Question

Assuming that 70 percent of the Earth’s surface is covered with water at average depth of 0.95 mi, estimate the mass of the water on Earth. One mile is approximately 1.609 km and the radius of the earth is 6.37 × 10^6 m. Answer in units of kg. Your answer must be within ± 20.0%

Answer 1

Answer:

The mass of the water on earth is $5.4537\times10^{20}\ kg$

Explanation:

Given that,

Average depth h= 0.95 mi

$h=0.95\times1.609$

$h =1.528\ km$

$h=1.528\times10^{3}\ m$

Radius of earth $r= 6.37\times10^{6}\ m$

Density = 1000 kg/m³

We need to calculate the area of surface

Using formula of area

$A =4\pi r^2$

Put the value into the formula

$A=4\pi\times(6.37\times10^{6})^2$

$A=5.099\times10^{14}\ m^2$

We need to calculate the volume of earth

$V = Area\times height$

Put the value into the formula

$V=5.099\times10^{14}\times1.528\times10^{3}$

$V=7.791\times10^{17}\ m^3$

Now, 70 % volume of the total volume

$V= 7.791\times10^{17}\times\dfrac{70}{100}$

$V=5.4537\times10^{17}\ m^3$

We need to calculate the mass of the water on earth

Using formula of density

$\rho = \dfrac{m}{V}$

$m = \rho\times V$

Put the value into the formula

$m=1000\times5.4537\times10^{17}$

$m =5.4537\times10^{20}\ kg$

Hence, The mass of the water on earth is $5.4537\times10^{20}\ kg$