1
$\begingroup$

Imagine having two boxes side by side out in the sun. Both boxes have cardboard as the floor and side surfaces. The top surface of the box is a piece of glass to allow in heat and insulate it. Each box contains a thermometer. If one box has the inside of it lined with mirrors (inside walls and bottom surface) and the other has the inside of the box lined with black paint (inside walls and bottom surface), which box will be hotter on the inside if left out for 5 minutes, as an example?

$\endgroup$
  • $\begingroup$ Do you have any thoughts? What do you expect, why? $\endgroup$ – JMac Mar 27 '19 at 18:57
  • $\begingroup$ I expect the inside of the mirror box will be cooler, simply because it's an opaque surface with higher reflectivity and lower absorptivity. However, it sounds like both boxes form a black body cavity if the top hole is small, in which cause they'd both have the same temperature. $\endgroup$ – Drew Mar 27 '19 at 19:21
  • $\begingroup$ I would guess the air in the mirrored box would have a higher temperature because the air will be constantly excited by the radiation bouncing back and forth. The black cardboard box will absorb radiation lowering the temperature of the air. $\endgroup$ – Cinaed Simson Mar 28 '19 at 23:34
0
$\begingroup$

When a photon interacts with an atom, three things can happen:

  1. elastic scattering, the photon will keep its energy and change angle

  2. inelastic scattering, the photon will give part of its energy to the atom and change angle

  3. absorption, the photon will transfer all its enregy to the absorbing atom (electron)

You are saying out on the sun, but let's just assume first we are talking about visible light, that interacts with the surfaces. Higher energy photons will go deeper into the material of the box and will do 2. inelasstic scattering, but it is not clear from your setup how thick the material of the box will be.

The black inside of the box will do 3 mostly, that is absorption. Most of the visible light will be absorbed by the black surface's atoms. In this case most of the visible light's energy will be transferred into the vibrational energy of the atoms and molecules of the black surface material (heat up).

Now the mirrored inside wall box will do mostly 2., that is inelastic scattering, where the energy of the photons will be kept, so mostly no energy will be transferred to the vibrational energies of the surface of the material of the mirrored inside walls. Now that is not completely true, because photons do exerpt pressure on the walls of the inside of the box, and transfer some of their momentum to the walls. Still, most of the photons' energy will be kept and will eventually (since the walls inside are mirrored) leave the box back to the outside via the glass on the top (I assume the glass is both ways seethrough).

So eventually the black inside box will have higher temperature.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ That makes sense, but I just had a thought from your last point. What if that glass at the top of the box was mirror/glass. Meaning see through where light goes in and mirror from the perspective of inside the box? Which box would have warmer air? I assume the mirror box because the photons are constantly exciting the air atoms in the chamber. In the instance of the black paint box, the air would be cooler but the walls would be warmer, correct? $\endgroup$ – Michael Minniti Mar 29 '19 at 4:40
  • $\begingroup$ Correct. The mirror wall box's walls would not heat up, but the air inside the box would heat up because photons would constantly elastically scatter on the air molecules, heating up the air. The black wall box would have heated up walls, that would dissipate somewhat outside (their outside walls would be cooled by the outside air), but they would still be much warmer walls. The air inside the black wall box would not be heated up so much. So you are right. $\endgroup$ – Árpád Szendrei Mar 29 '19 at 15:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.