Question

Estimate the changes in the density of water in the ocean at a depth $400m$ below the surface. The density of the water at the surface $= 1030kg{m^{ - 3}}$ and the bulk modulus of water $= 2 \times {10^9}N{m^{ - 2}}$.

Answer 1

The density of water is $1000kg/{m^3}$. The density of ocean water is more than the normal water due to the presence of salts in it. The density of ocean water is affected by two factors, namely, the temperature and salinity of the water.

Complete step by step solution:
Let us first write the information given in the question.
Depth of ocean $h = 400m$, the density of water at the surface $\rho = 1030kg/{m^3}$, and bulk modulus $B = 2 \times {10^9}N/{m^2}$.
We have to estimate the change in the density of water at the depth $\rho '$.
We have the following formula for the bulk modulus.
$B = \dfrac{{{\text{Pressure}}}}{{{\text{strain}}}}$
We can rewrite the above expression as below.
$B = \dfrac{{\rho gh}}{{\dfrac{{\Delta V}}{V}}}$
So, the volume change is given below.
$\Delta V = \dfrac{{\rho ghV}}{B}$ ……………..(1)
Here, $\rho$ is density, $g$ is the acceleration due to gravity, $h$ is the depth, $V$ is the volume, and $B$ is the bulk modulus.
Now, we can write the following relation between density and volume.
$\rho V = \rho 'V'$……………………..(2)
Here, $\rho '$is the density at depth and $V'$is compressed volume.
$V' = V - \Delta V$ ……………………(3)
Let us combine the equations (1), (2), and (3).
$\rho ' = \dfrac{{\rho V}}{{V - \dfrac{{\rho ghV}}{B}}} = \dfrac{{\rho B}}{{B - \rho gh}}$
Let us put the values in the above equation.
$\rho ' = \dfrac{{1030 \times 2 \times {{10}^9}}}{{2 \times {{10}^9} - 1030 \times 9.8 \times 400}}$
Now, let us further simplify it.
$\rho ' = \dfrac{{2060 \times {{10}^9}}}{{2 \times {{10}^9} - 0.004 \times {{10}^9}}} = \dfrac{{2060}}{{1.996}}$
$\rho ' = 1032.06kg/{m^3} \sim 1032kg/{m^3}$
Therefore, the change in density is given below.
$\Delta \rho = 1032 - 1030 = 2kg/{m^3}$.
Hence, the change in density is $2kg/m^3$.

Note:
The density is defined as mass per unit volume.
As the temperature increases the density of ocean water decreases and when temperature decreases the density increases.
Similarly, when the salinity of ocean water increases the density also increases and vice-versa.