Equipped with these formulae we can tackle the problem set by Hawking’s teacher as follows.
We first need to estimate some rough masses and temperatures for the tea and milk and surroundings as well as a value for λ .
We might estimate a typical cup of water to be 250 ml and the amount of milk to be 10 ml, which, given that the density of both water and milk are close to 1 g/ml, means that the mass of the initial tea before being mixed with milk might be estimated at 250 g and that of the milk as 10 g.
We might estimate the starting temperature for the tea without milk as a little below 100 ºC, perhaps 97 ºC, that of the milk as being 3 ºC, and the surrounding air in a typical room being 20 ºC.
We can quote from standard sources that the specific heat of water is roughly 4190 J / kg °C and that of milk is roughly 3930 J / kg °C.
Estimating a value of λ from first principles turns out to be quite hard. An interesting section of a publication by the Open University titled ‘Introduction to Mathematical Modelling’ shows how hard it can be, using a cup of tea as an example. It first estimates λ by considering heat loss through just the circular surface of the tea in the cup, then by including loss through the side of the cup too, in both cases coming up with wildly inaccurate figure which would indicate that the cup would take hours to cool (something the publication admits openly). A better way to estimate λ is by working backwards from the time taken by a typical cup of tea-like substance to cool, e.g. supposing that it takes roughly 10 minutes for a fluid close to 100 ºC to reach a drinkable temperature of around 60 ºC. Substituting these values in to the second formula in (2) indicates that a sensible value of λ might be 0.001. If the tea starts at 97 ºC and we add the milk immediately then by the formula in (1) this brings the temperature down to 93.59 ºC. If we let this cool for another ten minutes then the formula in (2) shows that the temperature reduces to 60.39 ºC. On the other hand, if we first let the tea cool for ten minutes without milk this brings its temperature down to 62.26 ºC, and if we now add the milk at 3 ºC then this bring the tea to 60.11 ºC which is less than with the milk added at the start, as predicted by Hawking. As a final note, however, I believe that there is a slight problem with the above, namely that it assumes that when the milk is added at the later time, the milk is still at 3 ºC. Assuming that the milk has been standing around during the ten minutes that the tea has been cooling, then using a value for λ of 0.001 we find that the milk will have warmed to 10.67 ºC. If we let the tea cool and then add the milk at this temperature then we find that the resulting temperature is 60.39 ºC – exactly the same temperature found when the milk was added at the start. Of course, I am not saying that Hawking wouldn’t have realised all of this, even as an A-level student, just that there is actually quite an important condition missing from the original anecdote, namely that the milk is added at the same temperature whether added early or later on. While this might seem a sensible assumption, it potentially makes the conclusion even more obvious and is possibly another hint at how Hawking came to see the answer so quickly. If the milk is kept in the fridge while the tea is cooling and then added at the end then this will of course cause a lower final temperature, in just the same way as if any part of the final tea mixture were kept in the fridge during the ten minutes. We now have enough information to answer the question posed by Hawking’s physics teacher. This all seems to vindicate Hawking’s answer, along with his reasoning, i.e. the rate of cooling is greater when the temperature is higher, and hence it ends up slightly cooler if the milk is added later.
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