Here's a little more data. If we use one SSR to control a 5500W/240V element, then 27.5W of heat will be dissipated by the SSR. That's when the PID is running the heater at 100% power. If at 50% then the heat dissipation drops accordingly, to 13.75W. This is not a whole lot. If your control box is very crowded and there is no room for air flow, then an external heat sink may be necessary, it depends.
For those that prefer two SSRs per load then you will have twice the heat dissipation to deal with.
I based this calculation on a 1.2V effective voltage drop across the SSR, as suggested by this paper on
SSR Thermal Considerations.
If we now had the thermal resistance of the heat sinks in question we could calculate what the maximum temperature inside the panel is allowed to be. My prediction is that it will be quite high, say 50°C, depending on the heat sink. Of course we do not want to run it at the limit, but knowing the limit allows us to apply a safety margin.