THis is for steady-state only (the act of switching can also causes a LOT of losses which should be accounted for by adding them into the conduction/steady-state losses).
THe voltage/resistance you use for your conduction losses is the voltage drop or resistance ACROSS the source-drain (the primary power terminals of the device). So you do not use the at the gate 10V since the other conduction and switching losses (rather than the gate capacitance charging losses) dominate, especially for power devices.
For example,
Ploss = RI^2 = (.23R)*(25A)^2 = 144W (for a 0.23R MOSFET which behaves like a small resistor when used as a switch)
Ploss = RI^2 = (0.7V)*(25A) = 17.W (for a 0.7V IGBT which behaves like a diode when used as a switch).
THe thermal resistance is how many degrees the device will rise above ambient for every watt of power that is dissipated in it (hence the units).
So without a heatsink, and if the heat path is from the silicon to the air (junction-ambient):
Trise = Ploss * ThermalResistance = 144W * 62.5C/W = 9000C
If the ambient was 25C, then the device temperatuer is 9025C. Which puts it well above the temperature limit which is probably 125-150C.
If you have a heatsink then the heat must flow from silicon (junction) -> case-> heatsink. Then the thermal resistance you use is the thermal resistance of junction-to-case + heatsink-ambient thermal resistance.
Obviously 9000C is a bit ridiculous so you'd need to find a much better MOSFET and/or heatsink to bring it within the device limits. ANd remember, you aren't even accounting for switching losses yet which, when the device is sized properly should be about equal to the conduction losses.
If the max device temperature is 150C and your max ambient temperature is 40C, then you can only allow a 110C temperature rise at most. So you must use this 110C in your temp rise calculation and find the appropriate heatsink (and if need be, find another device). So if you can estimate your maximum ambient temperature, device losses, and thermal resistance you can estimate your operating temperature.
Now here is one difference between IGBT and MOSFET. MOSFETs act like a small resistors when used as a switch and IGBTs act like diodes. For a fixed, lower current, the losses in the MOSFET will be less (since the voltage drop across the resistor is less that that across the diode). But as you increase the current, so too does the voltage drop across the resistor and so do the losses. But the IGBT's diode voltage drop stays the same. Since the power loss in a resistor increases by I^2 and the power loss in a diode only increases by I, this means that at a certain point (usually a very high current since you can get MOSFET with very very low resistance) the losses in the MOSFET become greater than those in the IGBT and continue to skyrocket from there.
In reality, the reason IGBTs are used over MOSFETs is because IGBTs are able to handle the higher voltages used in industrial drives. Higher voltages also mean lower current for the same power and lower current = less losses.