1. The dimensions are converted from cm to m:
10 cm ×1 m100 cm=0.1 m
5 cm ×1 m100 cm=0.05 m
The volume of the water in the tank is:
V=0.1×0.1×0.05=0.0005 m3
The mass of the water is:
m=ρV=1000kgm3×0.0005 m3=0.5 kg
2. g=9.81ms2
a. The force F exerted by the water is:
F=mg=0.5 kg9.81ms2=4.905kg∙ms2=4.905 N
b. The force F is exerted downwards, towards the centre of the Earth.
c. The water would want to move downwards because of gravity.
d. The height h is converted to meters:5 cm ×1 m100 cm=0.05 m
The potential energy U is:
U=Fh=4.905 N0.05 m=0.2453 N∙m=0.2453 J
3. The volume of the olive oil is assumed to be the same as that of the water. Thus,
V=0.0005 m3
The mass of olive oil in the tank is:
m=ρV=920kgm3×0.0005 m3=0.46 kg
4. g=9.81ms2
a. The force of the olive oil is:
F=mg=0.46 kg9.81ms2=4.5126 N
b. This force of the olive oil is directed downwards towards the bottom of the tank.
c. The oil seeks to move towards the bottom of the tank.
d. The height h is converted to meters:5 cm ×1 m100 cm=0.05 m
The potential energy U is:
U=Fh=4.5126 N0.05 m=0.2453 N∙m=0.2256 J
5. The available potential energy is:
U=0.2453+0.2256=0.4709 J
6.Tank Sketch:
7.
a. Tank Sketch Without Separator
b. Tank Sketch With Forces and Potential Energy (Without Separator)
c.
i. Assuming that all of the water is concentrated at a height of 0 cm, the height h = 0. Thus, the potential energy is:
U=Fh=4.905 N0 m=0 J
ii. Assuming that all of the water is concentrated at a height of 0 cm, the height h = 0. Thus, the potential energy is:
U=Fh=4.5126 N0 m=0 J
iii. The available total potential energy in the tank is:
Utotal=0J+0J=0 J
iv. The energy was converted to kinetic energy such that the oil would go up and the water would go down.
d. Oil has a smaller density compared to water. Therefore, its tendency is to go upwards when it is forced to be mixed to a higher density fluid such as water. The greater force and potential energy of the water above also shows that there is no equilibrium; thus, the water must go down when the separator is removed in order to achieve equilibrium.
8. With food colouring, the droplet of water will have a higher density and becomes naturally heavier than the oil and the water. Therefore, it will sink towards the bottom of the tank. When it reaches the water region, the droplet of water is initially unstable. The food colouring will dissipate and will be evenly distributed among the water particles until the solution becomes stable.
Upon insertion of a droplet of oil, the oil will remain on the oil-based region. It will remain stationary in this region and will stay stable.
The relative position of the water and oil-based fluids affect the stability because it dictates whether the droplets of fluid would move or dissipate. Naturally, the droplet of fluid would seek the region with the density similar to itself in order to achieve stability.
Application of the Model to Earth’s Atmosphere
1. The geography and geometry of the Earth account for a great difference of the Earth’s atmosphere from the oil-water physical model. The Earth is spherical in shape, and the surface is not perfectly level or flat; the oil-water physical model has a very flat surface. This significant difference between the two means also a significant difference when it comes to the potential energies of the systems.
2. Wildfires would introduce increased volumes of carbon dioxide, carbon monoxide, and other gases that are the by-products of the combustion of the wildfire. These gases are naturally hotter because they are the results of combustion. Therefore, these gases are lighter or have smaller densities. The tendency of these lighter gases would be to rise up to the atmosphere until the force and potential energies stabilize. Moreover, the temperature of these gases would eventually dissipate across the regions where they pass through. The atmosphere would fundamentally stabilize when the temperatures are levelled on the different regions.
