On Feb 19, 8:49*pm, wrote:

It is a most basic fact that objects in a room will all reach the same
temperature as the air in a room.

OK.

So how does the air transfer the energy to the solid substances such
as metal?

Conduction and radiation.

It is not denied that a piece of metal radiates according to Planck's
Blackbody Radiation Law, which describes the distribution of the
energy of the radiation. The total energy radiated from the surface
conforms with the Boltzman Stefan equation.

If the air brings the metal to it's temperature at equilibrium, the
metal will be radiating from it's surface ...

Sir... I am sorry to point out that it's (it is) _its_ temperature and
_its_ surface. _No_ apostrophe. The contrary is incorrect and
creates a bad impression on the reader.

the infrared energy
according to the Boltzman Stefan equation, which for 300K, (81F) is
460Wm-2. So a steel ball with 1 meter surface area will be radiating
this quantity of energy per second. If this quantity of energy is
leaving the surface of the metal at c and in this quantity per second,
the metal must be absorbing this quantity of energy at this rate from
the gases in the air.

Actually, the ball must be absorbing this power from the surrounding
EM field -- aka "thermal radiation bath" with which it is in
equilibrium.

A situation could be devised wherein it could be clearly shown that
the energy is not received as radiation from other solid surfaces.

That would be a non-equilibrium situation, and would violate the
principle of detailed balance. At equilibrium radiation out equals
radiation in, on any surface.

This energy cannot be transmitted to the metal merely by the
collisions or conduction through the air molecules.

No, it's transmitted via radiation, at equilibrium.

A quantification
of the energy of the gas molecules and the transfer of energy to the
metal must be achieved in order to have valid theoretical physics of
this most basic of phenomena of physics.

True. Something we are lacking!

This means the gas molecules are not transparent to the infrared as is
often believed, but are absorbing and radiating the full continous
spectrum of the infrared in the distribution similar to the Planck
Radiation Law.

That doesn't follow. At equilibrium the radiation in the room is in
equilibrium with the gas molecules, however weakly coupled
(transparent). We can make no deductions about the degree of
transparency or opacity from the mere fact of thermal equilibrium, and
equilibrium with the radiation bath. Your argument breaks down here.

...