RBM rbm2(remove wrote:

... Garage is 24' X 24' with 12' ceiling. There is sheetrock and insulation
in the ceiling and three walls...

Thermal mass C = 0.5Btu/F/ft^2x2304ft^2 = 1152 Btu/F
for 2304 ft^2 of 1/2" sheetrock.

Temperature is aprox 32 degrees. I want to raise the temperature to
60 degrees in about one hour. How many BTU's do I need?

At least (60F-32F)x1152Btu/F = 32,256, if we could raise the temperature
instantly, with no loss of heat to the outdoors.

With (say) R30 insulation, we have something like this, in a fixed font:

--- R30/2304ft^2 = 0.013 F-h/Btu
|--|--|------------------www--------------- 32 F
--- |
I Btu/h |
--- 1152 Btu/F
---
|
|
-

This is equivalent to:

0.013
-----------www---------- 32-60 F
| |
| Tt = 32+0.013I |
--- --- 1152 Btu/F
- ---
| |
| |
- -

RC = 0.013x1152 = 15 hours and 60 = Tt + (32-Tt)e^(-1/RC)
make I = 33,397 Btu/h, ie 9.8 kW.

Nick