat transfer, but the dominant cooling process for an uncovered cup of tea will be evaporation. The details of the rate of evaporation will depend on how tricky things like air circulation, but the cooling power typically increases faster than linearly with the temperature. That faster-than-linear dependence means you'll win big by adding the milk while the tea's hot (dmckee's point notwithstanding).

Improve this answer
$\endgroup$
3
  • $\begingroup$ +1 Were You around when this question was stirring up a lot of debate? : physics.stackexchange.com/q/5265 $\endgroup$
    – Georg
    Commented Aug 2, 2011 at 17:45
  • $\begingroup$ @Georg: I was around, but that question looks significantly harder than this one. I agree with your analysis of that problem: that evaporation will be the dominant effect. But it's hard for me to ascertain which of stirring vs. spoon in & out would be most effective at enhancing the cooling. Depending on how vigorous one was, I could see either one winning. $\endgroup$
    – Anonymous Coward
    Commented Aug 2, 2011 at 20:06
  • 1
    $\begingroup$ Thanks. I think I agree as long as the room isn't say, 80°C, as the chilled milk would then have a much higher temperature difference than the boiled water, and so would provide the most significant heat transfer if left unpoured. $\endgroup$
    – Tom Viner
    Commented Aug 2, 2011 at 20:49
2
$\begingroup$

The whole thing is simple. Compare the rate at which hot tea cools and the rate that cold milk warms up. The hot tea is going to cool down faster than the milk is going to warm up. (Milk will only be about 30 degrees F lower than room temperatue.) The higher the temperature difference, the faster the change in temperature. Putting the milk in early slows down the cooling of the tea by bringing the temperature closer to room temperature.

Something to note is that the milk in the container is not cooling very quickly because of its large volume. This would cause a high loss of temperature if added after the tea had been standing for a while. So add the milk early and bring the mixture to a temperature that's closer to final.

Hope this makes sense.

Improve this answer
$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.