This is a bit of dispute between work colleagues. An answer would be greatly appreciated.

My argument is as follows:

If you add X amount of milk at a temperature of M to a mug at room temperature R before adding X amount of water at temp W, the result would be a cooler cup of tea than if you'd added the hot water first. This would be due to the milk reducing the overall temperature of the mug in the time that has been in it resulting in the addition of boiling water having less of an effect and resulting in a slightly cooler cup of tea/coffee.

Conversely, if you add the hot water first, due to the mug being heated, the effect of adding the cool milk will be less and therefore the end result will be a hotter cup of tea.

My Colleagues arguement is simply that it would not make a difference but without any form of justification.

If anyone can give real scientific light to this issue it would be greatly appreciated :)


3 Answers 3


We can make the following simplifying assumptions:

1)The the milk/water cool according to Newton's law of cooling or Fourier's Law.

2)The effect of adding milk is an instantaneous drop in temperature of solution by a fixed amount $\Delta T$.

Let the initial temperature of the water be $T_0$ and the temperature of the solution(water or water+milk) as function of time be T.

Case 1:Milk is added at the end of the experiment

$\frac{dT}{dt}=k(T_{env} - T)$

where $T_{env}$ is the temperature of the environment.

The solution for $T$ is $T(t) = T_{env} + (T_0 - T_{env}) e^{-k t}$

After adding milk at time $\tau$ according to our assumption 2,we get the temperature of the solution as:

$T_1(\tau) = T_{env} + (T_0 - T_{env}) e^{-k t} - \Delta T$

Case 2:Milk is added at the beginning of the experiment

The only change from case 1 would be that the initial temperature of the solution would now be $T_0 - \Delta T$ instead of $T_0$.

So the solution fot $T$ will be :$T(t) = T_{env} + (T_0 -\Delta T- T_{env}) e^{-k t}$

Therefore after time $\tau$ the temperature of the solution will be:

$T_2(\tau) = T_{env} + (T_0 -\Delta T- T_{env}) e^{-k t}$

So finally we have,$T_2(\tau)-T_1(\tau)=\Delta T (1-e^{-k t})$

Now for all $t>0$,$(1-e^{-k t})$ is always positive.

So $T_2(\tau)>T_1(\tau)$ always.

Moral of the story:"If you want hot tea,add milk in the beginning!"

Note:Here we assumed Newton's Law of Cooling was valid which was somewhat a simplistic assumption which may not be true in the real world.

  • $\begingroup$ The second assumption, that the effect of adding milk is an instantaneous drop in temperature of solution by a fixed amount Δ𝑇, is not justified. Suppose the tea is already ice cold. Adding cold (but not ice cold) milk will raise the temperature of the ice cold tea. Suppose the tea has been chilled to the same temperature of the milk. Adding the milk won't change the temperature of tea one iota. A better assumption is that the heat capacity of the tea remains constant over the temperature range $\min(T_{\text{env}},T_{\text{milk}})$ to the initial temperature of the tea. (continued) $\endgroup$ Jan 16, 2020 at 17:07

I would side with your colleague.

When you mix, the energy is constant so the following is fulfilled (first law of thermodynamics) $$\Delta Q_{water} = \Delta Q_{cup} + \Delta Q_{milk}$$

So, there is a flow of energy (heat $\Delta Q$) from the water to the cup and milk. This flux will stop when all temperatures are the same.

In this picture (closed system of water, milk and cup and long time), it wouldn't matter how you mix and you would get always the same temperature.

Now, a caveat. Here there are no losses and in practical terms the transfer of energy is instantaneous (which would be equivalent as waiting "infinite" time). My bet is that such effects are very small and shouldn't change the conclusion unless you have a very strange cup (material-wise) or very little water and milk.

  • $\begingroup$ Regardless of the size of the effect, there would be an effect though correct? $\endgroup$ Oct 23, 2013 at 13:02
  • $\begingroup$ No, because in the end the amount of energy transferred among the different "bodies" is the same. $\endgroup$ Oct 23, 2013 at 13:03

Hah, you wouldn't be satisified with the answer. Here's why.

Their hypothesis is right in the sense that the $cup$ will indeed be cooler, but they are wrong in the sense that this comes at the expense that the $liquid$ will actually be hotter. The mug will be easier to hold though, which is why I suspect such a hypothesis even exists.

The act of putting cold milk into a room temperature mug and letting it sit for awhile makes the mug an important third body in what was the ideal case of a two body thermodynamic problem of just hot tea and cold milk. In the two body case it obviously does not matter which goes in the receptacle first since the receptacle doesn't affect the problem .

Ok, let's consider three bodies.

Case A: Letting cold milk sit in a mug for a period of say 3 minutes allows the mug to cool a little bit below room temp. Consider the heat flow between these bodies due to a temperature difference of about $20K$. In the first few moments the hot tea is added, and the mug warms up, but because it was allowed to cool below room temp, it is not too hot to handle. For now.....

Case B: Now let hot tea sit in the mug for a same period, allowing the mug to warm a large amount over room temp. Consider the heat flow between what could be about $70K$ worth of temperature difference. This is equivalent to three minutes worth of heavy cooling in the liquid (by using the mug as a heatsinking mass -70K relative.)--three minutes that the liquid in Case A would still be spending in the kettle. So while the mug is a lot warmer than in Case A, the liquid is a lot cooler. Now add in the cold milk with sass. Enjoy a tea drinkable immediately without threat of third degree burns. Who cares if the mug itself is too hot to touch? That's what the handles are for.

  • $\begingroup$ not even considering heat loss to atmosphere...which if we did, would make Case B even better... $\endgroup$
    – gregsan
    Oct 23, 2013 at 13:04
  • $\begingroup$ So in dumb persons terms, my colleague is right? $\endgroup$ Oct 23, 2013 at 13:07
  • $\begingroup$ oh wait I thought your colleagues made the hypothesis... $\endgroup$
    – gregsan
    Oct 23, 2013 at 13:15
  • $\begingroup$ the conclusion is the liquid and the mug will be at different temperatures, which one is higher depends on the order of milk/tea poured in, with a larger difference depending on the time the first fluid was allowed to sit in the cup. $\endgroup$
    – gregsan
    Oct 23, 2013 at 13:17
  • $\begingroup$ He said - It would make no difference, I Said, adding the milk before would result in it being slightly cooler.. $\endgroup$ Oct 23, 2013 at 13:18

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