Question

A glass of milk left on top of a counter in the kitchen at $15^{\circ} \mathrm{C}, 88 \mathrm{kPa}$, and 50 percent relative humidity is tightly sealed by a sheet of \(0.009\)-mm-thick aluminum foil whose permeance is $2.9 \times 10^{-12} \mathrm{~kg} / \mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{Pa}\(. The inner diameter of the glass is \)12 \mathrm{~cm}$. Assuming the air in the glass to be saturated at all times, determine how much the level of the milk in the glass will recede in \(12 \mathrm{~h}\). Answer: \(0.0011 \mathrm{~mm}\)

Short answer

Answer: The level of the milk will recede 0.0011 mm in 12 hours.

Step-by-step solution

Calculate the partial pressure of water vapor in the glass

First, we need to calculate the partial pressure of water vapor in the glass. Since we have 50% relative humidity, this means the actual water vapor pressure is 50% of the saturation pressure at the given temperature. We can use the Antoine equation to find the saturation pressure at \(15^{\circ} \mathrm{C}\): \(P_{sat} = 10^{8.07131 - \frac{1730.63}{233.426 + T}}\) where \(T\) is the temperature in Celsius. Plug in the given temperature: \(P_{sat} = 10^{8.07131 - \frac{1730.63}{233.426 + 15}} = 1.705 \times 10^3 \, \mathrm{Pa}\) Now, find the partial pressure of water vapor \(P_{v}\) in the glass by using the relative humidity (50%): \(P_{v} = 0.5 \times P_{sat} = 0.5 \times 1.705 \times 10^3 \, \mathrm{Pa} = 852.50 \, \mathrm{Pa}\)

Calculate the mass flow of water vapor through the aluminum foil

Now, we will use the permeance (\(\Pi\)) of the aluminum foil to determine the mass flow (\(\dot{m}\)) of water vapor escaping through the foil. The equation relating these variables is given by: \(\dot{m} = \Pi \times A \times (P_{v} - P_{ext})\) where \(A\) is the area of the foil and \(P_{ext}\) is the external pressure. We also need to convert the external pressure from kPa to Pa: \(P_{ext} = 88 \, \mathrm{kPa} \times 1000 = 88 \times 10^3 \, \mathrm{Pa}\) Calculate the area of the aluminum foil in contact with the air using the inner diameter of the glass (\(D = 12 \, \mathrm{cm}\)): \(A = \pi \times \left(\frac{D}{2}\right)^2 = \pi \times \left(\frac{12}{2}\right)^2 \mathrm{cm}^2 = 113.1 \, \mathrm{cm}^2\) Convert the area to m²: \(A = 113.1 \, \mathrm{cm}^2 \times 10^{-4} = 1.131 \times 10^{-2} \, \mathrm{m}^2\) Now, calculate the mass flow of water vapor through the aluminum foil: \(\dot{m} = 2.9 \times 10^{-12} \, \mathrm{kg/s} \cdot \mathrm{m}^2 \cdot \mathrm{Pa} \times 1.131 \times 10^{-2}\,\mathrm{m}^2 \times (852.50 \, \mathrm{Pa} - 88 \times 10^3 \, \mathrm{Pa}) = -3.511 \times 10^{-9} \mathrm{kg/s}\)

Calculate the amount of water vapor lost in 12 hours

Since we have the mass flow rate (\(\dot{m}\)) of water vapor escaping the container, we can now calculate the total mass (\(m\)) of water vapor lost in 12 hours. To do this, multiply the mass flow rate by the elapsed time (in seconds): \(m = -3.511 \times 10^{-9} \, \mathrm{kg/s} \times 12 \, \mathrm{h} \times 3600 \, \mathrm{s/h} = -1.511 \times 10^{-4} \, \mathrm{kg}\)

Calculate the receding level of the milk

Finally, we can determine the change in the level of the milk in the glass by using the mass of water vapor lost and the density of water (\(\rho_{water} = 1000 \, \mathrm{kg/m}^3\)). First, find the volume of water vapor lost (\(V_{lost}\)): \(V_{lost} = \frac{m}{\rho_{water}} = \frac{-1.511 \times 10^{-4} \, \mathrm{kg}}{1000 \, \mathrm{kg/m}^3} = -1.511 \times 10^{-7} \, \mathrm{m}^3\) Now, divide the volume by the cross-sectional area of the glass to find the change in the level of the milk (\(\Delta h\)): \(\Delta h = \frac{V_{lost}}{A} = \frac{-1.511 \times 10^{-7} \, \mathrm{m}^3}{1.131 \times 10^{-2} \, \mathrm{m}^2} = -0.0011 \, \mathrm{mm}\) The level of the milk will recede \(0.0011 \, \mathrm{mm}\) in 12 hours. This confirms the given answer.