# Energy required to evaporate given mass of water

1. Mar 3, 2013

### autodidude

1. The problem statement, all variables and given/known data

Calculate how much energy is required to evaporate 3.6x10^14kg of water.

(Original question is that if the oceans cover about 70% of the Earth's surface and an average of 1m evaporates per year, find the the energy required to evaporate the calculated amount of water (given radius: 6400km).)

2. Relevant equations
The assumption is that the evaporation is a simple function of heat.

3. The attempt at a solution
By dimensional analysis, I came up with:
$$E=m \delta T C_p$$
where E is the energy required, m is the mass, ΔT is the change in temperature and C is the heat capacity of water.

But I'm not sure if I'm supposed to assume that water evaporates at 100°C. A quick search showed that water evaporates at all temperatures greater than 0°C and there are a bunch of other factors like pressure and such. It's more involved than what I have but we haven't gone into all that yet and I don't think we will for this course.

So is the equation correct and if so, what should I be using for ΔT as I'm not given the average temperature of the oceans.

2. Mar 3, 2013

### Staff: Mentor

Have you heard about heat of vaporization?

3. Mar 3, 2013

### autodidude

Yes, it's defined on my sheet as 'the amount of energy in joules required to transform one gram of liquid water into water vapor at the boiling point of water'..

So should I be just looking at this energy rather than include the energy required to bring the surface of the water up to the boiling point?

4. Mar 3, 2013

### Staff: Mentor

Yes.

The question is slightly ambiguous, as heat of vaporization is a function of temperature, so you will need to assume some average temperature of the ocean surface. Or you can completely ignore it and use the value listed for boiling water (you will be off by about 10%). It all depends on how accurate you are expected to be.

5. Mar 3, 2013

### robphippen

You need to add two amounts of energy together. First, the amount of energy required to bring that amount of water to boiling point (which is mass x heat capacity x required change in temperature), second the amount of energy required to turn that much liquid water at boiling point to gas (ie stream) which is the latent heat of evaporation x mass

6. Mar 3, 2013

### Staff: Mentor

As OP already mentioned, you DON'T need to heat up the water to the boiling point.

Have you ever seen poodles drying out? Do you think they boiled out, or did they evaporate at the ambient temperature (whatever it happens to be)?

Or, if the poodles don't work for you, leave an open glass of water in your room, and report back in a month. Is there water in the glass? Have it boiled out?

7. Mar 3, 2013

### robphippen

Damp poodles and vaporizing oceans...

OK, very good point.

In fact this just makes the whole thing rather more interesting.

Of course the reason why my poodle dries out is because the air around is is not saturated with water vapor. If it was, then my poor poodle would not dry out.

Applying this to the ocean: if we start to heat up the ocean, we'd initially see more evaporation, but I think its fair to say that this would result in rather a large amount of water vapor entering the atmosphere, and so we would see the air get saturated with water vapor (at a molecular level, this means that the rate at which water molecules evaporate from the liquid water to become gas is exactly balanced by the rate at which water molecules leave that vapor (gaseous) state and become liquid water). If we heat up both the water AND the air above the water, then it will hold more water vapour, but at some point we get to thermal equilibrium (at this point I think my poodle and is getting distressed) and no more water will evaporate.

However, despite all of this, I think we can indeed say that the total energy required is 'just' the latent heat required.

8. Mar 3, 2013

### SteamKing

Staff Emeritus
You should not mistreat your poodle by waiting for him to dry out.

On the other hand, if you have a puddle,it is OK if it dries out on its own.

9. Mar 3, 2013

### Staff: Mentor

Oops. Or perhaps uups.

10. Mar 3, 2013

### autodidude

So would it then just be ~2.3kJ/g multiplied by the mass evaporated?

If so, does this assume that the forces on the water molecules at the surface is equal to the intermolecular forces on water molecules when they're boiling...which is why they require the same energy to evaporate?

11. Mar 4, 2013

### Staff: Mentor

There is no need to assume anything, you should be able to google tables of values of vaporization heat (or vaporization enthalpy) for any temperature.

12. Mar 4, 2013

### autodidude

What if I don't take an average temperature for the ocean and use the value listed for boiling water? Wouldn't this assume that the water at the surface is 'behaving' like boiled water which is why 1 gram is evaporating for every 2.3kJ or fenergy absorbed?