Question

Ships and sailing vessels often carry lead ballast in various forms, such as bricks, to keep the ship properly oriented and upright in the water. Suppose a ship takes on cargo and the crew, a total of 0.500 m^3 of lead blocks into water 2.00 km deep. Calculate (a) the change in the pressure at that depth and (b) the change in volume of the lead upon reaching the bottom. Take the density of sea water to be 1.025 x 10^3 kg/m^3, and take the bulk modulus of lead to be 4.2 x 10^10 Pa.

Answer 1

The problem involves calculating the change in pressure at a depth of 2.00 km in the ocean and the change in volume of a volume of 0.500 m^3 of lead due to this pressure increase. The density of seawater and the bulk modulus of lead are used to perform these calculations.

Step-by-step explanation:

The student has asked to calculate (a) the change in the pressure at a depth of 2.00 km in the ocean and (b) the change in volume of the lead upon reaching the bottom, given that 0.500 m3 of lead ballast is submerged. We will use the information that the density of sea water is 1.025 x 103 kg/m3, and the bulk modulus of lead is 4.2 x 1010 Pa.

Change in Pressure

To calculate the change in pressure at a depth of 2.00 km, we can use the equation for pressure due to a fluid column, P = ρgh, where ρ is the density of the fluid, g is the acceleration due to gravity, and h is the depth. Plugging in the density of seawater and the depth, we can calculate the pressure increase due to the added 2.00 km depth.

Change in Volume of Lead

The change in volume of lead due to the increased pressure at the bottom of the ocean can be calculated using the definition of bulk modulus, B = -ΔP / (ΔV/V), where B is the bulk modulus, ΔP is the change in pressure, ΔV is the change in volume, and V is the original volume. From here, we can solve for ΔV to find the change in volume of the lead.