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Re: My Answer to a Common Question - Fish!
Posted By: Don, on host 209.91.94.242
Date: Friday, December 1, 2000, at 10:41:33
In Reply To: Re: My Answer to a Common Question - Fish! posted by Wolfspirit on Friday, December 1, 2000, at 06:17:12:

> Okay. Let me state, speaking out of experience as a cook who's cleaned and prepared whole fresh fish of various sizes, that /dead/ gutted fish -- without their insides -- will fall right to the bottom of a pail filled with water. In other words, fish fillets and fish flesh in itself is denser than water.

This is indisputably true. No argument.

> Live fish float at variable depths, as we've already noticed, only because of their swim bladder control. Dead fish float to the top because after death their internal organs start to break down and fill the abdominal cavity with gases, which increases the buoyancy of the dead fish.

OK, that makes sense (why they float to the surface after dying, that is.) I will discuss the nature of buoyancy further below.

> Increased buoyancy is not the same as decreased real density, but (and here's the part that confuses me) the *average* bulk density of the fish in the bucket is, in effect, decreased whenever it diffuses gas into its swim bladder to swim upwards.

Yes, very true. The change in average bulk density is presumably accomplished in a fashion similar to how a submarine's ballast works. Basically, a mass of gas is held in a chamber of variable volume. When the chamber decreases in volume, the extra space it had occupied is filled with water, so basically, the total volume of the unit (fish or submarine as a whole) is reduced. The decrease in volume compresses the gas, so the total mass of the gas (and hence the entire system) does not change, but the volume does decrease. This DOES result in an increase in average bulk density, because the same mass is now contained in a smaller volume. When the object is in water, it will rise or sink to a level where the density of the surrounding water is equal to that of the object, because at that point, the forces on the object cancel (are equal and opposite in all directions) one another.

> However, if we consider the fish and the water in the bucket to be 1) a miniature closed system, and 2) we know that fish flesh, pound for pound, is heavier than the equivalent volume of water -- then even a small live fish swimming in a given volume of water should have a small, but definite, increase in the weight of the total system. That is, when compared to a fish-less bucket containing the same total volume of contents.

This is where I disagree. The fish _flesh_ is heavier than water, yes, but the gases in the fish (which come with the fish if it is alive and whole) are lighter than the water, and the combination results in a fish whose mass is exactly equal to the mass of a volume of water equal to the volume of the fish. There is one necessary premise for this: We must assume that the fish can remain motionless at any given depth of water without exerting any forces on the system around it. What I mean by this is that birds fly, but are not lighter than air; they use lift provided by airflow around their wings and pressure generated by flapping to provide an additional force to keep them in the air. However, I have seen sleeping fish which were utterly motionless maintain a position in the water, so I assume by induction that this is true.

Taking this premise as fact, we can use some deductive reasoning to confirm that a motionless fish has equal (bulk average) density to the water around it. We know that if an object has BA density less than that of the medium it is in, it will rise until it reaches a point where the density is equal; similarly, we know the reverse to be true of an object with BA density greater than that of it's medium. Hence, we can assume that the motionless fish must have BA density EQUAL to that of the water at the depth of the fish, or the fish would rise or sink until that were the case.

Hence, in the closed system described above (and assuming that the fish is alive and healthy and a closed system in itself), we can see that the bulk average density of the fish (including all gases contained within the fish) must be equal to the bulk average density of the water at the fish's current depth, and the total system (bucket and water) must have the same mass (and hence weight) as the system where a volume of water is replaced by an equal volume of fish.

> You'd probably need an analytical weighing balance to see the difference, though.
>
> Wolf "thinks the confusion was due to not understanding the effects of buoyancy" spirit

One final comment to clear up another point which may confuse some readers (it confused me for a while until I thought it through): The depth of the fish in the water makes no difference. As long as we are dealing with a closed system, the mass of the system cannot change. The fish's depth is a function of its density, and its changes in density are accomplished through changes in volume, _not_ mass, so the total mass of the system remains unchanged regardless of the depth of the fish.

Well, I should wrap this up. I originally intended to do a simple counterpoint, and ended up with a lengthy (for me) scientific discussion of the principles involved. I hope somebody gets SOMETHING out of that. BTW, I welcome any further discussion on the subject-- I wrote this out mostly to see if I could reason out an argument based on scientific principles, but some of my science is a little rusty, and I would appreciate any further discussion or corrections.

Don

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