James,

how much heat is dissipated depends on voltage drop across input and output terminal! So, only saying that output current is '60-80mA' does not tell much.
I just had a look at datasheet (Philips) and to my surprise didn't found any specifications about thermal resistance. VERY BAD, people from Philips!!

Maximum power dissipation is 150mW and maximum output current 150mA. Not 300mA as you told! So, we can make a simple estimation about maximum input voltage:

150mW divided by output current of 80mA (max.) yields 1.875V. This is maximum voltage drop across regulator. Means, in combination with output voltage of +5V, input voltage must be smaller than about 7V!! Ok, this value is valid for maximum ambient temperature of 85°C. But unfortunately there's absolutely no information about derating. So, further calculations can't be done.

From experience I can tell you, that thermal specifications are often refered to rather unrealistic mounting conditions. Means, you will only achieve maximum power, when soldering certain pins to large copper area. But, again, Philips does not tell anything about it!

From my feeling, I would state, that with standard mains power supply, you will get big problems. Reason: For surely delivering 5V + 0.2V even under undervoltage conditions of mains, typical input voltage at input of regulator will rise up several volts higher. Then your regulator will enormously heat up. So, use this regulator only, when you can guarantee that input voltage will not rise up to more than about 7V, or so. Then you will have some headroom, when ambient temperature is less than about +40°C.
LP2985 is a very good choice, when application is battery powered, and input voltage is less than 7V. But 80mA seems to be a bit too high for battery powered application...

Kai