Conservation of energy (Waterfall)

[mk9898]

Homework Statement

In a waterfall about 20 billion liters of water per hour drop about 110 m. What is the increase of the water temperature, assuming that the entire gain of heat is transferred to the water?

Homework Equations

I'm more curious about the question in itself. Perhaps I'm missing something but how could there be a heat difference or a heat transfer when the water is falling, increasing kinetic energy and actually due to the speed the water is being cooled? I don't think I'm understanding the question properly or the question is badly worded but can someone explain what exactly is going on here?

The Attempt at a Solution

The solution I believe is straight forward and it just involves potential energy = heat energy BUT I want to thoroughly understand the exercise/the physics behind it before just plugging in some numbers into a few formulas and getting a right answer.

 

[phyzguy]

When the water falls off the edge, it gains kinetic energy as it falls. When it hits the bottom, this kinetic energy gets converted into turbulent fluid flows and ultimately into kinetic energy of the water molecules, which is heat. So the temperature of the water increases when it hits the bottom of the waterfall.

 

[mk9898]

Ah ok. Got it thanks alot! And I assume we will have to use the constant pressure specific heat capacity to solve the problem.

 

[phyzguy]

Ah ok. Got it thanks alot! And I assume we will have to use the constant pressure specific heat capacity to solve the problem.

Yes.

 

[mk9898]

I have a general question that perhaps you would be able to answer. So there is a specific heat capacity and is given the letter "c". But then there are two other specific heat capacities that deal with constant volume and constant pressure. So, what is the point of c if it is somewhat useless? c depends on the temperature, pressure and the volume. So how could someone use c to gather useful data? c = deltaQ/(M*deltaT) but that is never the case since we also have to consider pressure and volume.

Or am I confusing something here? Is c used for liquids and solids and c_v and c_p are used for gases?

 

[phyzguy]

I'd have to see the tables you are referring to, but I think the answer is that the distinction between Cp and Cv applies primarily to gases. For liquid water, the heat capacity is (by definition of the calorie) 1 Cal/(g-degree C), where 1 calorie is 4.18 Joules. It is only weakly dependent on temperature at room temperatures.

 

[mk9898]

But then why do we use c_p for water in the exercise above?

 

[phyzguy]

But then why do we use c_p for water in the exercise above?

The pressure is constant, so Cp makes sense. What values(numerically) do you have for these specific heats and what do you propose to use?

 

[mk9898]

True but it is a liquid and we use c_p, c_v for gases.

 

[haruspex]

Or am I confusing something here?

I believe you are.
The specific heat of pretty much anything, including water, depends on both temperature and pressure. Many tables are published regarding these.
For compressible fluids, there is the additional complication that, at any given temperature and pressure combination, it will give one value if you hold the pressure constant and a different one if you hold the volume constant.

 

[mk9898]

Thanks for the post. If I assume that water is a incompressible liquid, would it suffice to use c instead of c_p in the above exercise? c_p and c_v are primarily used for gases since the difference of a solid or fluid when gaining or losing heat energy is relative small compared to what happens when a gas loses or gains heat. To be precise all states of matter (gas,liquid,solid etc) have c_p and c_v but the difference between the two is so small for liquids and solids, that we can ignore them and use just c. <-- Is this the right logic here?

 

[haruspex]

Thanks for the post. If I assume that water is a incompressible liquid, would it suffice to use c instead of c_p in the above exercise? c_p and c_v are primarily used for gases since the difference of a solid or fluid when gaining or losing heat energy is relative small compared to what happens when a gas loses or gains heat. To be precise all states of matter (gas,liquid,solid etc) have c_p and c_v but the difference between the two is so small for liquids and solids, that we can ignore them and use just c. <-- Is this the right logic here?

Sounds reasonable.