- #1

Richie Smash

- 293

- 15

## 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

^{-1}K

^{-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 mv

^{2}

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 = 1/2mv

^{2}

So v

^{2}=(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 = 1/2mv

^{2}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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