How does the timing of adding milk affect the temperature of my tea?

I hope this isn’t a really dumb question.

Let’s say I make some tea in two cups. One cup I leave to cool for 5 minutes, then add milk from the fridge. The second cup I add milk to immediately, then leave it to cool for 5 minutes.

At the point when both teas first have milk, is one hotter than the other, or are they the same?

Assume everything is pretty much as standard (the tea is hot, the milk is cold etc.) and everything is identical unless otherwise stated.

There is no sugar. There will be no sugar.

Oh and I’ve just realized: the milk added *after/i] five minutes is taken out of the fridge at the same time as the milk added before the five minutes.

There are two effects. Most obvious, when you add the milk first, the mixture is cooler, so there will be less heat transfer through the sides and top of the cup. The milk has fat in it, which I believe can slow evaporation of water from the surface of the coffee, which will also tend to slow the transfer of heat.

Since you’re starting with the same total amount of heat in the two cups, the one with less heat transferred out of it will be hotter, so the one with the milk added first should be hotter.

I’m assuming there’s more coffee than milk, and that the mixture is always hotter than room temperature (i.e. you’re not adding really cold milk to coffee barely above room temperature). ETA: Ummm, like you already said…

Martin Gardner addressed this in a “Mathematical Games” column. He cited Newton’s Law of Cooling, which is that the rate of cooling is proportional to the temperature difference. If your tea is 120 degrees (F) and room temp is 70, the rate of cooling is only half of what it would be if the same cuppa was outdoors in 20 degrees in the winter (or in a 20 degree fridge.)

So…let the tea cool faster while it’s hotter, and then add the milk, cooling it further. Overall, the result is cooler than if you do it the other way around.

I agree it first glance but in the case where the milk is added later, you need to also account for the milk warming for 5 minutes at room temperature. The difference between the milk temperature and room temperature is probably smaller than the difference between the tea and room temperature, so you it is probably not warming as fast as the tea is cooling, but it would still be interesting to run numbers (I am not sure what the equations for heat transfer are).

In this case, it’s going to depend on how rapidly the milk warms relative to the rate at which the tea cools. Basically, the temperature of an object that is cooling/warming to an ambient temperature goes exponentially with time, but the rate at which it decays depends on how well insulated it is, how much convection is going on, etc. The rate at which this decay occurs is governed by a decay constant λ:

T(t) = (T[sub]i[/sub] - T[sub]0[/sub]) e[sup]-λt[/sup] + T[sub]0[/sub]

where T[sub]i[/sub] is the initial temperature and T[sub]0[/sub] is the ambient temperature. A larger value of λ means less insulation and more rapid cooling to room temperature, while a smaller λ means more insulation and less heat loss.

Let’s assume that the temperature decay constant is precisely the same for the tea with milk (in scenario #2) as it is for the tea without milk (in scenario #1); call this λ[sub]t[/sub]. Assume, however, that the decay constant for the milk left out on the counter is different; call this λ[sub]m[/sub]. It can be shown that the difference between the two temperatures T[sub]1[/sub] and T[sub]2[/sub] in Scenario #1 and #2, respectively, is proportional to

T[sub]1[/sub] - T[sub]2[/sub] = K (e[sup]-λ[sub]t[/sub]t[/sup] - e[sup]-λ[sub]m[/sub]t[/sup])

where K depends on the temperature difference between the fridge and the room and also on the proportion of milk to tea. (Perhaps surprisingly, it doesn’t depend on the temperature difference between the tea and the room.) If the decay constant for the teacup, λ[sub]t[/sub], is greater than the decay constant for the milk, λ[sub]m[/sub], then T[sub]1[/sub] will be less that T[sub]2[/sub]. Otherwise, the converse will be true. As an example, suppose that I take the milk out of the fridge and put it in a thermos for the five minutes before adding it. Then it’ll barely warm up at all in the five minutes, and it’ll be like it came straight out of the fridge; in this case Scenario #1 results in cooler tea, as noted by Trinopus. On the other hand, if I take that tea and put it in the thermos, while leaving the milk on the counter, the milk will absorb a bit of extra heat from the room while the tea will lose very little heat to the room. So there will be more heat energy present in the final mixture than there would have been if I had put the milk into the (insulated) tea at the beginning. Thus, Scenario #1 results in hotter tea in this case.

My guess is that in “real life”, λ[sub]t[/sub] will be larger. An open cup of hot tea will have substantial amounts of convection and evaporative cooling going on, which would tend to make λ[sub]t[/sub] larger. The milk, meanwhile, is probably still sitting in a closed carton before it’s added, with a volume as large or larger than the teacup. This means that the surface-to-volume ratio of the milk container is smaller, which results in a smaller decay constant and thus slower warming. Convection and evaporative cooling are, of course, non-issues for the milk.

In conclusion, Scenario #1 still results in cooler tea in realistic scenarios where the milk sits on the counter for five minutes. Also, I have too much time on my hands.

The other contributor is that the cup with the milk added has a greater surface area, which will increase the total energy transfer rate. However change in surface area versus the volume of liquid depends upon the cup geometry. You can construct cups where the change in surface area varies almost linearly with volume (tall thin), right down to cups where it varies almost not at all (short wide). Thus the temperature decay constant above is not necessarily the same before and after adding milk. However the constant will always larger, and never smaller. It should be possible to design a pathological cup whose surface area increases rapidly enough as liquid is added that it can swap the cooling rates around. But any ordinary cup is going to behave as described above.

This is not correct. There may be a greater surface area on top, but you’ve got to include the other surfaces: the sides and the bottom, plus the surfaces of the milk in its own container. The side of the teacup gets hot, doesn’t it? That means that the tea is losing heat through the side.

Um, I’m at a loss to understand the comment - this is what I said. The surface area cannot drop. The area underneath and on the sides cannot fall when you add milk to the teacup. The area of the fluid in the cup increases when you add milk.

There will be sugar, and it will be added right after I take the tea bag out of the cup and stirred. The sugar gets added as soon as possible, because it will dissolve better. Then I add the cream. Yes, cream. Then I take it over to the computer desk, where I will stir it again and let it cool down.

You’re not considering the surface area lost by the milk.

Consider an idealised cup of tea with dimensions 9x10x10 in a 10x10x10 cup (HxWxD) and an idealised jug of milk 4x5x5. The tea has a total surface area of (2x10x10) + (4x9x10) or 560, and the milk has a total surface area of (2x5x5) + (4x4x5) or 130. Total surface area for both is 690. Mix the two and you get a volume of 10x10x10 with a surface area of 600. The total surface area has dropped by 15%.

OK, I was only considering the OP position, where the milk was not warming outside of the fridge. My point was simply that the preceeding analysis had not factored in surface area changes, and that the shape of the containers matters, but probably not enough to make a huge difference. One could construct pathological container geometries that swap the relationship around, but ordinary sane jugs and cups won’t. That is all. (Mind you, there are some pretty wild tea cup designs out there.)

Surface area isn’t the entire answer either. Because you then need to consider the mechanism of heat loss. Convective losses to moving air will increase with the area of the container, as will heat gain. Radiative loss and gain is limited by the apparent cross section seen, and a fluted surface doesn’t gain you anything as the flutes radiate between themselves and not away. Again, not exactly common for a real world container, but feasible.

FWIW, this Dutchman was trained by his British lady-wife to add the tea to the milk; a question of taste, apparently:confused: