Awesome question. To test the feasibility of your idea, we start with the heat equation, which in our case is (from these notes)
$$\frac{\partial T}{\partial t}=D_H\frac{\partial^2T}{\partial z^2}\tag{1}$$
where $T$ is temperature, $t$ is time, $D_H$ is the thermal diffusivity, and $z$ is elevation. $D_H$ can be calculated simply as
$$D_H=\frac{\lambda}{c_v}$$
where $\lambda$ is the thermal conductivity and $c_v$ is the volumetric heat capacity of the soil. Exact values for $c_v$ can be calculated if the soil’s porosity, volume fraction of organic matter, and volume fraction of water are known; soils that are not homogeneous (i.e. that contain a mix of things, as is the case with most real soils) involve slightly more complicated calculations for $c_v$. Slide 10 of the notes gives values of $c_v$ for the four individual components of soil: minerals, organic matter, water, and air. $\lambda$ must also be calculated from the various values of $\lambda$ of each of its constituents.
Solving this gives us
$$T(z,t)=T_0+(T_s-T_0)\text{ erfc}\left[\frac{z}{\sqrt{4D_Ht}}\right]\tag{2}$$
where
$$T_0=T(z,0)=T(\infty,t),\quad T_s=T(0,t)$$
and $\text{erfc}(x)$ is the complementary error function.
The example given in the notes finds a $D_H\approx8.907\times10^{-7}$ m2s-1; if we take the initial temperature at the top of the underground ocean to be that of deep-ocean water (around 32-38 °F, or 0-3 °C, or 273-276 K), and assume that the crust is around 100 meters thick (I’m just taking a random value of this), with the planet having an effective temperature $T_s$ of about that of Earth, at 252 K as per the equation), then we get
$$T(\text{top of ocean},t)=252\text{ erfc}\left[\frac{100}{\sqrt{4\times8.907\times10^{-7}\cdot t}}\right]=252\text{ erfc}\left[52979t^{-1/2}\right]$$
Notice that as $t\to\infty$, $T(\text{top of ocean},t)=T_0$, so the heat will eventually reach the water. The big problem, though, is that it’s going to take a long time for changes in temperature to propagate, because of the large depth of the crust. As we can see by taking the derivative of $T(x,t)$ with respect to $t$, heat will move relatively slowly.
This is made worse because the surface of the ocean will cool. I assumed this value of $T_0$ to be true at the start of the heat conduction (i.e. at the very beginning of the planet), so the ocean will slowly start to heat up. It might have been better to study the change in temperature $\Delta T$, $T-T_0$. Other things I’ve ignored include the actual effective temperature of the planet, which may be nothing like that of Earth, and circulation of water from the depths of the ocean, however thick it is.
Here’s the big problem when you’re trying to help life survive: The energy source isn’t too good. You’ll of course have the equivalent of a geothermal gradient (see also here), but in reverse (i.e. heat traveling downwards from the surface). The geothermal gradient is the change in temperature over the change in depth, or
$$\text{Geothermal gradient}=\frac{\partial T}{\partial z}=-252\frac{2}{\sqrt{\pi}}\frac{1}{\sqrt{4D_Ht}}\exp\left[-\frac{z^2}{4D_Ht}\right]\tag{3}$$
After about one year, I find a $\frac{\partial T}{\partial z}$ of about 5.96$\times$10-35 K/km - a tiny, tiny, tiny fraction of the geothermal gradient at Earth’s surface (roughly 25 K/km). This means that geothermal energy is not good, if the surface reaches thermal equilibrium.
So where’s the energy source? That’s the real problem. There will be a larger gradient between the top of the ocean and the far depths, but most organisms likely won’t move that far. I feel like the eventual lack of a change in temperature will be an enormous problem.
One thing I don’t know about is whether or not the crust - a shell, really - will transfer heat into the ocean via radiation. If we treat it as a black body, then it it should emit a power of
$$P=4\pi\sigma(R_\text{planet}-100)^2T_s^4\tag{4}$$
as per the Stefan-Boltzmann law; the power per unit area comes out to about 229 W/m2, less than the amount recieved on Earth’s surface (which, by comparison, is about 340 W/m2). However, I’m a little shaky on whether or not all of that radiation will actually be emitted by the crust and absorbed by the water.
