a gas stove flame emits light at 475nm, what is the temperature in kelvin?

Answer 1

If this is black-body radiation, the temperature is given by Wein's displacement law

T*l = 2.898*10^6 K*nm

Where T= temp in deg K and l = wavelength at the peak intensity

So T for l = 475nm = 6.1*10^3 deg K