Time it takes for Temperature Change

I have just been thinking about it for a while and would like to see if there is a way to do this problem.

The Setup:

We have an insulated cup with mass $m_c$ and specific heat $s_c$. The cup is at temperature $t_c$. We put a liquid with mass $m_l$ and specific heat $s_l$ in the insulated cup. The liquid is at temperature $t_l$. Assume the freezing/boiling point of the liquid is extreme enough so that it does not undergo any phase change when in the cup. Also, assume that these two objects are the only things affecting the temperatures of the liquid and the cup. (that is, ignore the atmospheric temperature)

The Classic:

We can easily find the equilibrium temperature by using the familiar $Q=mc \triangle t$ and summing, setting the sum equal to zero.

The Catch:

Is there a way we can find the equilibrium time? Of course, since we are given definite masses, specific heats, and temperatures for all the objects in question (in the form of variables) it seems that the answer should be expressed as a function of all those (with the addition of possible some unnamed constants). Is this known? If not, how could you find it?

-
You need to know the thermal conductivity of the objects/liquids and their geometry. (If you want to consider radiation, it becomes more complicated) –  Manishearth Apr 19 '12 at 11:08
It also depends on the dimensions/shape of the object. It requires solving differential equation for diffusion to get the time. And by the way - total equilibrium happens at infinite time! –  Pygmalion Apr 19 '12 at 11:32
See also problem physics.stackexchange.com/questions/24443 for the exact solution of simplified problem. You can also see experimental results that prove the theoretical view. –  Pygmalion Apr 26 '12 at 14:11

To solve such a problem, you have to solve diffusion differential equations of type

$$\frac{\partial T}{\partial t} = a \Delta T$$

where $a = \frac{\lambda}{\rho c}$ is diffusion constant, $\lambda$ is thermal conductivity, $\rho$ is density of material and $c$ is specific heat capacity. It is pretty mathematically obvious, that $\frac{\partial T}{\partial t} = 0$ for $t=\infty$.

The other possible way to come to the same conclusion is observing Fourier's law (which is by the way the ingredient you need to derive the differential equation above):

$$\vec{\dot{q}} = -\lambda \nabla T$$

where $\dot{q} = \frac{\dot{Q}}{A}$ heat flow density. Temperature gradients (or differences, right size) are inducing heat flow that is equalizing temperature. However, as temperature gradients become smaller, so is the heat flow density becomes smaller, by which we come to the same conclusion, that temperature completely equalizes for $t=\infty$.

EDIT: I forgot to mention natural convection, in case you have gasses and fluids in your problem. Theoretically, this is much tougher question, but the result is essentially the same. Natural convection is induced by temperature gradient and as temperatures slowly equalize, so is convection slower and slower... essentially, the process never stops.

-