The density of ice is 9.20x10² kg/m³. An ice cube which is 20.0mm high floats in water. What height of the side of the ice cube is above the surface of the water?
what is .92 times 20mm? That is the height below the water surface. Why does that work? Think the submerged height creates a bouyant force due to the displaced volume...Why is 90 percent of icebergs beneith the surface?
Hb = Di/Dw * H = Height below surface.
Di = 920 kg/m^3.
Dw = 1000 kg/m^3.
H = 20 mm.
Hb = 920/1000 * 20 = 18.4 mm.
20 - 18.4 = 1.6 mm above surface.
P.S. So, as bob said, 92%(18.4mm) is below surface.
To determine the height of the side of the ice cube above the surface of the water, we need to compare the densities of ice and water.
First, let's convert the height of the ice cube from millimeters to meters:
20.0 mm = 20.0 × 10^(-3) m = 2.0 × 10^(-2) m
Since the ice cube is floating in water, it is in equilibrium, which implies that the weight of the ice cube is equal to the buoyant force acting on it.
The weight of the ice cube can be calculated using its volume and density:
Weight of the ice cube = density of ice × volume of the ice cube
= density of ice × area of the base × height of the ice cube
The volume of the ice cube is given by:
Volume of the ice cube = area of the base × height of the ice cube
= side length of the base × side length of the base × height of the ice cube
= (side length of the base)^2 × height of the ice cube
Now, let's assume the side length of the ice cube's base is y meters. Then the volume of the ice cube can be written as:
Volume of the ice cube = y^2 × (2.0 × 10^(-2) m)
Since the ice cube is in equilibrium, its weight is equal to the buoyant force, which is given by:
Buoyant force = weight of the water displaced by the ice cube
= density of water × volume of the ice cube × acceleration due to gravity
= density of water × y^2 × (2.0 × 10^(-2) m) × acceleration due to gravity
Setting the weight of the ice cube equal to the buoyant force, we have:
density of ice × y^2 × (2.0 × 10^(-2) m) = density of water × y^2 × (2.0 × 10^(-2) m) × acceleration due to gravity
Canceling out the common terms and rearranging the equation, we get:
y = sqrt(density of water / density of ice)
Substituting the given densities:
y = sqrt(1000 kg/m³ / (9.20 × 10² kg/m³))
Calculating this, we find:
y ≈ sqrt(1.08695652174) = 1.04
Therefore, the height of the side of the ice cube above the surface of the water is approximately 1.04 meters.
To find the height of the side of the ice cube that is above the surface of the water, we need to compare the densities of ice and water. If the density of ice is lower than the density of water, the ice cube will float with a portion of it above the surface.
Given:
Density of ice (ρ_ice) = 9.20 x 10² kg/m³
Height of the ice cube (h_ice) = 20.0 mm
First, let's convert the height of the ice cube from millimeters to meters, since the density is given in kg/m³:
h_ice = 20.0 mm = 20.0 x 10⁻³ m
Now, we need to determine the density of water. The density of pure water is approximately 1000 kg/m³.
Comparing the density of ice and water, we have:
If ρ_ice < ρ_water, then the ice cube will float.
Now, let's calculate the height of the ice cube that is above the surface of the water:
The ice cube displaces a volume of water equal to its own volume when it floats. The volume of a rectangular prism (the ice cube) is given by:
Volume = length x width x height
Since we only need to find the height above the water surface, we can disregard the length and width.
Volume of the ice cube (V) = h_ice x 1 x 1 = h_ice
The weight of the displaced water is given by:
Weight = density x volume x g
where g is the acceleration due to gravity (approximately 9.8 m/s²).
Weight of the ice cube = ρ_water x V x g
Since the weight of the ice cube is equal to the weight of the displaced water, we can equate the two:
Weight of the ice cube = Weight of the displaced water
ρ_ice x V x g = ρ_water x V x g
We can cancel out the volume (V) and the acceleration due to gravity (g) from both sides of the equation:
ρ_ice = ρ_water
Now we can solve for h_ice, which is the height of the side of the ice cube above the surface of the water:
h_ice = ρ_water / ρ_ice
Substituting the values we have:
h_ice = 1000 kg/m³ / 9.20 x 10² kg/m³
Performing the calculation:
h_ice = 1.09 x 10⁻³ m
Therefore, the height of the side of the ice cube above the surface of the water is approximately 1.09 mm.