Approximatley how much energy does it take to vaporize one kilogram of steel?

Question

I fyou happen to know anythinga bout much energy is required to vaporize any amount of steel, please let me know. If you don't, please don't answer.

Answer 1

It depends on if the steel is already molten and at its vaporization temperature, or if it's at room temperature. If the former, it's just the latent heat of vaporization, which I can't find any references for, but it's a certain amount of energy per mass. (See below for a very rough estimate based on iron.)

If the latter, then you first need to heat it to its melting point. One particular steel I looked up melts at 1410°C. From room temperature you have to heat it 1390°C. The steel has a specific heat of 0.5 J/g-°C, so that's (1390°C)*(0.5 J/g-°C)*(1000g/kg) = 695000 J/kg. Then you need to melt it by applying the latent heat of fusion, which one reference suggests is 277000 J/kg. Then you need to heat it to the boiling point, which I also couldn't find, and apply the latent heat of vaporization. If you have a reference resource for properties of steel, hopefully you can find the temperature and heat of vaporization.

For a rougher estimate, I can use the values for iron, which is the main component of steel. Iron boils at 2861°C and has a heat of vaporization of 267000 J/kg (So that's how much energy you'd need if the steel was already molten and at its boiling point.), and I'll assume the liquid steel has the same specific heat capacity as the solid, which is unlikely. That's an additional temperature increase of 1451°C, and a heat input of (1451°C)*(500 J/g-°C) = 725500 J/kg. So, adding all of the terms, from room temperature you'd need 695000 + 277000 + 725500 + 267000 = 1964500 J/kg, or almost two megajoules per kilogram.

Answer 2

There are many different kinds of steel, but if you want to assume that it will be similar to pure iron, Fe has a heat of vaporization of 349.6 kJ/mol. A mole of iron has a mass of 55.845 g, so there are 17.9 mol in a kg.

Thus, it takes 6260 kJ to vaporize a kilogram of liquid iron at its boiling point. If you want to get it up to that point, you'll have to add in the energy to heat the solid (specific heat of solid iron), melt it at its melting point (heat of fusion) and heat the liquid to its boiling point (specific heat of liquid iron). These terms will likely be much smaller than the heat of vaporization, but they will add to it.