According to that calculator jremington was kind enough to provide a link to, considering worst case scenario of having an outside temperature of 5ºC and needing to maintain an even 30ºC, there would be 33W dissipation through the walls.

I found this calculation somewhere, checked the math and it does look ok.

Say 70% of 50 watt of Halogen Bulbs converts into heat (hypothetically)

i.e. = 0.7*50 = 35 watt of heat*

Bulb is on for 5 minutes = 605 = 300 sec

1 watt = 1 joule/ sec, thus 35*300 = 10500 joules = 10.5 KJ

Specific Heat of Dry Air (Cv) = 0.716 KJ/kg.K

Thus, 10.5KJ/0.716KJ/kg.K = 14.6 Kg.K

Density of Air = 1.3 Kg/m3

Thus 14.6 (Kg.K)/1.3 (Kg/m3) = 11.23 K-m3

This means temperature rise will be 11.23 K per m3 of dry air.

Converting Kelvin to Celsius

11.23 C temperature rise.

So, to keep that calculated ~11ºC rise in 5 minutes, considering the 33W loss from wall dissipation, I would have to have 2x50W lightbulb to get roughly 37W of heat.

Am I correct or completely wrong?

**EDIT:** No, wait, 50W to get 11.23º/5 minutes is for 1 m^{3} air mass. Considering my box is considerably smaller than that, I’d need less wattage to achieve the same.

Damn, can’t think, too sleepy… Will get back to this tomorrow morning.