if i have 33000kcal of heat, how much steam can i produce at what pressure?water consumption?

Answer 1

If you start with water at about 65 degrees F. you can bring 100 gallons of water to a boil then you will run out of heat energy (Kilocalories). You will not produce any pressure if you boil the water in an open barrel. The amount of water you would boil off is not enough to waste time calculating.

Answer 2

There are several things you need to know to calculate this:

First is the heat of vaporization, usually expressed in energy units per mole of substance. This site gives:

http://en.wikipedia.org/wiki/Standard_enthalpy_change_of_vaporization

40.65 kJ/mol

You probably want to know what a mol is, and the math for that is simple enough:

http://en.wikipedia.org/wiki/Atomic_weight
1 mol H2O = 18.016g (add up atomic weights for exact number)
2*1.008+16.00 = 18.016

Next you need to convert cals and kJ. The ratio is:
http://www.volker-quaschning.de/datserv/faktoren/index_e.html
1kcal = 4.1868 kJ

So lets finish this. Assuming the water is already at the boiling point, you want to calculate how much water vaporizes when that much energy is added to it.

Heres the math:

33000 kcal * 4.1868 kJ/kcal = 138164.4 kJ

how many mols are vaporized?
138164.4kJ / 40.65 kJ/mol

18.016 g/mol = 3398.9 mol

Finally you probably want grams, so:
3398.9 mols * 18.016 g/mol = 61234 grams vaporized

This is a lot (unlike the other person would have you believe), it would weigh 61.234 kg, or (x2.2) about 135 pounds of water. This also correlates to roughly 61.234 litres, or about 16.2 gallons.

33000 kcal is quite a lot of heat. To put it in perspective. "calories" when eating is actually kcals (check the label if you don't believe me). And, since a normal diet is about 2000-2200 "calories" (kcal), thats enough energy to feed a normal person for 15 days!

Here is some more for your interest:

http://atoc.colorado.edu/~toohey/A3500ans14.html
44kJ/g
www.glencoe.com/qe/science.php?qi=2695
A gallon weighs about 2800g

So, a gallon contains 44*2800kJ = 123,200kJ so we are talking about burning over a gallon of gasoline to get the kind of energy we are talking about.


As for the rest of your question ...at what pressure?

The pressure is going to depend on the container you put it in, but assuming you want the MAXIUM possible pressure, you would have to put the energy into increasing the temperature of the water vapor, not into boiling away as much as possible of it. Your pressure will depend not only on the size of the container, but how much water is in it in the first place.

Since you don't say either of these, the answer can't be given. However, its not too hard to calculate:

Water vapor at boiling point reaches 1 atmosphere (760 mm Hg). As the temperature rises, the pressure rises, and there are tables to calculate the pressure. If the pressure rises higher without a corresponding increase in temperature, you will get (very hot) liquid water. What you probably want to look up is a "triple point" or phase diagram table, which shows the state of the substance (water) for various temperatures and pressures. Heres a few:

http://www.lsbu.ac.uk/water/phase.html
http://en.wikipedia.org/wiki/Phase_diagram

The table I mentioned earlier is a "vapor pressure" table. Here's two links for that:
http://s-ohe.com/Water_cal.html
http://en.wikipedia.org/wiki/Vapor_pressure

Notice the pressure becomes VERY large as the water gets hotter. This means that steam can be under tremendous pressure even if the water vapor is only a few hundred degrees. This principle is used for steam engine propulsion, and is known as "live steam".