At the floating ice floe above the water there is a part with a volume of 2 cubic meters.

At the floating ice floe above the water there is a part with a volume of 2 cubic meters. What is the mass of the entire ice floe?

The given tasks: V1 (volume of the above-water part of the floating ice floe) = 2 m3.

Reference values: according to the condition ρw (water density) = 1000 kg / m3; ρl (ice density) = 900 kg / m3.

1) Let’s calculate the volume of the underwater part of the floating ice floe: Ft = Fa; m * g = ρw * g * V2; m = ρw * V2; ρl * (V1 + V2) = ρw * V2; ρl * V1 + ρl * V2 = ρw * V2; V2 = ρl * V1 / (ρw – ρl) = 900 * 2 / (1000 – 900) = 18 m3.

2) The total volume of the ice floe: V = V1 + V2 = 2 + 18 = 20 m3.

3) The mass of the entire ice floe: m = ρl * V = 900 * 20 = 18,000 kg (18 t).