So what do you all think is the correct size wire and breaker ? With multiplying by 125% and derating 70% for 8 ccc
First, the branch circuit conductors needs to have a table ampacity of 25A * 125% (continuous factor) = 31.25A at 75C, assuming 75C terminations at both ends. So that would be #10 Cu with an ampacity of 35A.
Second, the circuit conductors have to have a table ampacity of 25A / 0.7 (derating factor) = 35.7A at 90C, assuming 90C insulation. So again #10 Cu.
Lastly, the breaker has to be at least 25A * 125% = 31.25A, so a 35A breaker minimum But the small conductor rule limits #10 Cu to 30A breakers, with no exception for this application, so you'll need to use #8 Cu.
If you use #8 Cu with 90 C insulation, that has a table ampacity of 55A. So the derated ampacity is 38.5A. That means you can use a 40A breaker on it, if desired.
But if you used #8 Cu with 75C insulation, that has a table ampacity of 50A, giving a derated ampacity of 35A. For that you'd need to use a 35A breaker.
Cheers, Wayne