Rate of conversion of P.E. to heat for a waterfall

  • Thread starter Richie Smash
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In summary: You have to look at the energy at the start and the energy at the end. The difference between the two is what is being converted. Where is that energy at the start and where is it at the end?
  • #1
Richie Smash
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Homework Statement


There is a waterfall 230m high, take acceleration due to gravity to be 10 m s-2 and the specific heat Capacity off water to be 4200 J Kg-1K-1
First find the speed of the water hitting before htitting the ground using the Principle of Conservation of energy
Second assume all energy is going into raising the Temperature of the water, calculate the temperature rise.
And Third if the mass of water moved over the falls is 200Kg/s, what is the rate of conversion of potential energy.

Homework Equations


K.E = 1/2 mv2
P.E = mgh
Eh= m x c x dt

The Attempt at a Solution


So for the first question I did, mgh = m *230*10
= 2300m.

Now the P.E given is = to the kinetic energy of the water before hitting the ground
So I did 2300m = 1/2mv2
So v2=(2300*m)/(0.5*m)
so v = squareroot of 4600
v= 67.8m/s to three significant digits.

Now for the second part I used the formula for Specific heat capacity
Eh=m*c*dt

Eh being heat energy
dt being temperature raise so

dt= Eh/(m*c)

Eh is equal to the P.E given from the start
So dt= (m*g*h)/(m*c)
dt= (g*h)/c
dt = (10 *230)/4200
dt = 0.5 Kelvin.

For the Last part, this one I'm unsure of, but I did

If P.E= mgh then P.E= 200 * 230*10=460000J
and K.E = 1/2mv2 so K.E= 1/2 (200)(67.8)
K.E= 459,684 J

So the rate of conversion would be 460000-459684/s
=316 J/s

I just wanted to know if these were correct assumptions?
 
Last edited:
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  • #2
Richie Smash said:
P.E= 200 * 230*10=460000J
If you specify "PE per second", that would be right. I.e. what you have calculated is the power, not energy.
Richie Smash said:
So the rate of conversion would be https://www.physicsforums.com/tel:460000-459684/s
That calculation makes no sense. You calculated the landing velocity based on the gained KE equalling the lost PE. Subtracting one from the other would have given you zero had you not rounded the velocity to three sig figs. Your final answer is just rounding error.

The question is asking for the rate at which PE is being converted to heat energy (via kinetic).
 
  • #3
haruspex said:
If you specify "PE per second", that would be right. I.e. what you have calculated is the power, not energy.

That calculation makes no sense. You calculated the landing velocity based on the gained KE equalling the lost PE. Subtracting one from the other would have given you zero had you not rounded the velocity to three sig figs. Your final answer is just rounding error.

The question is asking for the rate at which PE is being converted to heat energy (via kinetic).

I understand what you mean, where do I start with calculating rate of conversion?

I'm thinking it might actually be

Energy supplied = m*c*dt
= 200*4200*0.5= 420,000J/S
 
Last edited:
  • #4
Can anyone verify this answer?
 
  • #5
Richie Smash said:
I understand what you mean, where do I start with calculating rate of conversion?

I'm thinking it might actually be

Energy supplied = m*c*dt
= 200*4200*0.5= 420,000J/S
No.
In post #1 you wrote
Richie Smash said:
P.E= 200 * 230*10=460000J
That's 460000J in how much time? What is happening to that energy, taken over the whole process? What rate is it happening at?
 
  • #6
haruspex said:
No.
In post #1 you wrote
That's 460000J in how much time? What is happening to that energy, taken over the whole process? What rate is it happening at?

Yes, that is 460000 J in one second. That energy is being converted to kinetic energy which is being converted into heat.It's happening at a number of Watts or Joules per second I believe.
 
  • #7
Richie Smash said:
Yes, that is 460000 J in one second. That energy is being converted to kinetic energy which is being converted into heat.It's happening at a number of Watts or Joules per second I believe.
Right, not the 420,000W you had in post #3. Maybe that was a typo.
 
  • #8
So 460,000J/s is the answer already?
 
  • #9
Richie Smash said:
So 460,000J/s is the answer already?
Yes. Are you puzzled?
 
  • #10
I am a bit puzzled yes, But I do see how it's not 420,000 that was for the energy required to raise the temperature by 0.5 K, I applied the wrong formula.
 
  • #11
Richie Smash said:
I am a bit puzzled yes
In what way?
 
  • #12
Saying 460,000J/S is the rate is saying that all the energy that is being produced per second is being converted to heat? Would't some of that go into Kinetic energy as well?
 
  • #13
Richie Smash said:
Saying 460,000J/S is the rate is saying that all the energy that is being produced per second is being converted to heat? Would't some of that go into Kinetic energy as well?
You have to look at the whole process. It all starts as GPE and finishes as heat etc. It just spends a while as KE along the way.
 
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Related to Rate of conversion of P.E. to heat for a waterfall

1. What factors affect the rate of conversion of P.E. to heat for a waterfall?

Some factors that can affect the rate of conversion of P.E. to heat for a waterfall include the height and volume of the waterfall, the temperature and humidity of the surrounding air, and the type of material the water is falling onto.

2. How does the height of a waterfall affect the rate of conversion of P.E. to heat?

The higher the waterfall, the greater the potential energy and therefore the greater the rate of conversion to heat. This is because there is more gravitational potential energy that is being converted to kinetic energy and then to heat as the water falls.

3. Does the volume of water in a waterfall impact the rate of conversion of P.E. to heat?

Yes, the volume of water in a waterfall can impact the rate of conversion of P.E. to heat. A higher volume of water means there is more potential energy being converted to kinetic energy and then to heat, resulting in a higher rate of conversion.

4. How does the temperature of the surrounding air affect the rate of conversion of P.E. to heat for a waterfall?

The temperature of the surrounding air can impact the rate of conversion of P.E. to heat for a waterfall. Warmer air has a lower density, allowing the water droplets to evaporate more easily and at a faster rate. This evaporation process requires energy, which can decrease the rate of conversion of P.E. to heat.

5. How does the type of material the water falls onto affect the rate of conversion of P.E. to heat for a waterfall?

The type of material the water falls onto can impact the rate of conversion of P.E. to heat. Materials that are more porous, such as rocks, can absorb some of the kinetic energy of the falling water, resulting in a slower rate of conversion to heat compared to materials that are less porous, such as concrete.

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