How Much Warmer Is Water at the Bottom of Niagara Falls?

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SUMMARY

The discussion focuses on calculating the temperature increase of water at the bottom of Niagara Falls compared to the top, which is determined to be 0.118°C. The calculation utilizes the principle of energy conservation, specifically the equation Δthermal + Δpotential = 0, factoring in the specific heat capacity of water (4187 J/(kg·K)) and the height of the fall (50.6 m). The conversation also touches on the conversion of kinetic energy to heat upon impact with the water surface.

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i can't even figure out where to start with this one

The water going over Niagara Falls drops 50.6 m. How much warmer is the water at the bottom of the falls thatn it is at the top? Disregard any possible effects of evaporation of water during the fall.

any help would be appreciated and the final answer is .118 C
 
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Energy is conserved:

\Deltathermal + \Deltapotential = 0

mc \Delta T + mgh = 0

Factor out the m

(4187) \Delta T + (9.8)(-50.6) = 0

\Delta T = 0.1184
 
thanx soo much i guess i thought it was soo hard that i believed myself
 
the unsettling thing is, what about the kinetic energy?
 
What kinetic energy? The water as it falls, increases speed (so kinetic energy also) as it loses potential energy. However, when it hits the bottom, it stops going down! There will be some small motion of water but it will be so turbulent that eventually it will be converted to heat. In the long run all the energy is converted to heat.
 
where did u get 4187?
 
joehoy41 said:
where did u get 4187?

The c term, the specific heat capacity of water in J/(kg.K).

BTW, this thread is nearly 8 years old!
 

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