Model tidal heating by comparison
What the equation actually means
I don't think that attempting to solve the equations to calculate tidal heating is going to get you anywhere useful. Instead, I think you should estimate tidal heating by comparison with known objects.
To boil down the first equation in the linked answer to just the variables:
$$E = R^5n^5e^2\cdot constants$$
where $R$ is the radius of the secondary, $n$ is the mean orbital motion (average velocity of its orbit) and $e$ is the eccentricity of the orbit (of the secondary).
The tidal heating in the secondary depends on the change in gravitational force from the primary over the course of an orbit. If the secondary orbits faster ($n$) or has greater motion to and away from the primary ($e$), then the tidal forces do more work pulling on the planet, increasing the tidal heating.
Ultimately, the force that matters is the force of gravity of the primary on the secondary. Mean orbital motion is given as
$$n = \sqrt{\frac{G(M+m)}{a^3}},$$ where $G$ is the gravitational constant, $M$ is the mass of the primary, $m$ is the mass of the secondary, and $a$ is the semi-major axis of the orbit. In most cases , $M >> m$, so we can drop $m$. Then we can look at the comparison between mean orbital motion and the equation for gravity. With this comparison we can see that $n^5$ from the tidal heating equation is roughly equal to
$$n^5 = \left(\frac{F_g}{m\cdot a}\right)^{5/2}.$$
The takeaway here is that, ignoring orbital eccentricity, the energy of tidal heating is proportional to
$$E \propto \frac{R^5F_g^{2.5}}{a^{2.5}m^{2.5}} $$
Comparisons for your system
To get an idea of the magnitude of these forces, the force of gravity of the Earth on Luna (i.e. our Moon) is $1.98\times10^{20}$ N, and the mass of Luna is $7.35\times10^{22}$ kg; the radius of Luna is 1734 km, the semi-major axis is 384,400 km. Combining these according to the proportionality equation above gives a 'score.' To make the number more tractable, we'll take the log base 10 of that score, this will show us how many orders of magnitude apart various tidal heatings should be. The log score for Luna in the Earth-Luna system is 50.8.
For an example of high tidal heating take Jupiter's force on Io; the force is $6.36\times10^{22}$ N and the mass of Io is $8.93\times10^{22}$ kg; radius is 1822 km, and semi-major axis is 421700 km. The log score for Io in Jupiter-Io is 56.9. This is six orders of magnitude more heating that Luna gets! No wonder Io is so volcanic.
For your listed problem statement, the force of gravity from the moon on the planet is $9.46\times10^{20}$ N, while the mass of the planet is $3.60\times10^{24}$ kg. The primary's radius is 5542 km; while the semi-major axis around the barycenter is 39464 km. The log score for your primary is 50.3, just a bit below what Luna is getting from Earth.
Note that Luna does not have a molten core, volcanoes, or anything else exciting that might come from tidal heating. Therefore, I conclude that the planet in your system will not be appreciably tidally heated.
The effects of oceans
The Love number is a parameter that measures the rigidity of the body, and thus the amount of energy that is transformed into heat by friction when the satellite is flexed by tidal forces. For the Earth, the Love number is relatively low, because of the oceans. You see, much of the energy of the moon's tides is used to move the oceans. The oceans do not generate very much heat when they are worked, because they are a liquid and the molecules easily slide past each other. On the other hand, rocky solids or even highly viscous liquid magma generates much more friction when it is acted on by a force.
Conclusion
Tidal heating of planet by its smaller satellite is insignificant. Even with the maximized conditions in your problem statement, tidal heating won't even reach what the Earth's (geologically dead) moon is getting. Consider that this heating is spread out within a planet two orders of magnitude more massive than the moon is, as well.
Also, having a liquid ocean on your primary won't help matters either.
