Answer to Question #265794 in Electric Circuits for Hab

Question

The deepest point in the ocean is in the Mariana Trench, about 11 km deep, in the Pacific. The pressure at this depth is huge, about 1.13 x 10^8 Pa. (a) Calculate the change in volume of 1.00 m3 of seawater carried from the surface to this deepest point. (b) The density of seawater at the surface is 1.03 x 10^3 kg/m3. Find its density at the bottom. (c) Explain whether or when it is a good approximation to think of water as incompressible.

Answer 1

Solution;

Given;

Depth =11km

Pressure=1.13×10⁸Pa

Density of seawater=1.03×10³kg/m³

Volume is 1.00m³

(a)

Change in volume;

$B=-V(\frac{\Delta P}{\Delta V})$

$2.1×10^9=-1(\frac{1.13×10^8-1.01×10^5}{\Delta V})$

$\Delta V=-0.053m^3$

(Negative means volume reduced)

(b)

Change in density is calculated as;

$\Delta \rho=\frac{m\Delta \rho}{V}=\frac{0.053×10^3}{1}$

$\Delta \rho=0.053×10^3kg/m^3$

Density at the bottom is calculated as;

$\Delta \rho=\rho_2-\rho_1$

$\rho_2=(1.03+0.053)×10^3kg/m^3$

$\rho_2=1.083kg/m^3$

(c)

Yes it is a good approximation because in the 11km the density of water changed by only (5-6)%.