If the RMS power dissipation is more than the heat dissipation...
This statement technically does not make any sense, but what I think you might mean is when the heat generation exceeds the heat dissipation.
When current flows through a resistor electrical power is converted to heat inside the resistor. This is heat generation. Consequently the temperature of the resistor rises, and the resistor dissipates the heat by conduction and radiation. There is some temperature where heat generation and heat dissipation are in balance, and this is the steady state operating temperature of the resistor.
If the resistor starts out cold and you apply a current to it, it will start to heat up. The rate which it heats up is related to the heat capacity of the resistor. Simply stated, big resistors have a bigger heat capacity than small resistors, meaning that bigger resistors will heat up more slowly from cold than small resistors.
The heating process in a resistor has a time constant, or characteristic time. This thermal time constant is analogous to the RC time constant of electrical theory.
Where this gets interesting is you can compare the thermal time constant to the frequency of operation of your electrical circuit. If the electrical frequency is much faster than the thermal time constant (it usually is), then you can forget about peak currents and just consider average heat dissipation produced by RMS currents or RMS voltages.
For instance, suppose your circuit frequency is 60 Hz. This gives a characteristic time of 1/60 second. Now suppose your thermal time constant is 1 second. This is much slower than the electrical time constant, so you can safely consider average heat generation and not worry about peak currents. If your circuit frequency was 1 kHz the statement about ignoring peak currents would be even stronger.
You may ask about how to calculate the thermal time constant of the resistor? This bit is hard, since the data is not readily available. You need the thermal resistance, measured for example in kelvins per watt (K/W). Then you need the thermal capacitance, measured in joules per kelvin (J/K). You multiply the two together to get the thermal time constant in seconds (K/W x J/K = J/W = s).
As a rule of thumb, if your circuit frequency is measured in kHz you can assume within reason that you only need to deal with average power dissipations in big elements like resistors. (But if you consider the microscopic bonding wires inside an IC, then all bets are off...)