3
$\begingroup$

The following thought experiment will help understand my confusion. Note that I'm only considering thermal equilibrium in the context of radiation heat transfer.

Let's consider an oven which is an enclosed cavity and contains a vacuum with walls that are a blackbody being held at a fixed and homogeneous temperature. Then consider another blackbody inserted into the oven (let's assume zero gravity so the inner blackbody does not touch the oven and there is no conduction). Sufficient time is allowed to elapse after insertion of the inner body such that a steady-state condition is obtained. My question comes down to: is the inner blackbody temperature at the same temperature as the walls of the oven? I'm 99% sure the answer to this is "yes", but I don't understand how this can be proved. But here's how I try to prove it ...

My understanding is that the most fundamental definition of thermal equilibrium is that there is no net heat transfer. My thought was that, from this, I should be able to show that the two bodies in my example would have to be at the same temperature since I've also seen thermal equilibrium described as the bodies under consideration must have the same temperature.

So let's apply this fundamental definition (no net heat transfer) to the inner blackbody in my example. $T_0$ and $A_0$ are the temperature and surface area of the inner blackbody, respectively, and $T_1$ and $A_1$ are the temperature and surface area of the oven, respectively. Using the Stefan-Boltzmann law for a blackbody:

$$Q_{net} = 0$$ $$Q_{emitted} = Q_{absorbed} $$ $$\sigma A_0 T_0^4 = f\sigma A_1 T_1^4$$

where $0<f<1$ is the view factor from the oven to the inner blackbody, i.e. the fraction of energy emitted by the oven and absorbed by the inner blackbody. Note that the oven can "see" itself and hence only a fraction f of the energy it emits is absorbed by the inner blackbody. Simplifying the last equation above yields:

$$A_0 T_0^4 = f A_1 T_1^4$$

And this is where I'm stuck. There is nothing here that tells me the temperatures must be equal ... unless $f=A_0 / A_1$. But I can't assume that anymore than I can assume the temperatures are equal. Unless I can derive $f=A_0 / A_1$ from some other means? But when deriving view factors, one assumes the temperatures must be equal, so I've reached a paradox.

Can somebody clarify what I'm missing here?

Update
As I pointed out in my comment, the only answer given so far reinforces my point: reciprocity is derived assuming the temperatures are equal, so this is circular logic. Still searching for an acceptable answer ...

$\endgroup$
1
$\begingroup$

Unless I can derive $f=A_0 / A_1$ from some other means?

If you take the view factor reciprocity theorem as a given, it's easy. Your $f$ is the view factor of the outer oven to the inner sphere. The view factor of the inner sphere to the oven is one, meaning all of the radiation exiting the sphere hits the oven walls. By the reciprocity theorem, $F_{0\to 1}\,A_0 = 1\,A_0 = F_{1 \to 0}\,A_1 = f A_1$, we have $f = A_0/A_1$.

$\endgroup$

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.