Does falling water chill due to losing gravitational energy?

Quote: Pacomartin

I must assume you misheard this person, or he got it backwards.

Quote: Wizard

I think that effect is negligible.

Quote: Wizard

Or I'm not asking the question properly.

Quote: Pacomartin

Potential energy has to be converted into some other kind of energy. That would imply water temperature goes up.

Quote: Doc

Wizard,

Reading your reply, I thought maybe I had misunderstood your question. I went back and looked at the thread again. First thing I noticed was that either my memory is bad or your first post has been edited. I think it initially had a quote that seemed related to running or cycling, with the waterfall question being a follow-on issue. However, that quote isn't there any more.

I also noticed that you have more recently restated the question as, "If we ignore the effect of air resistance and equal air temperature and pressure the entire time, does the water temperature change between at the top and half way down?"

For that question, I would repeat my earlier comment that if we also ignored evaporation, there is no reason to anticipate any temperature change of the falling water prior to impact. If we include evaporation, then the liquid water will be cooled during its fall. Upon impact and loss of velocity (kinetic energy), the temperature will rise, but I still suspect that the evaporative cooling will have a bigger effect that the conversion of potential energy to kinetic energy to internal energy.

In support of this suspicion, I will point to the first comment appended in that physics article that AZDuffman linked to at the beginning of the thread, the article that suggested the 0.1 degree temperature increase. That comment said:

Quote:

Having grown up in Niagara Falls, I can tell you that the water at the bottom is definitely colder than the water at the top. He must have forgotten to include convective cooling as the water falls through air.

That sounds right to me, except the commenter should have said "evaporative" cooling instead of "convective" cooling. (Convective heat transfer could either heat or cool the water, if the surrounding air is warmer or cooler than the water, but I suspect that is a very minor issue, too.) As for the original question in post #1 -- "... somebody said that that water temperature goes down after falling down the waterfall because it loses potential gravitational energy. Is this true?" -- that is definitely false reasoning.

     Doc

Quote: Doc

O.K., maybe now I see the source of the confusion. I think it lies in your understanding the meaning of gravitational potential energy. I think that several people participating in the thread understand this well, but others may not. That difference in understanding could create a barrier to effective discussion. If, however, I have misinterpreted your comments and underestimated your understanding of potential energy, I apologize. For now though, I will try to explain.

Gravitational potential energy of a body is not a property that can be determined by any examination of the body itself. It has nothing to do with the atoms/molecules inside the body, other than how much mass they represent all together. It is instead the energy that is related to the position/location of that body within a gravitational field. The term "potential" refers to how much energy we could get from the body by letting its position change; i.e., how much mayhem could we potentially create by letting the brick drop from the top of the wall to the ground below?

It is usually more useful to talk not about the absolute value of a body's potential energy but about how much that energy changes when the position/location changes. For example, the potential energy of a body decreases as its elevation is reduced and increases as the elevation is increased.

If we are considering a uniform gravitational field, such as near the surface of the earth, we can quantify the change in potential energy as the product of the body's mass times the acceleration of gravity times the change in elevation, mg(∆h). However, we cannot specify the value of its potential energy in absolute terms at either location unless/until we specify a "reference" location, such as the surface of the ground nearby (call that elevation zero). If we specify that reference state, then we can say that the body has a potential energy at any other location equal to the amount that its potential energy is greater than it would be down on the ground, mg(∆h) = mgh. Of course, if the body were to move lower than the surface of the ground, it would have a negative potential energy, based on that reference state. A negative value of potential energy means that we would have to add energy (i.e., do work) in order to lift the body back to the surface to a potential energy level of zero.

If we let the body fall from some height, the elevation and the potential energy reduce, but the force of gravity accelerates the body, giving it velocity and kinetic energy, which is equal to 1/2 the mass times the square of its velocity. If there are no other forces involved and the body starts from rest, it will achieve a kinetic energy equal to the loss of potential energy, (1/2)mV2=mg(-∆h). While it is falling, potential energy is being converted to kinetic energy, but this does not have any effect on the temperature of the body.

This answers the second version of your waterfall question, neglecting evaporation and interactions with the air. From the top of the falls to half way down, the water speeds up, but it does not change temperature.

When the water hits the pool at the bottom, it eventually comes to rest (very nearly). Its kinetic energy is again zero (neglecting the velocity of the stream at the top and at the pool), but it has lost a bunch of potential energy. That energy cannot have been destroyed. Instead, it has been converted to internal energy, which is represented as an increased level of molecular motion within the body (the droplet of water you have been following down the falls). That increased molecular motion can be detected as an increase in temperature.

This is what I have been describing as conversion of potential energy to kinetic energy to internal energy, and the physics article that AZDuffman linked calculated this as giving a temperature increase of 0.1 degrees.

I repeat again my comment that in the real world, a small portion of the falling water will evaporate as it is greatly exposed to the air during the fall. (In the theoretical waterfall in a vacuum, there would be even more evaporation.) This evaporation will absorb a lot of energy, and I highly suspect that it will cool the remainder of the water by more than the 0.1 degree temperature increase associated with the conversion of potential energy, resulting in a temperature decrease rather than an increase.

Quantifying the portion of the water that will evaporate and the portion of the absorbed energy that will come from the remaining liquid water (as opposed to coming from the surrounding air), would be a very challenging problem, so I cannot do a thorough job of predicting what the final temperature of the water would be.

Doc