About my imp

[Gabriele]

We have a container full of water with a mobile wall linked to a counterweight that mantain the pression actin inside the container to te wall. We bring water temperature stealing ENERGY from 4°C to near 0°C. Density of water decrease,pression on the wall increase, we Catch gravity potential energy with an addittional weight and take that apart. We gained a potential energy. Now we increase again the temperature of the water moving by near 0°C to 4°C again and we move the wall returning to original point. Considering water head increas of 8-9% we are don't talking about violating 2nd conservation law,but the FIRST. We can manipulate matter to Create HEAT... muhahahaha.

I'm waiting your answers. Thank you.
If you don't have understand well the experiment that can be done i will answer you here.
Thanks Again. Your friend Gabriele

my email is gabriele.citossi@virgilio.it

P.S. I'm waiting for a Nobel

[vasik041]

Just make calculations and see how unpractical such device will be 

[sm0ky2]

The density of Water does not change very much over such a small change in temperature.
we'll take the standard situation, but similarity applies to varying altitudes and pressures....

At Sea level, and normal Atmospheric Pressure::

at 4-degrees (c) the density of water is: 1 gram per cubic centimeter. 

at 0-degrees (c) the density of water is : 0.99987 grams per cubic centimeter.

or if you prefer the Standard conversion, it amounts to a difference of about 0.008 pounds per cubic foot.

or in terms of liquid-volume, a difference of:  0.000192 kilograms per liter.

you would need a LOT of water to make a very small difference in density.

Energy of a volume of liquid applies to the ENTIRE VOLUME of liquid, not just the temperature itself.
therefore, the more you increase the volume, the more energy is required to change the temperature of that volume of water.

The British Thermal Unit describes the amount of energy required to heat 1 pound of water by 1 degree (f) as: 1059 Joules
two pounds of water requires twice this amount of energy to heat.
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Now, let's examine what happens if you bring the volume of water down below 0, it will freeze,
then yes you experience a larger decrease in density ( close to 8 or 9 % as you predicted)
and the volume will increase. This would be sufficient to move your "wall".
(note that the Thermal Conditions of the device require an equal amount of energy extracted,
as put back in to raise the Temp. back to 4-degrees. Therefore: energy is not lost or gained by cooling and heating the water)
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NOW: We will make some assumptions (since you didn't describe this container in very much detail)

1) The container itself, is sufficiently rigid so as to not allow expansion in any direction, except that of the moving wall.
     - When water changes phases to ice, it expands in EVERY direction. This increases the pressure throughout the container.

2) The wall is oriented to expand in the vertical direction
    - This is the only way that makes sense, as any other orientation will result in a lowering of the volume of water-level,
       thus a loss in potential energy of the volume of water on the back end of the cycle.


Assuming the counter-pressure by the weight was not too heavy for this process, the weight will lift.
So, let's take a look at that process::

If we value the weight at let's say...   1 KG, and it is leveraged by the motion of the wall to lift a distance of 1 Meter.
E = mgh ;   1Kg * 1 m * 9.8 m/s = 9.8 Joules ;  ok, great we have now gained a potential of 9.8 Joules, by expanding the water.

The conclusion made here, is that when you raise the Temperature of the water, increasing its density once again,
and thus lowering the volume, you can "extract" energy by then lifting a second weight.

Logically, this makes sense. But only because you are missing one important factor::

The pressure

There is an important relationship between Temperature, Pressure, and Volume.
When you "extract" the energy gained by lifting the weight, you are also changing the pressure exerted on the volume.
And Thus you will lose exactly the same amount of energy, in form of Temperature.

[ modified for clarity:  The weight being lifted exerts pressure on the volume of water/ice inside the container, if you transfer this to the second weight, the pressure is relieved, thus lowering the Temperature of the volume.]

What this means is,. for every Joule of energy you "gain" in the form of Potential Energy of the second weight,
   You will have to put back into the water to heat it back up to 4-degrees.

Hope this helps.

                      Sm0ky2

[vasik041]

More practical would be phase transition e.g. liquid - gas as we see it in nature...

[Pirate88179]

More practical would be phase transition e.g. liquid - gas as we see it in nature...

But, to do that do you not need energy imported from somewhere?  In nature we see this as water evaporates but, there is that thing called the sun that inputs a lot of energy so this can happen.

Bill