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

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Homework Help Overview

The discussion revolves around rearranging equations to determine the temperature increase of water in a waterfall due to energy conversion into internal heat. The problem involves a height of 365 m and a mass of 1,000 kg, although it is noted that the mass may not be necessary for the solution.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the use of equations related to kinetic and potential energy, specifically ΔEk = 1/2 mv² and W = mgh. There is an exploration of how to connect these equations with the heat equation q = mcΔT. Some participants suggest working symbolically before inserting numerical values and question how to incorporate velocity into the heating equation.

Discussion Status

There is an ongoing exploration of how to combine the equations and clarify the role of energy conservation in the context of potential energy converting to kinetic energy and heat. Some participants express uncertainty about specific terms and their meanings, particularly regarding the use of zero in energy conservation.

Contextual Notes

Participants note that the question may be missing information or clarity, particularly regarding the definitions of terms like T and the implications of energy conservation in the context of the problem.

rose22
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Rearranging equations-need clarifying & pointing in the right direction! (long)

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:
ΔEk = 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!
 
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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.
 


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 ?
 


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:
V = mgh = mc\Delta T\\<br /> \Delta T = \dfrac{gh}{c}
Where the capital T is temperature.
 

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