Rearranging equations-need clarifying & pointing in the right direction!!!!!! (long)

rose22

1. For an assignment question, I need to rearrange equations to find the temperature increase of a waterfall, specifically energy that has been turned into internal heat of the water. The height of the drop is 365 m. Mass of water is 1,000 kg (though the question states answer can be found without it). There isn't any net transfer of energy between mass of water & kinetic energy gained in the fall.



2. Equations that I've used so far:
ΔE_k = 1/2 mv²
W = mgh

g = constant

I know that I need to use q = mcΔT



3. So far, this is what I have done:

(rearranged from 1/2 mv² = mgh)

v = √2gh
v = 2 x 9.8 x 365 = 7154 (7 x 10³)
√7154 = 84.58 = 85

Then, I did:
(rearranged from q = mcΔT)

ΔT = q divided by mc
85 / 1000 x (4.2 x 10³) = 357

Is this about right? How can I put the 3 equations together, or are they alright like that?

Thank you!!!!

dikmikkel

q is in fact an energy. It is the energy gained. Do it symbolically first and then insert your values.
How can you insert the velocity in the heating equation ?
You can find the decrease in potential energy which is equal to the gained kinetic energy and then the temperature. Maybe like this:
dE = 0 <=> T = V <=> T = mgh
And plug that into the T = mgh = q = mc dt <=> gh = c dt <=> dt = gh/c
Hope that i understood the problem.

rose22

i think i understand, but where does the 0 come in?? so, with T = mgh, is that deltaT?

so putting the equations together would result in:

T = mgh = dt = gh/c ?

dikmikkel

The T stands for kinetic energy, the V potential dE in- or decrease in energy.
The zero is because the energy is conserved(Or all potential energy turns into kinetic/Heat). I think something is missing in the question if it isn't right. The way i understand it is: All potential energy lost turns into heat and hence:
[itex] V = mgh = mc\Delta T\\
\Delta T = \dfrac{gh}{c}[/itex]
Where the capital T is temperature.