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.

Step 1
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
$$v_t=k\cdot <!--\; \cdot \; -->(v_{xx}+v_{yy}+v_{zz}),v(0,y,z,t)=20,v(10,y,z,t)=-<!--\; -\; -->10,v(x,y,z,0)=20-<!--\; -\; -->x.$$
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
$$v_t=k\cdot <!--\; \cdot \; -->(v_{xx}+v_{yy}+v_{zz}),v(0,y,z,t)=20,v(10,y,z,t)=-<!--\; -\; -->10,v(x,y,z,0)=20-<!--\; -\; -->x.$$
then the function $v(x,y,z,t):=u(x,t)$ solves our 3-dimensional problem.
Step 2
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 $w_t\equiv <!--\; \equiv \; -->0$.
Again, we look for a symmetric solution, so the problem reduces to the following 1-dimensional problem
$$0=k\cdot <!--\; \cdot \; -->u_{0xx},u_0(0)=20,u_0(10)=10.$$
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,⋅) converges to a harmonic function (as $T\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}$), we can assume the resulting error is negligible.