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).

solar oven

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?

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    $\begingroup$ Reflective walls are only helpful if they are shaped to reflect the incoming sunlight toward the food. Sunlight that reflects off two shiny walls and back out of the box is lost energy. Black walls, on the other hand, will absorb the sunlight, and then give up their heat to the air inside the box. The reason for the transparent lid (instead of just leaving the top open) is to keep the hot air inside. $\endgroup$ Commented May 7, 2018 at 21:28
  • $\begingroup$ > keep hot air inside ...also the lid should be a "cold mirror," reflecting radiant heat but with a spectral window for visible light. Given time, I think the (glass) lid's temperature would rise enough to eliminate any benefit wrt steel. But the rapid cooking could be complete before then. $\endgroup$
    – wbeaty
    Commented May 8, 2018 at 7:22

1 Answer 1


Think about where the blackbody radiation goes with an opaque lid. Half of the surface area is outside of the box, and so (at least- the inner surface may be cooler as well) half of the incident radiation is radiated away and wasted.

Compare this to an oven with a transparent top and insulated walls. It should be easy to convince yourself geometrically that more than half of the radiation from the walls ends up hitting another wall and so is not wasted. So the overall temperature of the box has to get higher to reach radiative equilibrium.

The box with an opaque lid works, so long as the lid is not insulated, but with a transparent lid it works better (i.e. it gets hotter).

Neither can get food hotter than the highest temperature steel can get in the sun. If it could, you could just put some steel in the oven as your "food" and, voila, you got steel hotter than it is possible to get steel. ;)

Incidentally, you get higher temperatures the smaller the lid is relative to the rest (the shape matters too). You reach the maximum possible temperature in the limit that the window size goes to zero. But the downside is that the oven takes longer to heat as the window gets smaller, so there's a balancing act here.

  • $\begingroup$ Re, "Neither can get food hotter than the highest temperature steel can get in the sun." If you want to build a practical solar cooker, it might help to remember that steel (or anything else) can get pretty ****ing hot if you use optics to concentrate the sunlight. $\endgroup$ Commented May 7, 2018 at 21:34
  • $\begingroup$ @jameslarge That is true. You really don't need (or even want) the steel to reach the maximum possible temperature. $\endgroup$
    – Chris
    Commented May 7, 2018 at 22:10
  • $\begingroup$ I think y’all are missing the question. I’m not building an ideal cooker. I’m asking: can a steel lid raise the food/air temp above the highest potential of the steel? (195F)?.... OR....would it only get as hot as the steel?.... is there an additive effect like there is with glass? With glass, it can get over 300F - I don’t suspect the steel version would get THAT hot, but, based on fundamental physical laws alone, will the food get hotter than the steel? will the steel trap heat like the glass? How much and why? Or will it maintain equilibrium regardless? $\endgroup$ Commented May 8, 2018 at 12:36
  • $\begingroup$ @Chris, objects inside a solar oven actually do get hotter than steel by itself, outside an oven, could get and Check out the link in my post for how and why metal in sunlight can only get to 195F —— $\endgroup$ Commented May 8, 2018 at 12:49
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    $\begingroup$ @ChristopherIves, A glass lid will not "trap" any radiation that enters through the glass, and is reflected back toward the glass by foil. Whatever can enter through the glass can exit through the glass just as easily. Soda-lime glass (ordinary window glass) is 80 to 90 percent transparent to wavelengths from visible down to about three microns. Longer wavelengths (e.g., black-body radiation from things inside the box) will be absorbed by the glass, making the glass itself hot. The glass will lose heat (by radiation and by conduction to air) on both sides. $\endgroup$ Commented May 8, 2018 at 14:13

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