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From this answer:

Actually, in the case of lead, if the water were deep enough, the lead would sink to the point where its weight equals that of the water under pressure at depths. As lead will compress as well as the water, that may never happen, but for other objects and/or fluids it might.

Is there a depth where a block of lead (say a 1" cube at sea level) would just suspend in the water?

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The density of lead is more than 11 times higher than the density of water. You will need the water density to increase due to pressure at the bottom by a similar factor. The pressure at the deepest ocean areas (about 11 km) is about $ 11 \times 10^7 Pa $. The bulk modulus of water being about 2 GPa, the change in volume produced by a pressure of $ 11 \times 10^7 Pa $. will be about $$ \frac{11\times 10^7 Pa}{2 \times 10^9 Pa} = 5.5 \times 10^{-2} $$ or about 5%. This is the order of magnitude of the density changes in the water. The effect is negligible compared with the goal. And compresibility of water at high pressures may be even higher that the value used in this estimation so the effect will be even smaller.

If you want some imaginary situation rather than realistic depths, we could estimate the pressure for water to reach density of lead. Assuming a still linear behavior of water pressure with depth, to have an increase in density of 11 times rather than 0.05, the depth should be about 11/0.05=220 times more than the 11 km. So about a couple thousand kilometers. However at this depth the gravity is significantly reduced (Radius of Earth 6400 km). The water compresibility may change significantly. And it way even suffer a phase transition. This should be looked up.
Yes, looking at phase diagrams of water, at pressures over about 10 GPa it will be in one of the solid phases so the question of buoyancy becomes irrelevant.

So the final point I think is that water will be solid before its density in liquid form becomes equal to the density of lead.

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  • $\begingroup$ So if I understand correctly - water has to be more dense. If we put water on top of the water surrounding the cube it's going to compress and become more dense, but at higher pressures it's even harder to make water more dense, so it's probably not even possible to compress water enough to make it dense enough for lead to be suspended? Is there a formula that we could use mathematically to figure out what depth it would need to be? I'm sure it's something insane, though I'm still curious how insane :) $\endgroup$ – Wayne Werner Oct 31 '16 at 16:12
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    $\begingroup$ 1. arithmetical error : "the depth should be about 11/0.05=220 times more than the 11 km. So about a couple a kilometers." 2. You are assuming the density of lead does not increase under pressure. $\endgroup$ – sammy gerbil Oct 31 '16 at 17:25
  • $\begingroup$ Correct, so it is more like 2000 km. Thank you for pointing this out. I should have said couple thousand kilometers. As this is what I had in mind when comparing with the radius of the Earth. I will try to correct if possible. The increase in density of lead will just act opposite to increased buoyancy. This was supposed to be an estimation of best case scenario, to show that is not something reasonable to expect. So neglecting the change in lead does not change the answer that you probably won't see lead floating in liquid water. $\endgroup$ – nasu Oct 31 '16 at 17:51
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For the weight to be counterbalanced, you need to displace the amount of water equal to the weight of the suspended mass. So the density of the displaced water would need to be at the same density as the lead. And you are not going to get there on the earth.

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  • $\begingroup$ Would the diameter of the column holding the water change the required depth, e.g. a column that's 2" wide vs. say the Pacific Ocean (only much, much deeper?) $\endgroup$ – Wayne Werner Oct 31 '16 at 16:03
  • $\begingroup$ Only if the piece of lead fits very tightly into the column so no water can go around it. Then you will have it sitting on the column of water, like a piston. But then the depth is not relevant as long is more than a few mm. $\endgroup$ – nasu Oct 31 '16 at 16:08
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Objects submerged in water (or any fluid, for that matter) experience an upwards force equal to the weight of the water that they displace. This is known as the buoyant force. It is important to note that the only thing that determines the amount of force is the amount of water displaced, and not depth.

When you drop a cube a lead into the water, it will experience a buoyant force, but since lead is more dense than water, the weight of the water displaced is less than the weight of the lead.

So, in short, it will effectively be lighter under water, but will still sink to the bottom, no matter how deep it is.

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