plantdoctor - 24-6-2004 at 23:44
If I wanted to create Magnesium Sulphate Anhydrous from Magnesium Sulphate Heptahydrate is there any method of calculating how much heat I would need
to drive the excess water off?
sanity gone - 25-6-2004 at 00:26
I don't know if that is what you are looking for:
heat = mass x specific heat x temperature change = heat capacity x temperature change
I don't know the specific heat for the dehydration but for water as a gas it is 1.86J/gK
hodges - 25-6-2004 at 14:28
I don't know the real answer, but my guess for an estimate would be at least the amount of heat necessary to vaporize the mass of water in
question (540 cal/g).
unionised - 25-6-2004 at 14:34
Do you mean how much heat or do you mean what temperature?
It loses about half of the water if dried at 100C and all the water if you dry it at 300C.
The best way is to keep drying it until it has lost about half its original weight.
t_Pyro - 25-6-2004 at 20:11
You need to know the enhalpy of hydration of anhydrous magnesium sulfate. If it is "- x" KJ/mole of MgSO<sub>4</sub> with liq.
H<sub>2</sub>O, then the heat required to dehydrate y moles of magnesium sulfate is y*x+{heat required to evaporate the water liberated}.