Sunday, September 30, 2018

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). 


Jacques Distler said...

Umh, no, you're actually cooling it.

If you weren't stirring it, there would be a thermal gradient in the liquid, with the coolest coffee near the surface and near the walls of the mug. By stirring, you bring hotter liquid closer to the walls and to the surface, thereby accelerating the heat loss to the environment.

Everyone knows that you can cool a cup of hot liquid by stirring it.

Anonymous said...

If you want to be precise, you have to indicate the temperature...:
Can you heat up a cup of coffee that is at ambient temperature by stirring?

DanM said...

Another question is: how much energy are you expending by stirring the coffee, and how does this compare to the "few microwatts" that you calculated? i.e., what is the efficiency of energy transfer from the engine (you) to the heat bath (coffee)? Let's suppose that the hand+spoon system has a mass of 1 kg. You are rotating it at about 1 Hz, in a circle of radius, oh, I dunno, say 10 cm. The work you are doing, per second, is (2*pi*R)*mV^2/R, and the velocity is 2*pi*R/(1 sec). This comes out to about 2.5 watts. Giving an efficiency of about 10^-6. I recommend a bunsen burner.

Douglas Natelson said...

Umm, Dan, the centripetal force (your mv^2/r) is perpendicular to the velocity, and so does no work. The viscous drag work you do while stirring should be exactly the viscous drag power dumped in as heat.

DanM said...

Damn, yea. Ok, fine. Get your workout one way or another.

Unknown said...

Great analysis. For a general education science class, I had groups of students investigate different ways heat could be generated (run a fan inside a cooler, etc). One group took a 2 oz. purell squeeze bottle and filled it with water, with a thermocouple stuck in through the dispensing hole. The bottle was then held in a small bar clamp ( whose end was machined to fit in a jigsaw. The students ran the jigsaw recording the temp every so often and got a temp raise of a few degrees (starting at ambient) in a minute or two.

Peter Armitage said...

I saw a demo in a Whole Foods some years ago for a super fancy blender that could blend at super high speed. The guy was showing that you could both blend and warm things up simultaneously via the internal friction. I wouldn't have believed it unless I saw him make tomato soup before my eyes. It was steaming in just a few seconds. So if you are putting enough power in, the heat transfer doesn't matter. Everyone knows!

I found a video and more info

Peter Armitage said...

Update: my memory must have been faulty. After reviewing the video, I see it must have been steaming in a few minutes.