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

## 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.

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?

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## Answers and Replies

haruspex
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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.
So the rate of conversion would be 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).

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

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Can anyone verify this answer?

haruspex
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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
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?

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.

haruspex
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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.

So 460,000J/s is the answer already?

haruspex
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So 460,000J/s is the answer already?
Yes. Are you puzzled?

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.

haruspex
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I am a bit puzzled yes
In what way?

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?

haruspex
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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.

• Richie Smash