Will a tall narrow cup keep a cup coffee warmer than a more evenly dimensioned cup? I noticed a colleague had a tall narrow cup for his coffee, and it got me thinking about whether it would retain heat for longer or not. 
Assume two cups, both are cylindrical, and both hold the same volume. One is taller and more narrow, and the other has even width and height dimensions. 
My first thought was that assuming the cup casing encompassed the liquid entirely, a perfectly spherical cup would retain heat best, as it has the smallest surface area.
However, this doesn't account for heat rising to the top of the cup, does that make a difference?
Also, the top surface and the sides and bottoms are different. The top of a cup has no casing, while the sides and the bottom are ceramic or porcelain. 
How would we determine which cup keeps the coffee warm longest?
 A: In general, yes, for a given volume a sphere has the least surface area and therefore the least heat loss to the environment.
But we have the complication that the top of the vessel must be open and uninsulated.
For a fixed temperature of the liquid and the ambient air, a simple model would say the three types of surfaces on a cylinder each have a fixed heat flux (power per unit area). Say these are $q_\mathrm{top}$, $q_\mathrm{bottom}$, and $q_\mathrm{side}$. The total energy per unit time lost to the environment is
$$ Q = \pi r^2 (q_\mathrm{top} + q_\mathrm{bottom}) + 2\pi rh q_\mathrm{side} $$
for a cylinder of radius $r$ and height $h$. If we fix the volume $V$ as a constraint, then $\pi r^2h = V$, so we have
$$ Q = \pi r^2 (q_\mathrm{top} + q_\mathrm{bottom}) + \frac{2V}{r} q_\mathrm{side}. $$
The optimal container has as its radius the $r$ that minimizes $Q$. This occurs for
$$ 0 = \frac{\mathrm{d}Q}{\mathrm{d}r} = 2\pi r (q_\mathrm{top} + q_\mathrm{bottom}) - \frac{2V}{r^2} q_\mathrm{side}, $$
or
$$ r = \left(\frac{Vq_\mathrm{side}}{\pi(q_\mathrm{top}+q_\mathrm{bottom})}\right)^{1/3}. $$
Qualitatively, we can note a few things. For instance, the ideal radius grows proportional to the cube root of the desired volume, which is not unexpected. In fact, for fixed $q$'s, the ideal container simply scales with $V$ - it's shape doesn't change.
Moreover the ideal container in this model is neither too wide nor too narrow. Too wide, and the $\pi r^2 (q_\mathrm{top} + q_\mathrm{bottom})$ term means all the heat is lost through the top (and bottom), while too narrow and the $(2V/r) q_\mathrm{side}$ term becomes larger than it needs to be.
I don't have an actual number, though, because I don't know the values for the $q$'s. One could presumably look up the properties of the ceramic involved for $q_\mathrm{bottom}$ and $q_\mathrm{side}$. Even then, though, I treat these terms separately because they could depend on the shape of the vessel: the sides will be more efficiently cooled by convection than the bottom, unless the cylinder is very narrow. On top of this, the heat transfer through the top needs to be modeled, and that itself is a difficult fluid dynamics problem.
A: Coffee would cool via heat being lost to the surroundings. This heat loss would be larger when the exposed surface area is larger. Now, whatever material makes up the cup, it would insulate heat somewhat, as against being held completely open, like the top surface. (That's one reason why ceramic cups are so common - if the cup doesn't insulate, its like holding the hot coffee right in your hand!)
So, the taller and narrower cup (an example here) exposes lesser area to air and heat loss through a larger volume of coffee, in the interior, would be reduced. 
In comparison, an ''even dimensioned'' one (for example, this) has a relatively larger surface area exposed to air, which means more cooling.
