Hottest tea during your breakfast problem Suppose you are having breakfast and you like to drink your tea with 50% milk, 50% tea. You like to drink your tea as hot as possible, but your friend already filled your cup with tea, so it's already cooling off. Drinking it immediately is not an option though, because before you drink your tea you always eat your cornflakes, which roughly takes 10 minutes. What will theoretically result in the lowest temperature drop after 10 minutes? Pouring the milk into the cup immediately and drink after 10 minutes or wait for 10 minutes before pouring in the milk. Or perhaps somewhere in between?
I think there is two sides to this problem. Pouring the milk into the cup immediately will result in a very big temperature drop, because, since there is as much milk as tea, the temperatures will be averaged out. However the temperature will not drop as much in the next ten minutes, because the temperature is closer to the room temperature it is in. This goes both ways.
So here is my question: Which method will result in the lowest temperature drop? Will this method always result in the lowest temperature drop? Why?
EDIT: Some extra information is needed to answer the question. The temperatures of the milk, room and tea are 5, 20 and 80 Celsius. Everything else is just normal. Just a normal room, a normal cup.
 A: Liquid in your cup will cool because of the following effects:


*

*evaporation

*convection

*conduction


The usual argument goes like this: the tea has heat capacity $c_t$ and initial temperature $T_t$. When you add an equal amount of milk at $T_m$ (assuming it has the same heat capacity for simplicity), the temperature immediately after mixing is $(T_t+T_m)/2$. After this, the mixture cools at a rate given mostly by evaporation. 
Now the rate of evaporation is a strong function of temperature: hot things evaporate more quickly, so lose more heat per unit time.  If I leave the milk aside, then the tea will lose a large quantity of heat per unit time (because it's hotter, and the surface area of the cup is the same). The milk will be heating up slowly (through convection with the air around it) while sitting next to the tea, but that is a very slow effect - also because the milk is close to room temperature.
In summary: if you keep them separate, the tea will lose more heat per unit time. When you finally mix tea and milk, the mixture will be cooler. If you mix them immediately, the rate of heat loss is lower and the tea will be warmer.
Possible factors that would change this:
- the milk is VERY cold, so it can gain heat by being exposed to the room
- the room is very hot - again, same effect on the milk
- the cup has a strange shape, in which the addition of the milk greatly increases the surface area. This would speed up the evaporation. Since rate of evaporation is proportional to the saturated vapor pressure and therefore approximately follows an exponential law in temperature (Clausius-Clapeyron), you would have to work hard to shape a cup such that this mattered. A cup with a very narrow opening at the "tea only" level, and much wider at the "tea plus milk" level, might just do it:

With the figures you gave, you would expect the temperature of the tea right after mixing with milk to be about 43 °C. The relative rate of heat loss between that and 80 °C is roughly given by the ratio of vapor pressures:
$$\frac{P_1}{P_2} = \frac{e^{-L/RT_1}}{e^{-L/RT_2}} = e^{-\frac{L}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)}\approx 0.18$$
So if the surface area is at least 5x bigger, adding the milk will cause the tea to cool faster... 
A: If we cool tea first, the heat transferred to air is larger than that if we pour milk first. This is due to large temperature difference. In terms of this, less energy remains in the cup. Thus it will be cooler. 
