I offer my colleague some milk in her coffee. The milk has just come out of the fridge.

"Not now," she says. "Not till after I've finished my sandwich, and I don't want it to go cold."

So: identical mugs and quantities of hot coffee and milk at same temperature; only difference is that the cold milk is added — straight from the fridge — immediately in one case and five minutes later in the second one.

My guess is that my colleague is mistaken, and that after the five minutes are up, and the milk added to the second one, that the milky coffee in it will be colder than in the first.

  • $\begingroup$ Since hotter objects cool faster, my gut says you are right. But I’ll get back to you with some calculations. $\endgroup$ Jun 4, 2020 at 16:21
  • $\begingroup$ This conundrum has been solved repeatedly, including here at Ph.SE. But I can't find the question right now. It's very tedious to develop. $\endgroup$
    – Gert
    Jun 4, 2020 at 20:56
  • $\begingroup$ It would be interesting to carry out a well designed experiment to put the matter to rest, 'for once and for all'. A simple theoretical treatment requires some assumptions that are always subject to debate. So theory alone probably cannot resolve the matter. $\endgroup$
    – Gert
    Jun 4, 2020 at 21:11
  • $\begingroup$ @SuperfastJellyfish 'Since hotter objects cool faster, [...]' But they also start from a higher temperature! $\endgroup$
    – Gert
    Jun 4, 2020 at 21:13
  • 1
    $\begingroup$ A very similar question once led to a breakthrough in statistics: en.wikipedia.org/wiki/Lady_tasting_tea $\endgroup$ Jun 6, 2020 at 13:50

2 Answers 2


This experiment (with tea and milk) seems to show that immediate addition of the milk gives the warmest drink:

Temperature v. time

But the experiment is far from perfect, especially because of the lack of replication.

And this simplified derivation, based on Newton's Law of Cooling finds the same. But it too is very open to well-reasoned critiques. It certainly relies on one rather questionable assumption.

And here's my own experiment. Experimental protocol is summarised at the end of this post.

I tracked both methods of drinking coffee over a $15\mathrm{min}$ period of time. At the end I found the coffee with immediate addition of milk to be about $8^\circ$$\mathrm{Celsius}$ hotter than when the milk was added at the end of the run. Even without replication that's likely to be highly significant.

The dimensionless temperature $\Theta$ v. time plots are quite revealing, Series $1$ is with immediate addition of milk, Series $2$ with addition at the end:

Theta v time

Assuming Newton's Law of cooling (see link for symbols used) applies, we can write:

$$\ln \Theta=\ln\Big[\frac{T(t)-T_{\infty}}{T_0-T_{\infty}}\Big]=-\frac{t}{\tau}$$


$$\frac{1}{\tau}=\frac{h A}{m c_p}$$

$\tau$ is the so-called characteristic time. For immediate addition of the milk, Linear Regression gave a value of $1/\tau=0.0123\text{ 1/s}$ and $1/\tau=0.0148\text{ 1/s}$ for addition at the end, from the two runs above. That's a difference of about $17\text{ %}$.

Firstly, strong linearity of the $(\ln\Theta, t)$ plots indicates that radiative and evaporative losses are probably negligible compared to convective losses. Newton's Law of cooling seems to hold.

Secondly, the greater cooling rate for Series $2$ indicates a higher value for $\frac{1}{\tau}$, so for $\frac{hA}{mc_p}$.

A possible cause of the difference in cooling rate may lay in the factor $A/m$. For a cylinder with (constant) radius $R_0$ we found:

$$\frac{A}{m}=2\pi \Big(\frac{R_0^2}{m}+\frac{1}{\pi \rho R_0}\Big)$$

As $m$ goes up, $A/m$ comes down and so does $1/\tau$ and the cooling rate. Surely a bucket of hot coffee would take longer to cool than a mere cup?

So maybe what we're observing is a mere mass effect.

The influence of the $A/m$ ratio on cooling rate was investigated empirically as follows.

The first two experiments were replicated but using coffee only. Another run was added in between the first two so we have three cooling rates for $A/m=0.83$, $A/m=0.87$, $A/m=0.91$:

Theta and A/m

Here the effect of $A/m$, although still present, is much smaller than in the first run.

Could the larger difference in cooling rate be caused by using milk? Milk has a lower heat capacity than water because it contains fat. That would increase $1/\tau$.

Various internet sources put the $c_p$ value for semi-skimmed milk at about $4.0\text{ kJ/(kgK)}$, as opposed to $4.18\text{ kJ/(kgK)}$ for pure water, a difference of about $4\text{ %}$. But we only used $27\text{ %}$ of milk in our first two runs, so it's unlikely to account for the $17\text{ %}$ observed difference in $1/\tau$ values.

Experiment protocol (summary)

  • $230\mathrm{g}$ of black, boiling instant coffee with $27\text{ %}$ of semi-skimmed milk.

  • Paper cup (take away), no lid. No insulation.

  • Slow magnetic stirring.

  • Environmental temperature $T_{\infty}=19.0^\circ\mathrm{Celsius}$, for the first two runs. For later runs that value had to be adjusted slightly.

  • $\begingroup$ Added my own experiment. $\endgroup$
    – Gert
    Jun 6, 2020 at 13:12
  • $\begingroup$ Hi, we've noticed that you have made a large number of minor edits to this post. Please be mindful that every edit bumps the post in the "active" tab of the site and try to make your edits substantial. If you foresee improving this post repeatedly, maybe collect several edits and make them in one go instead of submitting them individually. $\endgroup$
    – Chris
    Jun 8, 2020 at 10:37

The question is an old conundrum and can be found in various guises on the Internet and in handbooks. One can summarise it as follows:

If I add milk to my coffee and wait 5 minutes before drinking it and another person waits 5 minutes and then adds the milk to his/her coffee, who is drinking the hottest coffee?

A theoretical derivation of the end temperatures can be carried out based on the following assumptions/simplifications.

  • amounts of milk and coffee are identical in both cases
  • mixing of milk and coffee is adiabatic
  • the cooling of the liquid (coffee or coffee plus milk) follows Newton's Law of Cooling
  • radiative losses during cooling are negligible compared to convective losses
  • evaporative losses are negligible
  • material constants like density, specific heat capacity and convective heat transfer coefficients are independent of temperature
  • the above list of assumptions may not be fully exhaustive

A derivation based on the above is tedious and shows both end temperatures to be very close together. But the derivation is also useless for providing a definitive answer (but it can be useful to provide insights into the cooling process) because critics can always point to one or more of the assumptions not being met in reality. Further refining the model will likely not quell such critiques.

For those reasons a well designed experiment with sufficient replication (to allow statistical analysis) would be more interesting and insightful.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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