Question

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

Answer 1

`m_"water"approx5.75xx10^20 " kg"`

Explanation:

Surface area of the Earth =
`510.1xx10^6 " km"^2=5.101xx10^14 " m"^2`

70% of Surface area of Earth = `0.7xx5.101xx10^14 " m"^2=3.571xx10^14 " m"^2`

Depth of waters =
`1 " mile" = 1609.3 " m"`

Density of water = `1 " g"/"cm"^3=1000 "kg"/"m"^3`
`rho_"water"=(m)/(V) =>m_"water"=rho_"water"*V`

Volume = surface area `xx` depth
`V_"water"=(3.571xx10^14 " m"^2)xx(1609.3 " m")=5.75xx10^17 " m"^3`

Mass of the Earth's water =
`m_"water"=1000 "kg"/"m"^3xx5.75xx10^17 " m"^3=5.75xx10^20 " kg"`