I found this answer here which explains how to calculate the watts that are dissipated in a switching voltage regulator, but once you calculate the watts dissipated how do you know how much temperature your device will dissipate?
I am using an LM2576HVT-15. There will be an input voltage of 42 volts and the output voltage will be 15 volts. The load will draw somewhere between .2 amps and 1 amp so I calculated the watts dissipated being 15V * 1A = 15W. In the datasheet it shows an efficiency of about 87% so we get 15W / .87 = 17.24W which means about 2.24 watts will be dissipated.
Now that I know how many watts will be dissipated what do I do from here? The datasheet shows all of these temperature values, but I don't which ones I need to take into account for the heat dissipated? I'm just hooking it up to a breadboard with TO220 packaging.