Question

An airtight dispenser for drinking water is 25 cm × 10 cm in horizontal dimensions and 20 cm tall. It has a tap of negligible volume that opens at the level of the bottom of the dispenser. Initially, it contains water to a level 3.0 cm from the top and air at the ambient pressure, 1.00 atm, from there to the top. When the tap is opened, water will flow out until the gauge pressure at the bottom of the dispenser, and thus at the opening of the tap, is 0. What volume of water flows out? Assume the temperature is constant, the dispenser is perfectly rigid, and the water has a constant density of 1000 kg/m^3.

a) Calculate the volume of water that flows out.
b) The temperature variation affects the volume calculation.
c) The atmospheric pressure is irrelevant to the problem.
d) The dispenser's rigidity has no impact on the solution.

Answer 1

The volume of water that flows out when the tap is opened can be calculated using the density of water, the cross-sectional area of the dispenser, and the pressure difference between the bottom of the dispenser and the atmospheric pressure.

Step-by-step explanation:

To calculate the volume of water that flows out when the tap is opened, we need to find the difference in pressure at the bottom of the dispenser and the atmospheric pressure. We can use the equation ΔP = ρgh, where ΔP is the pressure difference, ρ is the density of water, g is the acceleration due to gravity, and h is the height difference.

First, let's find the pressure at the bottom of the dispenser. The gauge pressure is given by P = P₀ - Pₐ, where P₀ is the pressure at the bottom of the dispenser and Pₐ is the atmospheric pressure. We want the gauge pressure to be 0, so P₀ = Pₐ. Therefore, we can substitute P₀ with Pₐ in the equation and rearrange it to solve for h:

h = ΔP / (ρg)

Now we can calculate the volume of water that flows out. The volume of water is given by V = A * h, where A is the cross-sectional area of the dispenser at the bottom.

Substituting the given values into the equations, we have:

ΔP = 0 atm - 1 atm = -1 atm

ρ = 1000 kg/m³

g = 9.8 m/s²

A = (25 cm * 10 cm) / (100 cm/m)² = 0.025 m * 0.1 m = 0.0025 m²

h = (-1 atm) / (1000 kg/m³ * 9.8 m/s²) ≈ -0.01 m

V = 0.0025 m² * -0.01 m ≈ -0.000025 m³

The volume of water that flows out is approximately -0.000025 m³.