Will ice made with heavy water sink in a glass of regular H2O ?

If so, what ratio of regular to heavy water will be neutrally buoyant in water at 4 Celsius ?

I fancy an ice cube with layers of regular and heavy water in its composition - I assume it will float and sink by turns as each layer melts !

Cool idea!

According to the Wikipedia page, heavy water is 11% denser than regular water at 1.11g/mL. Meanwhile, ice at around -20C is around 8% less dense than regular 0C water at about 0.92 g/cc (presumably only the very outer layer of the ice cube would be at 0C where it touches the water; the interior would remain colder until the cube melted). So theoretically, deuterium ice should be 1.11*0.92 \approx 1.02 g/mL. That assumes, of course, that there's no air entrained in the ice cube. I'm not sure how much volume of air is in a typical ice cube, but hopefully it's less than 2%.

Ignoring the air, the neutral buoyancy point would occur when the fraction of heavy water was enough to bring the ice density to 1.0 g/mL. For a unit volume of ice, let's call the volume of heavy ice x. That means the volume of regular ice is 1-x. Knowing the densities of both, it's easy to solve the equation to find x for neutral buoyancy.

1.02x + 0.92(1-x) = 1.00

x = 0.8

Hence, for theoretical neutral buoyancy, you'd need about 80% heavy water and 20% regular water to make the ice. In reality, though, the proportion of heavy water will probably have to be somewhat higher to account for the entrained air.

If you do it, let us know how it turns out!