ok math smarty, can you calculate how much current would have to flow through the wood to warm it, say, to 100 degrees F?

I sure can.

First we need to estimate the rate of heat removal from the surface. To do this, we can assume a very conservative value of the natural convection film coefficient in air of 3 W/m2/K. (This is quite low; more likely values would be from 5-10 W/m2/K, but we want to assume a worst case scenario here.)

Next, we need the surface area for cooling. Since an AA cell is 14 mm diameter by 50 mm long, we get a surface area of:

A = pi/4 x 14^2 x 50 = 7700 mm^2 = 0.0077 m^2

Not all of this area is equally exposed for cooling since some of it is underneath or next to other spacers. To make a worst case estimate we will take only a quarter of this area, let's say 0.002 m^2.

You asked for our spacer to have a surface temperature of 100 F. I'm going to do the calculations in SI units since it works more conveniently with current and voltage, so that becomes 38 C. Let's assume the air temperature in the room is 20 C, which is an averagely comfortable 68 F.

We now find the heat loss from the spacer is estimated at:

Q = 3 W/m2/K x 0.002 m2 x (38 - 20) K = 0.1 W.

Next, we need to see how much current would generate 0.1 W of heat with a supply at 1.5 V. Since (power) = (current) x (voltage), we find the current would be:

I = 0.1 W / 1.5 V = 0.067 A.

Lastly, we need to see what resistance would produce a current of 0.067 A at 1.5 V. We find that by Ohm's law:

R = V / I = 1.5 / 0.067 = 22 ohms.

Having done the calculations, we then have to see what the answer means. It suggests that for a wooden spacer to heat up to 100 F by resistive heating from a 1.5 V supply, it would need to have a resistance no higher than about 20 ohms. In practice this is a very low resistance indeed, and unless the wood was impregnated with salt or some other conductive substance, it is extremely unlikely that it could have such a low resistance. We therefore presume that the use of wood for insulating dummy cells in a battery charger is not likely to be a problem. :thumbsup: