Lee, Allow me...
But Larry, We all agree that you cannot get more heat from a certain (pre-determined) amount of coal. Your equation is totally valid, but it doesn't apply to my question. Here's why:
To apply your equation directly, the Change in U would apply to a system like a toaster, or if a pile of coal were burning in a room with no exhaust. 100% of the heat generated is transferred into the room. Putting a fan on it cannot make more heat and your equation is obeyed and observed. If you DO get more heat out either example, you've used more energy ...and your equation applies and we'd all agree and lock this thread down.
But our system is different. It has heat from that certain amount of coal (or energy source) moving through a tube (piping/chimney and the stove itself) and we want to move the heat from that tube to the room. Your equation applies ONLY to the heat/energy INSIDE the tube (the coal)! We are conversely focusing on MOVING THAT HEAT INTO THE ROOM from inside the tube or Thermal Resistance. These are a totally different set of equations.
Thermal Resistance is how a material resists heat flow. If we encased your coal stove with glass, and asbestos, all of the heat generated by the coal would simply go right up the chimney and almost none would be transferred into the room. Notice that your 1st law of thermodynamics is true and applied only to the coal INSIDE the stove. But it DOESN'T apply to whether we transferred the heat (from the coal) to the room yet, or Thermal Resistance. If our stove and pipe was made of aluminum foil, I'm sure you'd agree that we'd get more heat from it than the duplicate glass/asbestos stove with the exact same amount of energy/coal in both examples. Yet we are not disobeying the 1st law of thermodynamics!
On this site: http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/heatloss.html#c1
Heat Loss Rate = Q/t = [(area) x (Temp Inside - Temp Outside)] / Thermal Resistance.
So Heat Loss (which is what we want) is greater as Temp Inside and Temp outside are further apart. Try some examples, like only 10 degrees apart will give a certain number, but 400 degrees apart will give a much larger number, or heat loss. Our fan makes the Temp Outside lower and gives us a bigger number in the numerator.
One more: http://en.wikipedia.org/wiki/Thermal_conduction
: and scroll down to "Cylindrical Shells" (e.g. pipes).
look for the words: "the rate of heat transfer is" and you'll see another equation with T2-T1 in the numerator where T2 and T1 are the temperature on the outer and inner wall. You'll agree that as T2-T1 is greater, the larger the number, or heat loss!These are the equations showing that the larger the difference in the temperature inside and outside of the stove, the greater the rate of heat transfer occurs. A fan makes the outside temp lower, so heat is transferred at a greater rate
If you still disagree with this: Ask yourself this fundamental question: Why do they put fans on car radiators? We say it's because the fan transfers (not create) the heat inside the material (radiator) faster by decreasing the outside temperature.
Please say you agree. We all love you Lar!