Combining two thermal conductivity (k) values

[morrobay]

In the case when heat is flowing at a steady state through two materials with different
thermal conductivities, the classic Q/t = A(T2-T1)/ L1/k1 + L2/k2 applies. In this case T2 is fixed and the two different k values are combined.
In the case of cooling , Newtons Law of Cooling applies dT/dt = -k ( T initial - T surroundings)
With solution T(t)= T surr +(T ini - T surr)e^-kt
Regarding Newtons Law of Cooling:
If the heat is being transferred through a container wall with a different (k) than the material
that is cooling how can two thermal conductivites be applied in the solution to Newtons Law of cooling ?
For example suppose a gas ,Neon, with a thermal conductivity (k) of .0116 cal/sec
is transferring heat through a copper walled container, Cu (k) = .92 cal /sec.

 

[Chestermiller]

This is not a straightforward problem. There is convective heat transfer in the Neon which may provide a resistance. One needs to know the convective heat transfer coefficient between the wall and the neon, or, equivalently, the convective boundary layer thickness. As a worst case, one could assume that there is no convection, and the neon could be treated as a rigid body, in which case one would be faced with a transient conduction heat transfer problem, including heat conduction throughout the neon.