The density of water is 1 g/cc. Ice floats on water with 90% of its volume

inside the water. What will be the mass of a cube of ice of side 3 cm?

This is pretty easy to calculate when you realize that the mass of the water displaced is equal to the mass of the object floating in it.

Take an iceberg of volume V. It has density d_i. So its total mass is V*d_i = m.

Now, it is displacing 90% of its volume. The sea water has a d_w. The total mass of water displaced is 0.9 * V * d_w, and it equals the mass of the iceberg, V*d_i.

Divide both sides by V. You get 0.9 * d_w = d_i. Now we just have to solve for d_w. Dividing both sides by 0.9, you get d_w = d_i/0.9.

Answer: 24.3 g

Explanation:

According to Archimedes principle,

buoyant force = weight of liquid displaced

Density of the object floating in a liquid = density of the liquid * fraction of the object inside the liquid

Mathematically represented thus:

ρ (object) = ρ (water) * fraction f

ρ (water) = 1 g/cc, fraction of the object inside the liquid = 10% = 0.1

ρ (object) = 1 * (1 - 0.1) = 0.9 g/cc

ρ (object) = 0.9 g/cc

To get the mass of a cube of ice, we use the formula M = ρV

ρ = 0.9 g/cc, L = 3cm, V = L^3 = 3^3 = 27 cc

M = 0.9 * 27

M = 24.3 g