# In the Niagara Falls, which factors prevent rise of T H2O falling a certain height?

I need some ideas on a problem.

The first part says: Whats the posible rise in the temperature of the water falling 49.4 m in the Niagara Falls? That one was easy, with answer 0.112 Kelvin. ($\Delta T = \frac{g*h}{c_{H_2 O}}$)

The second part asks what factors tend to prevent that rise in temperature? Im thinking kinetic energy, preassure, air conductivity but im not sure.

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The conservation of energy means that the potential energy liberated in falling must be present as heat. But there are precisely three ways for the water to lose that heat

• conduction/convection: by contact with air
• radiatively: by emitting predominantly infrared light.
• evaporation: by losing vapor into the air as the water is falling.

From experience sweating, it should be apparent that in a fast air-flow environment, evaporative cooling is much more important than the other two effects. But this list is exhaustive--- there is no other place you can put the energy liberated by the fall.

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Why does it all have to go to heat? What happens with increased kinetic energy of the droplet and the surrounding air? – Bernhard Feb 25 '12 at 8:20
@Bernhard--- that heats the air, since the motion in the kinetic energy of the air is in steady state. So the amount of kinetic energy in the air isn't increased by the water, except to the extent that the air conducts, convects, or radiates away the heat. All energy ends up as heat eventually. – Ron Maimon Feb 25 '12 at 15:09
Of course, but this dissipation to heat in the air phase is not explicitly in your list. Conduction/convection is with respect to heat itself, this is a different process. – Bernhard Feb 25 '12 at 18:41

The temperature rise is just the potential energy change divided by the specific heat capacity of water (which gives the answer you quoted).

Anyhow, the rise would be less than this if not all the potential energy change is converted to heat, but I'd be surprised if this were the case. I suspect the only other significant effect would be evaporation cooling the water as it falls.

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Although I also find his "dimensional analysis" wording problematic, his calc is right. link (so maybe that link doesn't work, the silly comments don't take the carrot and multiply characters well, you should be able to figure it out) – Alan Rominger Feb 24 '12 at 13:58
Heat transfer to the surroundings while falling is the biggie; will be a combination of water-air temperature difference and also evaporative cooling (a function of relative humidity). – Daniel Chisholm Feb 24 '12 at 19:23
Instead of downvoting, I deleted the dimensional analysis thing. It's not dimensional analysis. – Ron Maimon Feb 25 '12 at 1:32