How many watts to heat an enclosed space.

If I take a metal trash can and flip it upside down, what size heating element would it take to maintain 35 degrees Fahrenheit on a 0 degree day.

The trash can would be 18inches in diameter and 36 inches in hight so 5.3 cubic ft. Asumming this is dry air. The trash can is let’s say 12 gauge steel so it’s R value would be maybe .5

This isn’t homework it’s general information I’m looking for to calculate how much energy a friend is wasting.

I’m not looking for alternatives, just want to know how many watts this would take. I don’t know how to calculate this, the number I came up with is way too low.

I created a calculator one time for a company that produced cooling and heating units. You’d put in the size of the enclosure, the material it was made of, insulation, if it was painted, if it was in sun, the watts generated by the device inside the enclosure, and what temperature you wanted to maintain, and it would tell you how many BTU you’d need to keep that temperature. The math was very complicated.

They’ve rewritten it and it’s behind a “gimme your contact info” form, but still accessible. I can’t attest to it’s accuracy or usability any more, but here it is:

Volume and moisture are irrelevant once the desired temperature has been reached. Assuming no heat loss through the floor, I get 15.9 square feet of surface area times a temperature difference of 35 degrees divided by the R value of .5 … 15.9*35/.5 = 1113 BTU per hour. 1 KwH = 3412 BTU so 1113 / 3412 * 1000 = 325 watts. Here in Boston, that’s about $1.56 a day.

Thanks a lot I was a bit confused on the math. This makes perfect sense now.

the assumption of zero heat loss through the floor is very far off reality.

Another question is how well the base of the trash can seals against the floor. If the floor is uneven and there’s a gap between the trash can and the floor, the wind will blow in there and all the warmed air will be driven out very quickly.

It’d also depend on whether the can is exposed to wind or in a shelter. And whether it’s embedded in snow or in the clear.

So I think we can take bizerta’s calculation as a far lower bound. I’d WAG the actual losses are about double that number.