# If 70% of Earth's surface is water, on average 1 mile deep, what is the mass of the Earth's water in kilograms?

Mar 25, 2016

${m}_{\text{water"approx5.75xx10^20 " kg}}$

#### Explanation:

Surface area of the Earth =
$510.1 \times {10}^{6} {\text{ km"^2=5.101xx10^14 " m}}^{2}$

70% of Surface area of Earth = $0.7 \times 5.101 \times {10}^{14} {\text{ m"^2=3.571xx10^14 " m}}^{2}$

Depth of waters =
$1 \text{ mile" = 1609.3 " m}$

Density of water = $1 {\text{ g"/"cm"^3=1000 "kg"/"m}}^{3}$
${\rho}_{\text{water"=(m)/(V) =>m_"water"=rho_"water}} \cdot V$

Volume = surface area $\times$ depth
${V}_{\text{water"=(3.571xx10^14 " m"^2)xx(1609.3 " m")=5.75xx10^17 " m}}^{3}$

Mass of the Earth's water =
${m}_{\text{water"=1000 "kg"/"m"^3xx5.75xx10^17 " m"^3=5.75xx10^20 " kg}}$