Physicspunks

[Tim May]

On Friday, August 17, 2001, at 01:17 PM, An Metet wrote:

> It's not surprising when a hot body and a cold body approach the same
> temperature.
>
> But, in the case described the two bodies start out at the same
> temperature.  One radiates more energy towards the other one at that
> particular temperature.  So, you would expect that one body would
> become hotter than the other.

Radiative transfer goes as the fourth power of the temperature 
difference between two bodies. The hot object will radiate energy to the 
cooler object roughly as (delta T)^4. When the temperatures are equal, 
radiative transfer is equal in both directions.

Think of "radiance" as being a bit analogous to thermal conductivity. 
Heat transfer still depends on a driving force (temperature difference).

All heat transfer functions have something like this form:

Heat Transfer is proportional to a (power per unit area) x (area) x 
(temperature)^(some exponent)

This is of course the same form Ohm's law takes:

current = voltage/resistance

current = voltage x "conductance"

Here, current plays the role of heat transfer. Voltage is the potential 
difference, or delta T. Conductance plays the role of emissivity, 
radiance, etc.

This is just the very basic flux equation, seen in physics, geology, 
even economics.

The "perpetual motion machine" envisaged would no more work for two 
metals of differing emissivities than such a machine woudl work by 
placing two metals of different conductances together.



> From the hints you've dropped I see the general outline of the
> solution.  The photons going between the two blocks of metal will
> "thermalize" and the volume between the two blocks will look like the
> inside of a cavity.  The rate of energy transfer in each direction
> will then be the same.
>
> In fact, if the two blocks were contained in a large thermos which
> perfectly reflected the photons, the same effect would occur.  It
> would be like an inside out cavity.  Each block would end up at the
> same temperature.  (Perhaps slightly lower than the start temperature
> because it must take some energy to fill the space with thermalized
> photons.)
>

Yes, you've got it right.


--Tim May