Modeling a Heat PDE I am trying to answer the following question...

Consider a wall made of brick $10$ centimeters thick, which separates a room in a house from the outside.  The room is kept at $20$ degrees.  Initially the outside temperature is $10$ degrees and the temperature in the wall has reached steady state.  Then there is a sudden cold snap and the outside temperature drops to $-10$ degrees.  Find the temperature in the wall as a function of position and time.

I am okay executing the separation of variables technique, but I can't really reason through how to model this scenario.  The solution manual states that the Initial/Boundary Value Problem is...$$u_t = ku_{xx},\ u(0,\ t) = 20,\ u(10,\ t) = -10,\ u(x,\ 0) = 20 - x$$
This question comes in a section before higher dimensional heat equations are introduced, but to me, it seems that this should be modeled as a three-dimensional heat equation, because thin walls are two-dimensional, and the bricks are prescribed thickness.  How can I intuitively reason through this word problem to model it correctly?
 A: Strictly speaking, you are correct. But if you stay away from the corners, then you just have 1d heat conduction across the brick only - the transverse gradients can be neglected. Then, the equations are just as you decribe with an error function solution. HTH
A: Following user89699's answer, let me phrase the same thought differently.

In principle, you're right. The wall is 3-dimensional, so the 3-dimensional heat flow would be our first choice. For simplicity, let us assume that the wall stretches infinitely in directions $y$ and $z$, and has thickness $10$ in direction $x$. Then the temperature function $v$ solves
$$
v_t = k \cdot (v_{xx}+v_{yy}+v_{zz}),\ v(0,y,z,t) = 20,\ v(10,y,z,t) = -10,\ v(x,y,z,0) = 20 - x.
$$
But because of the symmetry of the setup, we expect that the solution is independent of $y$ and $z$. In other words, the heat only flows orthogonally through the wall (in direction $x$). Formally, this observation can be rephrased as follows. If $u$ solves the 1-dimensional heat equation
$$
u_t = k \cdot u_{xx},\ u(0,t) = 20,\ u(10,t) = -10,\ u(x,0) = 20 - x
$$
then the function $v(x,y,z,t) := u(x,t)$ solves our 3-dimensional problem.

Of course, the wall is not infinite, but this is why we call it modeling: the reduction to 1-dimensional case is not 100% precise, but it's useful. In reality, the thickness of $10$ cm is relatively small compared to the other two dimensions, so we expect the error resulting from our idealization to be small.

As requested in the comment - a few words about the initial conditions. Before the temperature drop, we had another solution $w$ with boundary conditions $w(0,y,z,t) = 20$ and $w(10,y,z,t) = 10$. Moreover, we assume this was a stationary state, which means that $w_t \equiv 0$.
Again, we look for a symmetric solution, so the problem reduces to the following 1-dimensional problem
$$
0 = k \cdot u_{0xx},\ u_0(0) = 20,\ u_0(10) = 10.
$$
The condition $u_{0xx}=0$ tells us that $u_0$ is a linear function, and so $u_0(x)=20-x$. This function then serves as the initial condition after the temperature drop.
The steady state assumption is yet another idealization. In reality, we should rather solve the heat equation with boundary conditions $20$ and $10$ and take $w(T,x,y,z)$ (with some large time $T>0$) as the initial condition in the next step. However, since $w(T,\cdot)$ converges to a harmonic function (as $T \to \infty$), we can assume the resulting error is negligible.
