Seawater has a density of $1,025 \mathrm{~kg} / \mathrm{m}^{3}$ at the surface. The bulk modulus of seawater is $2.100 \times 10^{9} \mathrm{~N} / \mathrm{m}^{2}$. What is the density of seawater at a depth of $10.00 \mathrm{~km}$ ? (Use $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ and ignore any change in density with depth when calculating the pressure.)
A) $1230 \mathrm{~kg} / \mathrm{m}^{3}$
B) $1764 \mathrm{~kg} / \mathrm{m}^{3}$
C) $1076 \mathrm{~kg} / \mathrm{m}^{3}$
D) $1574 \mathrm{~kg} / \mathrm{m}^{3}$
E) $1973 \mathrm{~kg} / \mathrm{m}^{3}$

Question

Quizplus · Accepted Answer

The pressure increase at a depth of 10.00 km can be calculated using $P = \rho g h$, where $\rho = 1025 \mathrm{~kg/m^3}$, $g = 9.8 \mathrm{~m/s^2}$, and $h = 10000 \mathrm{~m}$. This gives $P = 1025 \times 9.8 \times 10000 = 1.0045 \times 10^8 \mathrm{~N/m^2}$. The change in volume ($\Delta V/V_0$) due to this pressure can be found using the bulk modulus equation, $B = -P (\Delta V/V_0)$, rearranging gives $\Delta V/V_0 = -P/B$. Substituting the given values, $\Delta V/V_0 = -1.0045 \times 10^8 / 2.100 \times 10^9 = -0.047833$. The negative sign indicates a decrease in volume. The density change can be found from the relation $\rho = \rho_0 / (1 + \Delta V/V_0)$, where $\rho_0 = 1025 \mathrm{~kg/m^3}$. Substituting the values, $\rho = 1025 / (1 - 0.047833) = 1076 \mathrm{~kg/m^3}$.

Seawater Has a Density Of 1,025 kg/m31,025 \mathrm{~kg} / \mathrm{m}^{3}1,025 kg/m3 At the Surface

Seawater Has a Density Of $1,025 \mathrm{~kg} / \mathrm{m}^{3}$ At the Surface