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Consider a glass of water and a glass of coffee. Their contents differ by no more than a few grams of particles coming from the roasted and ground coffee, yet the former lets almost all visible light pass through, and the latter blocks most of it.

I am wondering what are the most efficient ways to turn water opaque by pouring matter in it, under normal temperature and pressure. By efficient I mean minimising the mass of added matter. By opaque, let's say 99% of daylight is blocked by two inches of liquid.

There is no other reason for my asking than sheer curiosity from the many hours I spend gazing fixedly through the coffee pot.

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put on hold as off-topic by Brandon Enright, Kyle, ACuriousMind, Qmechanic 10 hours ago

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1 Answer 1

I think you need to consider a suspension vs. a solution.

A suspension is a combination in solid and liquid matter wherein the solid is supported in the liquid and does not settle because Brownian motion provides random impulses in random directions that overwhelm gravity. Coffee is a suspension, but not a stable one. Quantum dots are small enough to be a semi-permanent suspension. By its nature, there are "gaps" between the solid material that are filled with water. Suspensions are "granular" at a physical size that is potentially much larger than a wavelength. So I'd say that a suspension is not an efficient way to create an opaque "form" of water.

A solution would appear to be much more efficient. A solution enables a mixture of water and the solute at a molecular level. Whereas a suspension has solid particles at the "center" of the suspended particle that are "wasted" because they don't absorb anything, a solution does not waste any molecules; every molecule of solute has a chance to absorb a photon.

Technically, I'd say that the most efficient way to produce an opaque material with water is to add a highly absorbing molecule (e.g. dye) at sufficient concentration.

In this context, I don't think there is much value in trying to imagine the "local" nature of a photon. Just think of it as how much light is absorbed per centimeter of distance. You will find that dyes will have an extinction coefficient (per cm) that is related to the molar concentration. (See http://www.exciton.com/ or http://www.exciton.com/pdfs/ABS473-091912.pdf )

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