posted by Albert on .
If the density of ice is 0.92 g/cm^3, what percent of an ice cube’s (or iceberg’s) mass is above water.
8% for an ice cube. The part below water (92% of V) will then displace a mass of water equal to the mass of the ice. This assumes that the surrounding water is fresh, which is not quite true for icebergs.
The density of sea water is typically 1.025 g/cm^3, because of the salt content. For an iceberg floating in sea water, the fraction f of the volume V above the water obeys the equation
(1-f)V*(density of sea water)= 0.92 V
(1-f)*1.025 = 0.92
1-f = 0.898
f = 0.102 or 10.2%
The teacher probably expects 8% for an answer, but it isn't all that simple.