FL 5 CHEM/PHYS Q40

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mgarc805
Posts: 35
Joined: Tue Jun 25, 2019 11:23 am

FL 5 CHEM/PHYS Q40

Post by mgarc805 » Tue Jun 25, 2019 11:28 am

What is the primary process responsible for the loss of latent heat and entropy from the ocean at the air-sea interface ?

A. Precipitation

B. Condensation

C. Evaporation

D. Melting

the correct answer is Evaporation.

can anyone explain to me how come if the ocean is loosing energy it will result in evaporation i thought evaporation was the result of introduction of energy in other words ice needs energy to melt.
NS_Tutor_Mathias
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Joined: Sat Mar 30, 2019 8:39 pm

Re: FL 5 CHEM/PHYS Q40

Post by NS_Tutor_Mathias » Wed Jun 26, 2019 7:11 pm

The vapor pressure of water at the temperatures the ocean is at is significant. While in a perfectly closed system (say you had a small container with water), you would have a consistent amount of water in the surrounding atmospheric air since the process is in equilibrium, the ocean doesn't really work that way: Newly formed water vapor can be carried away, and the ocean itself will lose some net amount of water that it won't regain until it is refilled indirectly by precipitation (this is the hydrlogic cycle in a nutshell).

So you are right, an input of energy is required to evaporate the water- but that energy is lost with the evaporated water (the input of energy in fact increases the mean molecular kinetic energy of these molecules of water, something we call an increase in temperature).

What also makes this effect even more dramatic is that not only do you lose some water molecules and their kinetic energy once they evaporate (since they will be carried away), it is also that those that evaporate are the ones with the highest kinetic energy (enough to break free of the intermolecular forces of water), basically skewing the distribution of those remaining towards just the ones with less kinetic energy. In simpler terms: What is left gets colder!

A big thing to conceptualize here is that you are right, evaporating things requires energy - and as water evaporates, some energy is actually lost from the remaining body of water. This is the idea behind evaporative cooling.

I realize this can be a little confusing and benefit from multiple possible perspectives, so let me know if you want me to give it another shot from a different angle!
Encrypted
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Joined: Sun Aug 04, 2019 10:21 pm

Re: FL 5 CHEM/PHYS Q40

Post by Encrypted » Sun Aug 04, 2019 10:23 pm

Another angle if you don't mind please!

I'm having trouble wrapping my head around this one.
NS_Tutor_Mathias
Posts: 280
Joined: Sat Mar 30, 2019 8:39 pm

Re: FL 5 CHEM/PHYS Q40

Post by NS_Tutor_Mathias » Mon Aug 05, 2019 2:53 am

How about this, for the least-theoretical take possible:
When water evaporates off the ocean, it is essentially lost - a lot of it rises, gets blown over land, precipitates back down, collects in some watershed as a stream, (or refills an aquifer), the stream mounds into a river, and eventually the river dumps back into the ocean (virtually all of them do this). But the water coming back is now significantly colder (hence why it is a liquid, not a gas!), so every time this happens, the ocean is losing heat overall despite retaining more or less the same net amount of water.

Or as a general perspective, so we can leave the hydrologic cycle out of this:
Throwing heat at a pot of water will raise the temperature of that pot. And the higher the temperature, the faster that body of water will be losing molecules to evaporation. However, the process of evaporation is clearly endothermic - the heat of vaporization is a positive constant after all (for water, about 40kj/mol). That means to change the phase of any single molecule requires energy input - meaning the substance itself is going to lose some every time a molecule changes from a liquid to gas.

This is the general concept & idea behind evaporative cooling: Even things that vaporize readily require energy to change phase, and so anything they are in contact with will actually become colder as they evaporate. Sweating works this way, as do for example treatments for fevers such as alcohol rubs, or the practice of methanol injection into a car intake. In the case of sweat, the water at the surface of your skin evaporates and cools your skin. When alcohol is used, evaporation proceeds more rapidly, allowing for a greater amount of cooling provided alcohol is consistently re-applied. In methanol injection into an intake, the spray of liquid methanol changes phase to a gas and cools the surrounding air (conveniently making it denser, allowing you to get more oxygen into the combustion chamber and ultimately go faster).


(I did my very best to leave any talk mean molecular velocities and Maxwell distributions out of this. While this whole idea of how and why evaporative cooling works goes pretty deep, you will be very well served by just knowing that evaporation always requires an energy input for the MCAT)

Addendum:
Your ice example is really the same thing. Try to differentiate adding energy to ice (this will just make the ice warmer, not melt it yet!) to the process of ice melting (this will actually LOWER the temperature of the remaining ice).

This is why on curves like this one, where heat is being added at a constant rate to a fixed amount of water, you'll notice that as the substance is undergoing a phase change, the temperature isn't changing - all energy put into it is going into the phase change, rather than becoming warmer.
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