After a shipwreck, a solid steel spoon lies at the bottom of the ocean, 5.75km below the surface. What's the water pressure at that depth? Find the fractional volume change in the spoon due to compression forces.
You will need the density (rho) and bulk modulus (E) of seawater.
Pressure = (rho)*g*H
where H is the depth
The density of sewater is about
1030 kg/m^3
The relative volume change is
deltaV/V = -P/E
You should be able to find E with a Google search. It may also be in your class notes or textbook.
The bulk modulus you want is that of steel, not seawater. (Duh)
To calculate the water pressure at a certain depth in the ocean, you can use the hydrostatic pressure formula:
P = ρ * g * h
Where:
P is the pressure
ρ (rho) is the density of the fluid (in this case, water)
g is the gravitational acceleration (approximately 9.8 m/s^2)
h is the depth
Let's calculate the water pressure at a depth of 5.75 km (which is equivalent to 5,750 meters):
First, we need to convert the depth to meters:
5.75 km = 5.75 * 1000 m = 5750 m
Now, we can substitute the values into the formula:
P = ρ * g * h
P = 1000 kg/m^3 * 9.8 m/s^2 * 5750 m
Calculating this equation, we find that the water pressure at a depth of 5.75 km is approximately 56.21 MPa (megapascals).
To find the fractional volume change in the spoon due to compression forces, we need to know the bulk modulus of steel.
The bulk modulus is a measure of how resistant a material is to volume change under pressure. For steel, the bulk modulus typically ranges between 160 and 200 GPa (gigapascals).
Assuming a bulk modulus of 170 GPa, we can use the equation:
ΔV/V = -P/K
Where:
ΔV/V is the fractional volume change
P is the pressure
K is the bulk modulus
Substituting the values:
ΔV/V = -56.21 MPa / 170 GPa
Calculating this equation, we find that the fractional volume change in the spoon due to compression forces is approximately -0.00033 or -0.033%. This indicates that the spoon would undergo a very small reduction in volume at such depths due to compression forces.