Working out a value for H!?


by echoindia756
Tags: working
echoindia756
echoindia756 is offline
#1
Feb12-09, 01:47 PM
P: 2
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



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)
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LowlyPion
LowlyPion is offline
#2
Feb12-09, 03:15 PM
HW Helper
P: 5,346
Welcome to PF.

4200 or 4186 the method looks sound.
echoindia756
echoindia756 is offline
#3
Feb12-09, 04:24 PM
P: 2
Quote Quote by LowlyPion View Post
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.

LowlyPion
LowlyPion is offline
#4
Feb12-09, 06:31 PM
HW Helper
P: 5,346

Working out a value for H!?


Quote Quote by echoindia756 View Post
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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