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

• rose22
In summary, the individual is asking for help rearranging equations to find the temperature increase of a waterfall, where energy is converted into internal heat of the water. They have used equations ΔEk = 1/2 mv2 and W = mgh, and are aware that q = mcΔT needs to be incorporated. They have rearranged their equations and calculated a value of 357 for ΔT. However, they are unsure if this is correct and how to incorporate the velocity into the heating equation. They are advised to try using the equations T = V = mgh = q = mcΔT, where the zero represents the conservation of energy.
rose22
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!

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:
$V = mgh = mc\Delta T\\ \Delta T = \dfrac{gh}{c}$
Where the capital T is temperature.

Hello,

I can provide some clarification and guidance on rearranging equations to find the temperature increase of a waterfall.

Firstly, it is important to understand the concepts and variables involved in the question. In this case, we are dealing with the energy of a waterfall and its conversion into internal heat of the water. The key variables are the height of the waterfall (h), mass of water (m), and the gravitational constant (g).

Based on the equations you have provided, it seems like you are on the right track. The equations for kinetic energy (ΔEk) and work (W) are applicable in this scenario. However, we also need to consider the specific heat capacity (c) and the change in temperature (ΔT) of the water.

To rearrange the equations, we can start with the equation for work (W = mgh) and substitute the value for gravitational constant (g) to get:

W = mg x h

Next, we can substitute the equation for kinetic energy (ΔEk = 1/2 mv2) into the equation for work, which gives us:

ΔEk = mg x h

Now, we can rearrange the equation to solve for the change in temperature (ΔT):

ΔT = ΔEk / (mc)

Substituting the values given in the question, we get:

ΔT = (1000 kg x 9.8 m/s2 x 365 m) / (1000 kg x 4.2 kJ/kg∙K)

= 85.24 K

Therefore, the temperature increase of the waterfall is approximately 85.24 Kelvin.

In summary, to rearrange equations, it is important to understand the concepts and variables involved and use the appropriate equations to solve for the desired variable. I hope this helps to clarify and guide you in the right direction. Let me know if you have any further questions. Good luck with your assignment!

## 1. How do I know when to rearrange an equation?

Rearranging equations is typically done when solving for a specific variable or simplifying a more complex equation. It is important to identify what variable you are trying to solve for and make that the subject of the equation.

## 2. What are the steps to rearranging an equation?

The first step is to identify the variable you are solving for and make it the subject of the equation. Then, use inverse operations (addition, subtraction, multiplication, division) to isolate the variable on one side of the equation. Finally, simplify and solve for the variable.

## 3. Can I rearrange equations with multiple variables?

Yes, equations with multiple variables can be rearranged. However, it is important to have enough information to solve for all of the variables in the equation. This may require using multiple equations or substitution.

## 4. How do I check if my rearranged equation is correct?

You can check your rearranged equation by plugging in your solution for the variable into the original equation. If the left and right sides of the equation are equal, then your rearranged equation is correct.

## 5. Are there any common mistakes to avoid when rearranging equations?

Yes, a common mistake is forgetting to perform inverse operations on both sides of the equation. It is important to maintain the equality of the equation by performing the same operations on both sides. Another mistake is mixing up the order of operations, so it is important to follow the correct order (PEMDAS) when simplifying the equation.

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