So you seem to have already worked out what information is required for calculating overheating: the max current and the ambient temperature.
The MOSFET has an on-resistance, called RDS(on). That's the resistance between drain (D) and source (S) terminals when driven on at a specific gate voltage or VGS. That gate voltage is almost universally 10V. Under those conditions, the MOSFET acts like an ideal resistor, so Ohm's law can be applied.
V = I*R
Given the resistance and the current, what is the voltage across the MOSFET? No, it's not 50V. You should get an answer under 1V.
Given the voltage and current through the MOSFET, how many watts of power is it dissipating?
Then you have to find the "thermal resistance" in the datasheet. It will be given in degrees per watt. Sometimes degrees C, more often degrees K, which makes no difference to us. This is highly variable depending on what heatsink you attached the MOSFET to. The datasheet doesn't know what heatsink you have so it will only give you one or two examples.
There's actually a whole lot of thermal resistances that can be important here. The 'junction' is the actual semiconductor element inside the MOSFET. You're interested in junction-to-ambient, since you've specified an ambient temperature. Or maybe you will be using a heatsink so you want junction-to-case. That is the thermal resistance from the junction to the outside heatsinking surface of the device.
Junction-to-ambient for a TO220 case MOSFET might be 50ºK/W. So hitting it with 2 watts of heating will heat it up by 100 degrees above ambient. Add the ambient of 32ºC and you're at 132ºC.
How hot is it allowed to get? That's in the datasheet too. Usually maximum junction temperature is 125ºC, so you've exceeded that. The thing will burn up.
The thermal resistance junction-to-case will be much lower. Maybe 2ºK/W. So putting it on a heatsink will help. Maybe you find a heatsink with 15ºC/W thermal resistance. That's actually a pretty small heatsink. Maybe $1 or $0.50 and it just clips over the TO220 case. So 2W power will heat up the heatsink to 60ºC above ambient. 92ºC total. The junction is then only 4 degrees hotter than that, so the device is safe and won't burn up with 2W power dissipation.
For manual switching, you don't need to consider switching losses unless you do something stupid like putting a capacitor on the gate to slow it down.