Can you heat up your coffee by stirring?

A fun question asked by a student in my class:  To what extent do you heat up your coffee by stirring it? 

It was a huge conceptual advance when James Prescott Joule demonstrated that "heat", as inferred by the increase in the temperature of some system, is a form of energy.  In 1876, Joule set up an experiment described here, where a known mass falling a known distance turns a paddle-wheel within a volume of liquid in an insulated container.  The paddle-wheel stirs the liquid, and eventually the liquid's viscosity, the frictional transfer of momentum between adjacent layers of fluid moving at slightly different velocities, damps out the paddle-wheel's rotation and, if you wait long enough, the fluid's motion.  Joule found that this was accompanied by an increase in the fluid's temperature, an increase directly proportional to the distance fallen by the mass.  The viscosity is the means by which the energy of the organized motion of the swirling fluid is transferred to the kinetic energy of the disorganized motion of individual fluid molecules.

Suppose you stir your coffee at a roughly constant stirring speed.  This is adding at a steady rate to the (disorganized) energy content of the coffee.  If we are content with rough estimates, we can get a sense of the power you are dumping into the coffee by an approach close to dimensional analysis.

The way viscosity \(\mu\) is defined, the frictional shear force per unit area is given by the viscosity times the velocity gradient - that is, the frictional force per area in the \(x\)-direction at some piece of the \(x-y\) plane for fluid flowing in the x direction is going to be given by \(\mu (\partial u/\partial z) \), where \(z\) is the normal direction and \(u\) is the \(x\)-component of the fluid velocity).

Very very roughly (because the actual fluid flow geometry and velocity field are messy and complicated), the power dumped in by stirring is going to be something like (volume of cup)*(viscosity)*(typical velocity gradient)^2.  A mug holds about 0.35L = 3.5e-4 m^3 of coffee.  The viscosity of coffee is going to be something like that of warm water.  Looking that up here, the viscosity is going to be something like 3.54e-4 Pa-s.  A really rough velocity gradient is something like the steady maximum stirring velocity (say 20 cm/s) divided by the radius of the mug (say 3 cm).  If you put all that together, you get that the effective input power to your coffee from stirring is at the level of a few microwatts.  Pretty meager, and unlikely to balance the rate at which energy leaves by thermal conduction through the mug walls and evaporation of the hottest water molecules.

Still, when you stir your coffee, you are veeeerrry slightly heating it!  update:  As the comments point out, and as I tried to imply above, you are unlikely to produce a net increase in temperature through stirring.  When you stir you improve the heat transfer between the coffee and the mug walls (basically short-circuiting the convective processes that would tend to circulate the coffee around if you left the coffee alone).