The Premise: Imagine a solar collecting oven and how it works. Like the one below. We know it works by the greenhouse effect, letting light through, reflecting light from other areas as well with its foil sides. Light and radiant energy is bounced around the reflective surfaces, some leaks back through the glass, but much of it bounces until it hits a nonreflective surface, which in the oven's case would be the food you are trying to cook. Thus the food heats up 'artificially' from it's normal 'food-in-sun' potential, because of this concentrated sunlight and energy. These ovens, made of household materials can easily exceed 250-300°F in direct sun in 80°F or higher temperatures (ie, the summer).
The Questions: If you were to replace the glass with a sheet of matte, black steel (which would have blackbody-like emissivity/absorptivity) instead of the glass, do you lose all the additive potential which made the solar oven work in the first place? Or can the oven work on radiation from the solar-heated black metal?
Reasoning: Obviously, the opaque steel will shield the food from direct sun. This will decrease the hottest possible temperatures, and probably by a significant amount. However the steel can get close to ~364K, or ~195F — since it is nearly a blackbody — and this will radiate both, back into the outside air, and also down into the solar oven. The reflective walls will bounce the radiant heat around until it hits the food, AND also back to the source; back to the steel lid. This is somewhat unlike in the glass version, which partially reflected the heat and absorbed relatively little compared to what the steel can. If you were to only analyze heat radiated at the steel, you'd see that the steel is being subjected to radiation from sunlight on the outside, and radiation from itself, reflecting back to itself on the inside.... is this additive?
If I'm understanding this dynamic correctly so far, this would mean that the steel lid would heat up more rapidly than if it were not facing a sealed air chamber lined with reflective surfaces, as it would be effectively insulated against 'giving up' its heat through conduction on the opposite side of the sun.
So the steel heats up more easily to its idealized potential equilibrium temperature of 195F. But, once it gets to that temperature would there be a 'build up' of heat in the oven? Or would it increase it's absorptivity/emissivity slightly more and readily absorb heat from the oven and emit it back to the outside as quickly as the steel could replenish the oven's 'heat' through its own radiation?
I know that a material's absorptivity/emissivity increases with temperature, but my comprehension falls apart with how that dynamic actually works.
In Summary: Will the radiated heat from the replaced steel lid heat the oven, or will it cease to be an 'oven' by any reasonable standards because that heat would be transferred out of the oven as quickly as it could enter and there would be no build up of heat within?
Will the food be able to get any hotter than the hottest potential temperature of the black steel in the sun?