Mental excercise.

[Ferret]

I got a funny thought in my head the other day, how high up would a litre of water have to be droped from to makes it evaporate? Baring it splashing out, wind resistance etc.?

I think I've got it right, backwards done KE= 1/2*M*V^2. And the one 4.2 joules thing

Worked out the

V = a * t

X = .5 * a * t^2

And the number that popped out was 428.5 Metres which seems low, am I right, am I mad or do I just have to much timwe on my hands?

 

[ph34r]

...or do I just have to much timwe on my hands?

I'll go with this option.

 

[Ankh-Morpork Guard]

Ow.

 

[Hand of Evil]

I think your number is off think it would be a lot higher, it would be ice?

 

[Pbartender]

You need to specify how hot your water is to begin with.

1 calorie is the energy required to raise the temperature of 1 gram (1 mL) of water by 1 degree Celcius. 1 calorie == 4.1868 joules.

1 Calorie is the energy required to raise the temperature of 1 kg (1 liter) of water by 1 degree Celcius.

So, assuming you want just enough energy to turn the water into steam (and not further heat the steam), the height you drop the water from depends entirely on the mass of water dropped, and the initial temperature of the water.

Someone can correct me, if I make a mistake, but...

[Napkin calculation]If the 1 L of water was at room temperture (25 C), and you wanted it to reach the boiling point (100 C), and it was falling in earth's gravity... Then you would need about 314 kJ of energy, it would have to be traveling about 800 m/s at impact, it would fall for about 81 seconds, and you would have to drop it from a height of about 32 km. Actually getting it to evaporate completely would require a bit more energy, but I don't have that figure on hand.[/Napkin]

But that would only work in the absence of wind resistance, and several other complicating factors.

 

[Stone Angel]

PBartender pointed out the very variables I was going to ask about.

So what brought about this thinking have a nasty Water Elemental you need to get rid of!


The Seraph of Earth and Stone

 

[tarchon]

Water doesn't have to be heated to evaporate. You would have to know the temperature/pressure/humidity profile of the air column in question to answer this, as well as the dispersion pattern of the water. The real answer is "it never evaporates completely" - it's an asymptotic process (if it was 500 m, we wouldn't have rain, right?), and you have to solve a differential equation to find the remaining fraction at time t or distance x since the evaporation rate varies with remaining volume. Not a very difficult DE if you approximate the water body as a sphere, but there's more to it than just using the heat of vaporization and the kinetic energy.

 

[tarchon]

So what brought about this thinking have a nasty Water Elemental you need to get rid of!

Oh, my answer as a DM is that water elementals can control their evaporation rate, so it won't happen.

 

[Mark]

Oh, my answer as a DM is that water elementals can control their evaporation rate, so it won't happen.

And so endeth the lesson...

 

[Ferret]

You need to specify how hot your water is to begin with.

1 calorie is the energy required to raise the temperature of 1 gram (1 mL) of water by 1 degree Celcius. 1 calorie == 4.1868 joules.

1 Calorie is the energy required to raise the temperature of 1 kg (1 liter) of water by 1 degree Celcius.

So, assuming you want just enough energy to turn the water into steam (and not further heat the steam), the height you drop the water from depends entirely on the mass of water dropped, and the initial temperature of the water.

Someone can correct me, if I make a mistake, but...

[Napkin calculation]If the 1 L of water was at room temperture (25 C), and you wanted it to reach the boiling point (100 C), and it was falling in earth's gravity... Then you would need about 314 kJ of energy, it would have to be traveling about 800 m/s at impact, it would fall for about 81 seconds, and you would have to drop it from a height of about 32 km. Actually getting it to evaporate completely would require a bit more energy, but I don't have that figure on hand.[/Napkin]

But that would only work in the absence of wind resistance, and several other complicating factors.

Thats what I missed out, trademark silly mistake.