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Heating water by dropping it from a high height.

  1. Apr 12, 2012 #1
    I was reading through my notes and this idea popped into my head:
    If you had to drop a liter of water from a certain height, could that potential energy before it's dropped be used to heat the water when it hits the ground?

    Like you got this equation ΔE = C ΔT

    Could one use that value of E in this equation Ep = mgh to work out the height that has to be dropped?
  2. jcsd
  3. Apr 12, 2012 #2
    This is just common sense (well, to me anyway). Assuming all the energy gets transferred into heating the water and none of that heat gets lost, then yes, all the kinetic energy heats up the water. But this never happens in reality due to the nature of materials and their properties.

    In real life, some of the energy is given out as sound (which is how you hear the splash), and as energy in the form of heat or other energy forms (can't think of any others).

    If you drop the water from a bowl into another bowl and none of the water splashes out then some of the heat will be lost into the bowl and then through the bowl and into the air, and so on. You'll lose some energy due to sound. In fact, I imagine most of the energy transfered into the bowl is kinetic energy which then runs through you if you're holding the bowl, or the table if the bowl is resting on a table. Then the kinetic energy the bowl gains immediately transfers into the ground, which is similar to my next point...

    If you drop water onto the ground then I think most of the energy will be lost in the Earth, i.e. your dropping of the water will heat the Earth up slightly and give it some kinetic energy. This "heating up" or "energy transfer" of water to the Earth is indirect and assumes the Earth's crust is rigid to some extent (I think). One litre of water dropped onto the ground might cause the Earth to heat up by somewhere between 10^-20 to 10^-50 degrees or so. i.e. not very much. However, most of the energy given to the Earth by the water will be kinetic. Similar to the transferred heat, the kinetic energy will dissipate into the Earth, actually moving the Earth slightly. However, since the Earth is very large and spherical, someone on the other side of the Earth will roughly balance out
    your actions, and even if they don't your effect on the Earth will be very insignificant.

    This is common sense...

    For precise calculations you'll need to know things like the material of the ground (and the material's density and the speed of sound through the material), the bowl, the Earth under your house, the height you drop the water from, the amount of water can be found by, and so on...
  4. Apr 12, 2012 #3
    Also, the water at the bottom of a waterfall has to have sufficient kinetic energy [itex] \frac{1}{2}mv^2 [/itex] to move away from the point of impact, or else a big lake would form at the bottom.
  5. Apr 12, 2012 #4


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    Basically, what you are considering is conservation of energy, where 1 J of energy is the same whether it be either kinetic, potential, electrical, thermal, ... Due to losses to other forms of energy that are not useful to you when acting out the conversion process as stated in the previous posts, 100% of the potential energy will not be converted to heating up the water.

    You can consider 1 litre of water falling into a hydroelectric generator, producing electrical energy which you could then use to heat 1 litre of water in a pot on your stove. C ΔT would be less than the mgh.
  6. Apr 12, 2012 #5
    Good point on surface tension. I guess that's as deep as we need to go for the purposes of answering OP's question.
  7. Apr 12, 2012 #6


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    You could arrange that most of the energy goes into heating the water. E.g. roll an insulated box of water down a ramp, with the wheels braked by having to drive paddles in the water.
    But it would be quite hard to detect the warming. A 100m drop would only warm it by about 0.23C, if I've calculated correctly.
  8. Apr 13, 2012 #7
    Yes but as soon as the water ends up in the tank, it moves the tank and then the Earth. I imagine most of the energy from the KE of the water gets transferred into the KE of the Earth. Think about it. If the Earth were a ball of 10cm radius then most of the KE of the water would transfer into KE of the Earth. You could more accurately model the situation by collision of two objects, rather than modeling the situation as a transfer of heat.

    Edit: of course the energy dissipates gradually with time. After the first 1/10th second after impact, most of the K.E. of the water will still be in the form K.E. (of the water)
  9. Apr 13, 2012 #8


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    No. While this is true for the momentum (as it is conserved), most of the energy can be used to heat the water. The earth has a really large mass, and E=p^2/(2m), therefore the energy per momentum is really small.
    The real limit is the isolation of your system and purely practical - your water needs some containment, this containment will heat up as well, this containment needs some external support which gives the possibility of heat transfer and so on.
  10. Apr 14, 2012 #9
    Unanticipated, Thomson and Joule met later that year in Chamonix. Joule married Amelia Grimes on 18 August and the couple went on honeymoon. Marital enthusiasm notwithstanding, Joule and Thomson arranged to attempt an experiment a few days later to measure the temperature difference between the top and bottom of the Cascade de Sallanches waterfall, though this subsequently proved impractical.

    Taken from http://en.wikipedia.org/wiki/James_Prescott_Joule
  11. Apr 14, 2012 #10
    James to Amelia

    Forget the honeymoon and chuck that thermometer over Ames.I'm off to meet Thomo so we can measure the temperature at the top and bottom of a waterfall.
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