energyfluid dynamicsthermodynamics
And will it cool off faster if you blow across the top of the cup or directly into the coffee?
Does it have to do with the fact that when you blow across the top of the cup the velocity of the air increases which causes an area of low pressure above the cup, resulting in steam from the coffee to be forced up and out of the cup? Or is that not it?
I We did the experiment. (Early results indicate that dipping may win, though the final conclusion is uncertain.)
There were four trials, each lasting 10 minutes. Boiling water was poured into the canning jar, and the spoon was taken from the ice bath and placed into the jar. A temperature reading was taken once a minute. After each trial the water was poured back into the boiling pot and the spoon was placed back into the ice bath.
Method: Final Temp.
1. No Spoon 151 F
2. Spoon in, no motion 149 F
3. Spoon stirring 147 F
4. Spoon dipping 143 F
Temperature readings have an uncertainty of $\pm1\,\mathrm{^\circ F}$.
Red line: no Spoon
Green line: Spoon in, no motion
Aqua line: Stirring
Blue line: Dipping
$$\begin{array}{|c|cl|cl|cl|cl|} \hline \text{Min} & \text{No Spoon} & & \text{Spoon} & & \text{Stirring} & & \text{Dipping} \\ \hline & \text{°F} & \text{°C} & \text{°F} & \text{°C} & \text{°F} & \text{°C} & \text{°F} & \text{°C} \\ \hline 1' & 180 & 82.22 & 175 & 79.44 & 175 & 79.44 & 177 & 80.56 \\ 2' & 174 & 78.89 & 172 & 77.78 & 171 & 77.22 & 173 & 78.33 \\ 3' & 171 & 77.22 & 168 & 75.56 & 167 & 75 & 168 & 75.56 \\ 4' & 168 & 75.56 & 165 & 73.89 & 164 & 73.33 & 164 & 73.33 \\ 5' & 164 & 73.33 & 162 & 72.22 & 161 & 71.67 & 160 & 71.11 \\ 6' & 161 & 71.67 & 160 & 71.11 & 158 & 70 & 156 & 68.89 \\ 7' & 158 & 70 & 156 & 68.89 & 155 & 68.33 & 152 & 66.67 \\ 8' & 155 & 68.33 & 153 & 67.22 & 152 & 66.67 & 149 & 65 \\ 9' & 153 & 67.22 & 151 & 66.11 & 150 & 65.56 & 146 & 63.33 \\ 10' & 151 & 66.11 & 149 & 65 & 147 & 63.89 & 143 & 61.67 \\ \hline \end{array}$$
The discussion of nucleation sites is very much to the point. Water at atmospheric pressure without nucleation sites will theoretically boil only at $320.7 {}^o C$. The bubbles act as nucleation sites that reduce the energy required for evaporation. In the case of a bubble, the effective contact angle between the superheated liquid and the bubble surface is $180 {}^o$ which reduces the superheat needed to evaporate the water to $0$. Interestingly, there is actually an impediment to bubble growth caused by the reduced temperature of the vapor inside the bubble and a corresponding lower superheat boundary layer of liquid surrounding it.
FYI: My information is based on Collier and Thome pages 138 and 549.
In that text, an equation is given for the rate of bubble growth as:
$$ R = \sqrt{\frac{12\alpha_f}{\pi}} \frac{\rho_f c_{pf} \Delta T}{\rho_g i_{fg}}\mbox{Sn} t^{1/2}$$
where
$$\mbox{Sn} = \left[ 1-(y-x)\sqrt{\frac{\alpha_f}{D}}\left(\frac{c_{pf}}{i_{fg}}\right)\left(\frac{\partial T}{\partial x}\right)_p\right]^{-1}$$
and the variables are:
$R$ - rate of bubble growth
$\alpha_f$ - thermal diffusivity of liquid
$\rho_f$ - density of liquid
$c_{pf}$ - specific heat of liquid phase
$\Delta T$ - temperature difference
$\rho_g$ - gas density
$i_{fg}$ - latent heat of vaporization
$t$ - time
$D$ - molar diffusion coefficient
It's been a while since I looked at this in depth, but I think the $x$ and $y$ variables refer to position relative to a uniformly heated tube coaxial to the $y$ axis. Honestly, I don't expect you to actually use this formula, but hopefully it will impress that there are people who have spent a great deal of time on this subject. If you find it interesting, you might have a promising career in power plant boiler engineering in general or nuclear power plant engineering in particular.
Best Answer
When a cup of coffee is hot, the air molecules directly above it get hot as well. After some time, they reach equilibrium and no heat transfer (or maybe very little transfer) occurs. By blowing, you disturb that equilibrium and replace the hot air molecules directly above the cup with colder air and therefore create once again a steeper temperature gradient. This fact makes the content of the cup cool faster via forced convection.
See also convective heat transfer