Seawater has a density of 1,025 kg/m³ at the surface. The bulk modulus of seawater is 2.100 × 10⁹ N/m². What is the density of seawater at a depth of 10.00 km? (Use g = 9.8 m/s² and ignore any change in density with depth when calculating the pressure.) A) 1076 kg/m³ B) 1230 kg/m³ C) 1574 kg/m³ D) 1764 kg/m³ E) 1973 kg/m³

Question

Quizplus · Accepted Answer

The pressure increase at a depth of 10.00 km can be calculated using the formula $P = \rho g h$, where $\rho$ is the density of seawater (1025 kg/m$^3$), $g$ is the acceleration due to gravity (9.8 m/s$^2$), and $h$ is the depth (10,000 m). This gives $P = 1025 \times 9.8 \times 10,000 = 1.0045 \times 10^8$ N/m$^2$. The change in volume ($\Delta V/V$) due to this pressure can be calculated using the bulk modulus formula, $B = -P \times (\Delta V/V)$, rearranging gives $\Delta V/V = -P/B$. Substituting the given values, $\Delta V/V = -1.0045 \times 10^8 / 2.100 \times 10^9 = -0.047833$. The negative sign indicates a decrease in volume, which means an increase in density. The new density can be calculated as $\rho_{new} = \rho / (1 + \Delta V/V)$, which gives $\rho_{new} = 1025 / (1 - 0.047833) = 1076$ kg/m$^3$.

Seawater Has a Density of 1,025 Kg/m3 at the Surface

Seawater Has a Density of 1,025 Kg/m³ at the Surface