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.
To determine the percentage of an ice cube's or iceberg's mass that is above water, we need to compare the density of ice to the density of water.
1. Determine the density of water: The density of water is approximately 1 g/cm^3.
2. Calculate the fraction of the ice cube or iceberg's volume submerged in water: Since density is mass per unit volume, we can equate the densities of ice and water to find the fraction of the ice's volume submerged.
Density of ice = 0.92 g/cm^3
Density of water = 1 g/cm^3
Fraction submerged = (Density of ice / Density of water)
Fraction submerged = (0.92 g/cm^3 / 1 g/cm^3)
Fraction submerged = 0.92
3. Calculate the percentage submerged: Since the fraction submerged represents the fraction of the ice's volume that is underwater, we can convert this fraction to a percentage.
Percentage submerged = (Fraction submerged) x 100
Percentage submerged = (0.92) x 100
Percentage submerged = 92%
Therefore, approximately 92% of the ice cube's or iceberg's mass is submerged below water.