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

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To calculate the height of a waterfall based on the temperature change of the water, the relevant equations involve energy conservation principles. The potential energy of the water at height h is equal to the thermal energy change, expressed as mgh = mCΔT. Simplifying this leads to h = C/g, where C is the specific heat capacity and g is the acceleration due to gravity. For a temperature change of 1°C, this formula provides a straightforward way to determine the height. Understanding these relationships is crucial for solving similar physics problems.
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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
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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