I don't know the whole solution, but I can show you a way you can probably go.
As @nluigi pointed out in his comment, the governing equations of your problem are given in
Modeling Transient Heat Transfer between two 1-D materials:
$$
\frac{\partial T}{\partial t} = \alpha \frac{\partial^2 T}{\partial x^2}
$$
with temperature $T$, thermal diffusivity $\alpha=\frac{k}{\rho \, c_P}$, time $t$, space $x$, thermal conductivity
$k$, density $\rho$ and specific heat capacity $c_P$.
The fundamental solution of this differential equation is given by
$$
H(x,t) = \frac{1}{(4 \pi \alpha t)^{\frac{n}{2}}} \, \exp \left( - \frac{|x|^2}{4 \alpha t} \right)
$$
where $n$ is the dimension of your space coordinate $x$ and $|x|^2 = \sum_{i=1}^n x_i^2$ is the square of the
Euklid norm (with $x_i$ being the coordinates of $x$ in a cartesian coordinate system). Please note that $x$ may be
a scalar value for one-dimensional problems ($n=1$), a two-dimensional vector for two-dimensional problems ($n=2$)
or a three-dimensional vector for three-dimensional problems ($n=3$).
An example for a one-dimensional problem is heat transfer through a wall. Your problem is (approximately)
two-dimensional. A three dimensional problem would be a hot or cold sphere within a large room.
To calculate a solution for a specific problem, you need the initial conditions, given as
$$
T(x,t=0) = T_0(x)
$$
and calculate the convolution between the fundamental solution and the initial conditions:
$$
T(x,t) = (H * T_0) (x,t) = \int\limits_{{\mathbb R}^n} H(x-y,t) \, T_0(y) \, \mathrm{d}y
\\
T(x,t) = \frac{1}{(4 \pi \alpha t)^{\frac{n}{2}}} \int\limits_{{\mathbb R}^n} \exp \left( - \frac{|x-y|^2}{4 \alpha
t} \right) T_0(y) \, \mathrm{d}y
$$
When I understand your problem correctly, you can use
$$
T_0(x) = \left\{ \begin{array}{ll} T_I & \mbox{for } |x| < x_B \\ 0 & \mbox{otherwise} \end{array}\right.
$$
with $T_I$ being your inner temperature at the beginning and $x_B$ being the coordinate of your wall. Note that since the differential equation is linear, you can add any baseline temperature to the solution as well as multiply the solution with any scale. Thus for the sake of mathematical solution, you could calculate with $T_I=1$ and scale the result. However, I'm an engineer and it doesn't hurt much to just drag $T_I$ along.
I recently had the one-dimensional problem, and this one can still be handled with some effort:
$$
T(x,t) = \frac{1}{\sqrt{4 \pi \alpha t}} \int\limits_{-\infty}^{+\infty} \exp \left( - \frac{|x-y|^2}{4 \alpha t}
\right) T_0(y) \, \mathrm{d}y
\\
= \frac{T_I}{\sqrt{4 \pi \alpha t}} \int\limits_{-x_B}^{+x_B} \exp \left( - \frac{|x-y|^2}{4 \alpha t}
\right) \, \mathrm{d}y
\\
= \frac{T_I}{2} \left( \mathrm{erf} \left( \frac{x+x_B}{\sqrt{4 \alpha t}} \right) -
\mathrm{erf} \left( \frac{x-x_B}{\sqrt{4 \alpha t}} \right) \right)
$$
with $\mathrm{erf}(x) = \frac{2}{\sqrt{\pi}} \int\limits_0^x \exp\left( -t^2 \right) \, \mathrm{d}t$ being the error function.
For $\alpha=1$, $T_I=1$, $x_B=1$ the following graph shows the solution for some $t$:
However, for higher dimensions of the initial conditions given above, the equation needs to be transformed into spherical coordinates first. For two dimensions you get
$$
T(x,t) = \frac{1}{4 \pi \alpha t} \int\limits_{{\mathbb R}^2} \exp \left( - \frac{|x-y|^2}{4 \alpha t}
\right) T_0(y) \, \mathrm{d}y
\\
\Rightarrow\quad
T(x,t) = \frac{T_I}{4 \pi \alpha t} \int\limits_{0}^{r_B} \left( \int\limits_0^{2\pi} \exp \left( - \frac{|x-y|^2}{4 \alpha t} \right) \, \mathrm{d}\varphi(y) \right) r(y) \, \mathrm{d}r(y)
$$
giving (at least me) considerable headaches when trying to evaluate. Probably somebody out there knows what
$$
\int\limits_0^{2\pi} \exp \left( - |x-y|^2 \right) \, \mathrm{d}\varphi(y)
$$
evaluates to, but I currently don't. Probably there also is a much cleaner approach to evaluate the convolution integral in two dimensions.
