How Do You Calculate the Height of a Waterfall Based on Temperature Change?

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Homework Help Overview

The discussion revolves around calculating the height of a waterfall based on the temperature change of the water, specifically a 1°C increase from the top to the bottom of the waterfall. The problem involves concepts from thermodynamics and gravitational potential energy.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to relate the temperature change to potential energy using the equation E = MCΔθ and discusses the relationship between mass, specific heat, and height. Some participants question the specific heat value to use and explore the implications of the equations presented.

Discussion Status

Participants are engaged in exploring the relationships between energy, temperature change, and height. Some guidance has been offered regarding the equations, but there is no explicit consensus on the next steps or the specific values to use.

Contextual Notes

There is a mention of uncertainty regarding the specific heat capacity value (4200 or 4186), which may affect the calculations. The original poster expresses difficulty in progressing beyond their initial setup.

echoindia756
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Working out a value for H!?

Hi there!

I'm a new student to physics and I'm still learning a lot about it.
Our teacher recently gave a question and I'm not sure how to work out the answer, can anyone help me?

So here it goes:
Here's a waterfall (See pic), the water at the bottom of the waterfall is 1C hotter than the water at the top of the waterfall.

How would I work out a value for h?

Please see picture below for diagram

http://www.gadgetireland.com/images/uploads/pic1.jpg

Any help much appreciated!

Here is my attempt at solving the problem:

E = MC (delta) theta

= (Mass of water) (4200) ( theta + 1)--------------------------------------------
Potential energy = mgh

= M(9.8)(h)
--> 9.8mh = 4200 (mass)( theta + 1)

--> 9.8h = 4200( theta + 1)
 
Last edited:
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Welcome to PF.

4200 or 4186 the method looks sound.
 


LowlyPion said:
Welcome to PF.

4200 or 4186 the method looks sound.

Right I understand that but I can't seem to get any further than the above.
 


echoindia756 said:
Right I understand that but I can't seem to get any further than the above.

Sorry I thought you had it.

Examine then the ΔE which by your equation can be written as

ΔE = m*C*ΔT

mgh = mC*ΔT

gh = C*ΔT

For a 1° change then

h = C/g
 

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