Re: Would black fire be the hottest? Black Fire?

The "color" of fire is so inherently a part of the physics of the electromagnetic radiation of substances according to their temperature that I can not imagine any way to make "black fire" that would be hotter than visible fire.

The heat of the gases that are the "fire" is obtained from the chemical reaction(s) that we call "combustion". The electromagnetic radiation emitted by the fire's gases is produced by the continual buffeting of the gas molecules with each other which causes them to vibrate in various ways. Photons are emitted at many different frequencies depending on the various vibration modes of the molecules, and we can predict the spectral content and the amount of energy emitted using equations that are well verified, and are collectively called the equations of blackbody radiation.

The basic equation is called the Planck Radiation Formula, and gives the "shape" of the spectral distribution of the radiation. From that formula others are derived, such as the Stefan-Boltzmann Law which gives the total radiated power at a given temperature, and the Wien Displacement Law which gives the wavelength of the peak of the radiation.

One thing to notice if you examine the equations and the graphs: If one were to increase the temperature of the gas so that the peak wavelength were well below ultraviolet wavelengths (in an attempt to make the flame "black"), the visible wavelengths would still be higher in power than at lower temperatures! So no matter how hot one were to make the flame it would always be emitting visible radiation. You can see this graphically at this very cool Applet.

I will assume that we will not try to make a cooler flame as that is not what you were asking! If we did, it would eventually cease to be a "flame" as its visible components of radiation became so weak that we could no longer see them, at which point it would no longer be incandescent. At that point the gases would be "black", but not hot!

John Link, MadSci Physicist