Electronic – Test current carrying capacity of conductor

In your edit, what's missing is that the rate of cooling will depend on the temperature. In general, the cooling rate will increase as the temperature increases. When the temperature rises enough that the cooling rate matches the heating rate, the temperature will stabilize.

But the actual cooling rate is very difficult to calculate. It depends on what other materials the copper is in contact with (conductive cooling), the airflow around the conductor, etc.

As an added complication, the heating rate will also depend on temperature, because the resistance of the copper will increase at higher temperatures.

So without much more detailed information about your conductor and its environment, its not really possible to give a precise answer to your initial question, how hot will it get?.

As for the second question, how fast will it heat up if there's no cooling, you can calculate that from the heat capacity of copper, which Wikipedia gives as 0.385 J / (g K), or 3.45 J / (cm^3 K).

The voltage rating is related to the breakdown voltage of the plastic between the pins. You shouldn't exceed the breakdown voltage even of theres no current at all.

The current rating is, as you say, relates to the pin resistance and how much the connector will heat up. You shouldn't exceed the current rating even at very low voltages.

Edit:

As KellenJB says in another answer, it looks like the key thing you're missing is that the power consumed in the connector (and so the self-heating which could damage the connector) is not related to the voltage between the pins, but to the current through the pin. This current, combined with the (very small) resistance of the pin or contact, generates a small voltage between one end of the pin and the other (or between one pin and the socket its mated to). This voltage, multiplied by the current, gives the heat generated in the connector.