Does water temperature affect hydrostatic pressure of fixed mass of water in open vessel?

Question

Lets say I have a pressure transducer sitting at the bottom of an open-topped 5-gallon bucket (assume its a vertical-walled cylinder). The transducer is measuring the gage pressure of the water column.

If the temperature of the water fluctuates (assume uniformly throughout volume, no evaporation) will I see a corresponding fluctuation in the output of the pressure transducer?

I understand that pressure is $p=\rho g h$, so pressure is proportional to density, which changes with temperature. But volume is also proportional do density for a fixed mass, so I think that with a vertical walled cylinder, where $h$ is proportional to volume, the measured pressure will not change because the increase in $h$ will compensate for the proportional decrease in $\rho$. Is this correct? If I had, say a conical vessel, would the answer be different?

Answer 1

The answer depends on the shape of the vessel.

For a vessel with constant section, the weight of the column of water above your transducer is unaffected. But when you have a conical vessel, some of the mass is no longer "directly above" your transducer as the liquid expands. If the cone gets wider, you will feel less weight; if it tapers narrower you will feel more weight.

Answer 2

Lets think of how you can derive $P = \rho g h$.

You can determine the hydrostatic pressure by considering the weight of the water column above the point you wish to measure pressure.

The weight would be the force felt on the bottom of the plate, and could be determined by $$W = mg =\rho V g$$ where $W$ is the weight $\rho$ is density, $V$ is the volume of water and $g$ is the acceleration due to gravity.

We also know that pressure is force over area, or $$P = \frac F A$$ where in this case $A$ is area perpendicular to the direction of the force (in this case vertical force).

Since the force in this case is the weight, we can do some substitutions to get to the equation $$P = \frac {\rho V g}{A}$$ and we can notice that if the area is the cross section, $$V = A \ h$$

You can substitute that relationship in to determine that $$P = \rho g h$$

Essentially, since mass doesn't change, weight doesn't change, and therefore the force/pressure should not change, due to the nature of how we derive hydrostatic pressure in the first place.