How to model an ideal kiln? I have not taken thermodynamics yet, and I am wondering how I would model an ideal kiln.

In the drawings above, two possible cases are illustrated, one very simple, and the other slightly more involved, but closer to the way a real kiln operates.
In both cases the kiln is made of a sphere of an insulating refractory of a given material M, such as a 50-50 mix of aluminum oxide and silica, with thickness It and a diameter of Id.
The radiator in the first case is just a sphere with a diameter Rd. In the second case, the radiator is a toroid with cross-sectional diameter of Rd and a ring diameter of Rring.
The charge is a sphere of material C to be heated, for example, steel, with a diameter of Cd.
Power, P, is supplied to the radiator. Eventually, the kiln will reach a steady state in which the temperatures are all different. The temperatures will be as follows:
To..... The outer temperature of the insulator
Ti..... The inner temperature of the insulator
Ta..... The temperature of the air inside the kiln
Tr..... The temperature of the radiator
Tc..... The temperature of the charge
The questions for the model are to figure out the steady state temperatures of the system and to compute how long it will take for the system to reach the steady state assuming it starts at an ambient temperature like 20C.
 A: I'm not going to propose a complete model for your problem, which would be mathematically very demanding. But it's possible to model parts of the problem, with some reasonable assumptions.
For the left hand case, if we assume only radiation as the heat transfer mechanism and consider the air to be completely transparent to the radiation, then the heat flux from radiator to kiln $\Phi_{r,k}$ can be calculated from Stefan-Boltzmann:
$$\Phi_{r,k}=C_{r,k}\Bigg[\Big(\frac{T_r}{100}\Big)^4-\Big(\frac{T_k}{100}\Big)^4\Bigg]\tag{1}$$
With:
$$C_{r,k}=\frac{1}{\frac{1}{C_r}+\frac{A_r}{A_k}\Big(\frac{1}{C_k}-\frac{1}{C_r}\Big)}$$
Where $C_k, C_r$ are the radiation factors in $\mathrm{Wm^{-2}K^{-4}}$ and $A_k,A_r$ the surface areas of the kiln and radiator respectively.
During the transient (heating up of the kiln) $T_k$ of course changes with time and I'll simply represent it as $T_k=T(=f(t))$.
Again for simplicity, we'll use lumped thermal analysis, that is we assume there are no temperature gradients in the kiln:
$$\frac{\partial T}{\partial r}=0$$
The kiln also loses heat to the environment and we'll use Newton's cooling law (convection only) to describe that:
$$\Big(\frac{dQ}{dt}\Big)_{cooling}=hA(T-T_{\infty})\tag{2}$$
Where $h$ is the heat transfer coefficent (kiln to air), $A$ the outside surface area of the kiln and $T_{\infty}$ the ambient temperature.
The overal heat flux then becomes:
$$\frac{dQ}{dt}=\Phi_{r,k}-\Big(\frac{dQ}{dt}\Big)_{cooling}\tag{3}$$
As we consider the temperature of the kiln to be uniform, an increase in temperature $dT$ requires an amount of heat $dQ$:
$$dQ=mc_pdT\tag{4}$$
Where $m$ is the mass of the kiln and $c_p$ its specific heat capacity.
Inserting $(1)$, $(2)$ and $(4)$ into $(3)$, we get:
$$mc_p\frac{dT}{dt}=C_{r,k}\Bigg[\Big(\frac{T_r}{100}\Big)^4-\Big(\frac{T}{100}\Big)^4\Bigg]-hA(T-T_{\infty})\tag{5}$$
$(5)$ is a first order differential equation that allows separation of variables and has analytical solutions.
The equilibrium temperature $T_e$ (steady state) is found for $t\to +\infty$. Then the net heat flux is zero:
$$\frac{dQ}{dt}=\Phi_{r,k}-\Big(\frac{dQ}{dt}\Big)_{cooling}=0$$
$$\implies C_{r,k}\Bigg[\Big(\frac{T_r}{100}\Big)^4-\Big(\frac{T_e}{100}\Big)^4\Bigg]=hA(T_e-T_{\infty})\tag{6}$$
From which $T_e$ can be extracted.
