Re: What is hotter, a small 'oven' or a larger one?

Hi there!

The short answer to your question is "maybe" -- the "maybe" comes from all that has been left out from the problem setup. At risk of boring you, please allow me to explain:

In general, the steady-state temperature distribution in an object (in this case, throughout the interior and walls of each kiln/oven) represents a balance between the heat energy flowing into and out of each location within the object. That is, for every piece of the object,

dQ-in = dQ-out,

where dQ-in is the energy flowing into (or being liberated -- by burning, etc) that piece of the kiln/over, and dQ-out is the energy flowing out. The flow of heat energy may be via conduction, convection, or radiation; in the everyday world, heat flow is typically via conduction within the object, and via convection of the air at the surface.

Allow me to explicitly assume what you have implied above: That the walls of each kiln are everywhere identical in their thermal properties (no difference in heat flow through long sides vs short sides, through walls vs floors, or through kiln/over doors vs walls); that the thermal conduction from the kiln/oven surfaces is also identical everywhere (the outside temperature is uniform, both ovens are standing on "feet" so the convection can work on all sides equally well, no fans blowing on only one oven, etc); and, that the temperature everywhere inside each kiln/oven is uniform. Under these assumptions, we can simplify the general heat flow differential equation to a simple algebraic relationship:

(dQ-in/dt) = C*A*deltaT

where (dQ-in/dt) is the rate at which heat energy is being produced inside the oven, C is the effective thermal conductivity of the oven walls, A is the total outer surface area of the oven, and deltaT is the difference between the inside and outside steady-state temperatures. Ignoring C for now, just look at what happens when you adjust (dQ-in/dt) and/or A: doubling the heat production rate in a particular oven will double deltaT, while doubling A without changing (dQ-in/dt) will halve deltaT.

However, getting back to your example, suppose we want to compare two ovens which have identical heat conductivities C. Then, write down the two equations for the ovens:

(dQ-in1/dt) = C*A1*deltaT1

and

(dQ-in2/dt) = C*A2*deltaT2

and divide one equation by the other, so that the C's drop out:

(dQ-in1/dt)/(dQ-in2/dt) = (A1/A2) * (deltaT1/deltaT2)

In your case, let's call the smaller oven #1 and the larger oven #2. You want to have the heat production in oven #2 to be twice that of #1, or:

(dQ-in2/dt) = 2*(dQ-in1/dt)

You want oven #1 to be a cube, with oven #2 to be an oblong box shaped like two cubes stuck together. So, if the outer area of oven #1 is 6 mumble-mumble-units, then the outer area of #2 is 10 mumble-mumble-units, and

A2 = (10/6)*A2

You can now see that the ratio (deltaT1/deltaT2) is simply:

deltaT1/deltaT2 = 5/6

In other words, the second oven has a deltaT (temperature difference from the inside to the outside) that is 6/5 of oven #1, or 20% higher. That is, it is the LARGER oven that is "hotter", with the higher temperature.

Of course, a little head-scratching can reveal that the larger oven (with twice as much heat being produced inside) doesn't have to be the hotter oven. The trick is in the ratio A1/A2: shape the second oven like a pizza box, with a huge surface area, and it will certainly be cooler than the cubic oven #1 -- even with a higher heat production rate.

Before I go, consider another case that one can pull from your question. Suppose that the two ovens are simply cubes, with sides of length L1 and L2, respectively. Suppose further that the heat production rates are simply proportional to the internal volume -- that is, if you double the volume, then you will double the heat production. Then, the combined equation for the two ovens becomes:

(L1*L1*L1/L2*L2*L2) = (L1*L1/L2*L2) * (deltaT1/deltaT2),

which reduces to

(L1/L2) = (deltaT1/deltaT2)

in other words, if you have two cubic ovens as I just described, then the larger one will be "hotter", and the ratio for the delta-T's is equal to the ratio of the side lengths. So, a cubic oven with twice the volume (and twice the heat energy production) will have twice the delta-T.

I do hope that this helps. If you're interested in heat flow problems like this (ovens and/or refrigerators), then I recommend reading the thermodynamics chapters in any calculus-based physics text, and then moving on to some basic thermodynamics text. I don't have a good recommendation handy, but you could probably ask a faculty or staff member at your local/favorite engineering school/department. Good luck!

Aaron J Redd

P.S. Note that the above equations involve delta-T, the difference between the temperatures inside and outside the ovens. If room temperature is taken to be 25C (or, say, 65F) then a delta-T of 100C (180F) would imply an absolute temperature inside the oven of 125C (or 245F). Doubling the delta-T would give you an absolute temperature in the oven of 225C (425F).