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1. Homework Statement
An open ended can, with height 0.3m is filled with air (P = Patm ~ 100kPa) and is submerged (open end downward) 3 metres below the water surface (bottom of can 3m below while top of can is 2.7m below). The pressure above the water's surface is also Patm. The air is initially at 20 C, and after it is submerged it remains at 20 C. Assume the air acts as an ideal gas. Estimate the height of the water that will fill the can. p(water) = 1000kg/m3 Where p = density; P = pressure
2. Homework Equations
P = pgh or P = pgΔh
3. The Attempt at a Solution
How I figure it, is that you calculate the pressure at the very bottom of the can first. P(bottom) = (1000 kg/m3)(9.81 m/s2)(3 m) = 29430 kPa. The air in the can will initially be the same as it was originally  Patm  but eventually the water pressure underneath the can will compress the air and the water will begin to rise into the can. What i'm not sure about is how the height of the can comes into play as there is no radius given to calculate volume. Would the problem involve the pressure difference between the downward pressure on the top of the can, and the upward pressure on the opening in the bottom of the can?
This is my first post and any feedback on anything I've done wrong in my post (and problem solving logic) would be appreciated, as well, the problem statement and the diagram are attached in case my explanation wasn't clear. Thanks everyone
1. Homework Statement
2. Homework Equations
3. The Attempt at a Solution
An open ended can, with height 0.3m is filled with air (P = Patm ~ 100kPa) and is submerged (open end downward) 3 metres below the water surface (bottom of can 3m below while top of can is 2.7m below). The pressure above the water's surface is also Patm. The air is initially at 20 C, and after it is submerged it remains at 20 C. Assume the air acts as an ideal gas. Estimate the height of the water that will fill the can. p(water) = 1000kg/m3 Where p = density; P = pressure
2. Homework Equations
P = pgh or P = pgΔh
3. The Attempt at a Solution
How I figure it, is that you calculate the pressure at the very bottom of the can first. P(bottom) = (1000 kg/m3)(9.81 m/s2)(3 m) = 29430 kPa. The air in the can will initially be the same as it was originally  Patm  but eventually the water pressure underneath the can will compress the air and the water will begin to rise into the can. What i'm not sure about is how the height of the can comes into play as there is no radius given to calculate volume. Would the problem involve the pressure difference between the downward pressure on the top of the can, and the upward pressure on the opening in the bottom of the can?
This is my first post and any feedback on anything I've done wrong in my post (and problem solving logic) would be appreciated, as well, the problem statement and the diagram are attached in case my explanation wasn't clear. Thanks everyone
1. Homework Statement
2. Homework Equations
3. The Attempt at a Solution
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