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?
Best Answer
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).