Basic Thermosdynamics Problem

I have a 100 watt light bulb inside a steel cube. The outside air temperature is 75 degrees. How how is the air inside the cube?

I actually want the general solutionm - I've just concocted a typical problem. Approximations are definitely appropriate - I'm looking for a method of quick estimation.

Also, I'm not I sure even asked the right question, I may have left out some needed info (which you can add, but don't make is too hard.)
Who is Participating?
Let's say your steel box is a 1 ft cube.
It has a total surface area of 6 ft² = 864 in²

Pretend all the surface area is vertical.
The average power loss is (100 w)/(864 in²) = 0.116 w/in²

From the Watlow chart:

The box temperature would be 100 deg F.

Make it a 6 inch cube.
The area goes down to by a factor of four to 216 in²

The average power loss is (100 w)/(216 in²) = 0.463 w/in²

The box temperatue goes up to 180 deg F.
For a 1 inch cube.

Average power loss is (100 w)/(6 in²) = 16.6 w/in²

The box temperatrue goes up to 1000 deg F.

How large is the cube?   Is the outside air free to circulate around the cube (this could make the problem complicated)?  Is the cube in thermal contact with anything besides the air?
You can find reference data you need for this  sort of calculation at the Watlow website:


Look in particular at the first few topics in the reference section:

            Heat Loss Factors & Graphs
            Quick Estimates of Wattage Requirements (PDF)
            Properties of Gases & Air (PDF)

The best way to solve this problem is by experiment.  Actually put a light bulb in a steel box.

But you can do a good job modeling thermal behavior with an electrical circuit.

     Power ==> Current            Temperature ==> Voltage

     The light bulb is a current source of 100 V
     The air is the box is a thermal resistance.
     The steel box is a series resistance and a large thermal capacitor.
     Convection between the box and the ambient as resistance.
     The ambient temperature is a voltage source of 75 deg.

The air in the box will still be hottest near the bulb (where the power is coming in)
and coolest near the wall (where the power is going out).

And the top of the box will be hotter than the bottom.

But the Watlow tables give you all the infor you need to start to solve this problem.

Upgrade your Question Security!

Your question, your audience. Choose who sees your identity—and your question—with question security.

First off, I'd call this a heat transfer problem, not a thermodynamics problem, though that name would make a lot of sense.  

There are lots of parameters here.

How thick is the surface of the cube?  How big is the cube?  What is the thermal conductivity of the steel used in the cube?  Are all of these parameters such that we can assume that individual faces are of a homogenous temperature, or the whole cube is a homogenous temperature?  Or do we have to numerically model in detail a whole octant of the cube?

Do we assume that the air inside is a homogenous temperature, or will we have to model convective effects?  How much of the heat of the light bulb is transferred to the cube via direct irradiation, and how much is transferred through convection?

Just making wild guesses, I'd say we could make the assumptions that we ignore convection inside the cube (so radiation is the only heat form we need to worry about), and the cube is thick enough that each individual face is at a homogenous temperature.  Each face is receiving and passing through 100/6 watts.  The face least effective at passing through the heat will be the bottom face, so it will be hottest and will be responsible for heating up the air inside the cube.

That watlow reference is good.  See

We're going to be emitting 100/6 Watts from a bottom surface of s^2 square inches.  Assuming s is 6 inches, that means we'll be emitting about .46 W/in^2.  Divide that by 0.63 to correct for being a bottom surface, so we're looking up the value of .73 W/in^2 on the graph.  They usefully have a curve for oxidized steel on that graph, and a temperature of about 210F emits at .73W/in^2.

So that means that the bottom surface will be about 210 degress on the outside, and a few more on the inside due to conductive resistance.  So, I'd estimate the air inside would be around 230F or so.

That watlow curve is for 70 ambient, not 75, but there are so many assumptions in my estimate that that won't be a significant source of error.  If this were real engineering I'd throw in a +/- 40 degrees margin on my estimate.  Heat transfer calculations are not very precise.  
Since the bottom face will be the hottest, there will be heat flow from the bottom to the sides and top in addition to emission from the bottom to the outside, so the bottom surface will be emitting less than 100/6 Watts.
Also, the 0.63 factor will vary with the size of cube.
There are a wide variety of assumptions one can make.  I can't strongly defend any of the assumptions I made.  I assumed that heat transfer between faces would be insignificant.
In this case, I believe convection and conduction inside the box are mcuh more important than radiation.

Although the bottom surface fo the box certainly has the highest thermal resistance due to convection,
I expect it will be the coolest.

Convection inside the box delivers far more heat to the top than the sides or bottom.  Hot air really does rise
rfr1tzAuthor Commented:
That was pretty good everyone. I now feel more comfortable baking a pie in my easy bake oven.
If the heat transfer between faces is insignificant, the temperature difference between the faces would also be insignificant.
According to the graphs, radiation transfers more heat outside than convection.
Radiation would be even more dominant inside, where the ambient air temperature will be much closer to the temperature of each surface.
I believe the air in an easy bake oven is able to flow between the outside and the inside.
Radiation is probably more important than convection in transfering heat from the bulb to the pie.
The 0.63 factor is only for convective heat loss, you shouldn't divide the radiation loss for the bottom surface.
Question has a verified solution.

Are you are experiencing a similar issue? Get a personalized answer when you ask a related question.

Have a better answer? Share it in a comment.

All Courses

From novice to tech pro — start learning today.