"So if I am figuring this right, the power dissipated is 5*0.093=0.465 watts."

The actual power dissipation calculation for a linear voltage regulator is the voltage drop across the regulator times the current. So if the input voltage is 15 vdc then the power loss would be 1 watt (10 X .1). So one way to lower the heat is to keep the input voltage as low as possible, but no lower then the drop out limit specification of the device, 8 volts would be nice.

As far as temperature Vs power dissipation, it is pretty complex but can be expressed mathematically. First the max temp spec is for the silicon chip inside the package, not the external package temperature. A heat sinks effectiveness is based on several things, it's size, how good the heat conduction is between the regulator package and the heat sink, the ambient temperature, etc.

A data sheet for the regulator should cover the topic if you are really interested in the math. Of all semi-conductors, linear voltage regulators are the one we usually don't worry too much about because they usually have internal protection circuitry that turns off the current flow if the temperature or current draw goes over specification.

So other then lowering the input voltage if possible, I wouldn't worry to much about it. If the heat really bothers you, or you are planning to use battery power then consider using a switching voltage regulator, they are much more effecient and therefore produce less heat for the same current draw.

Lefty