Question

An 8-cm-internal-diameter, 30-cm-high pitcher halffilled with water is left in a dry room at \(15^{\circ} \mathrm{C}\) and \(87 \mathrm{kPa}\) with its top open. If the water is maintained at \(15^{\circ} \mathrm{C}\) at all times also, determine how long it will take for the water to evaporate completely.

Short answer

Answer: The time required for complete evaporation of the water in the pitcher depends inversely on the value of K. If K is high, which means the room conditions favor faster evaporation, the time will be less. If K is low, the time required for complete evaporation will be higher.

Step-by-step solution

Calculate the volume of water in the pitcher

We are given that the pitcher is half-filled with water, so let's calculate the volume of the water. The volume of the cylinder can be calculated using the formula: \(V_{cylinder} = \pi \times r^2 \times h\) Here, the diameter of the pitcher is given as 8 cm, so the radius is: \(r = \frac{d}{2} = \frac{8}{2} = 4 cm\) The height of the pitcher is given as 30 cm. So, the total volume of the pitcher would be: \(V_{cylinder} = \pi \times (4)^2 \times 30 = 480\pi\: cm^3\) As the pitcher is half-filled with water, we need to find half of the total volume as the volume of water in the pitcher: \(V_{water} = \frac{1}{2} \times 480\pi = 240\pi\: cm^3\)

Calculate the mass of water in the pitcher

Now let's convert the volume to mass. We have the volume of water, and we can use the density of water, which is 1 g/cm³, to determine the mass: \(mass\: of\: water = density \times volume\) Since the density of water is 1 g/cm³, the mass of water in the pitcher will be: \(mass\: of\: water = 1 \times 240\pi = 240\pi\: grams\)

Calculate the evaporation rate

Now, we'll use the saturation pressure of water at the given temperature (15°C) to estimate the evaporation rate. The saturation pressure of water at 15°C can be found in steam tables, and it is approximately \(1.7 \mathrm{kPa}\). The evaporation rate can be calculated as: \(Evaporation\: rate = K \times (P_{sat} - P)\) Here, K is a constant depending on room conditions, \(P_{sat}\) is the saturation pressure, and P is the actual pressure in the room. The value of K is not provided, but we can go ahead with the formula by keeping K as it is. So, we have: \(Evaporation\: rate = K(1.7 - 0.87)\)

Calculate the time required for complete evaporation

Now that we have the mass of water and the evaporation rate, we can calculate the time required for complete evaporation. The equation will be: \(Time = \frac{mass\: of\: water} {Evaporation\: rate}\) Substituting our values for mass and evaporation rate, we get: \(Time = \frac{240\pi\: grams}{K(1.7 - 0.87)}\) Without knowing the value of K, we cannot get the exact time required for complete evaporation. However, we can conclude that the time will depend inversely on K. If K is high, which means the room conditions favor faster evaporation, the time will be less. If K is low, the time required for complete evaporation will be higher.

Key concepts

Saturated Pressure

Understanding saturated pressure is vital for calculating evaporation rates in exercises like the water evaporation from a pitcher. Saturated pressure, also known as vapor pressure, refers to the pressure exerted by a vapor in thermodynamic equilibrium with its condensed phases at a given temperature in a closed system. In simple terms, it's the pressure at which a liquid and its vapor can coexist at a particular temperature.

When a liquid is exposed to an atmosphere with a pressure lower than its saturated pressure, evaporation occurs more readily. In the scenario of the pitcher at a room temperature of \(15^{\textdegree} C\), we observe that the actual pressure in the room is \(87 \text{kPa}\) whereas the saturated pressure for water at the same temperature is \(1.7 \text{kPa}\). This disparity is crucial as the evaporation rate depends on the difference between the saturated pressure and the actual atmospheric pressure.

Saturation Temperature

Saturation temperature plays a parallel role to saturated pressure in the water evaporation process. It is the temperature at which a substance changes phase from liquid to vapor or vice versa, at a given pressure. It is commonly referenced in steam tables that list the properties of water and steam. In our given problem, the water in the pitcher is maintained at \(15^{\textdegree} C\), which is crucial because this temperature corresponds to a specific saturated pressure that, in turn, affects the evaporation rate.

It's important to maintain the water at this constant temperature to ensure a consistent evaluation of the evaporation rate. Any change in temperature would result in a change in the saturated pressure and therefore alter the evaporation dynamics.

Mass Transfer

Mass transfer is a core concept when dealing with evaporation, and is particularly relevant to the textbook exercise. It refers to the movement of masses from one location to another. In the context of evaporation, mass transfer involves the movement of water molecules from the liquid phase into the gaseous phase. Our exercise requires the knowledge of mass in order to calculate the time needed for the water to evaporate completely from the pitcher.

The beginning step of identifying the volume and corresponding mass of the water in the pitcher is a foundational aspect of mass transfer that sets the stage for understanding the duration of the evaporation process. Once the mass of water is determined, it can be used alongside the evaporation rate to estimate the time it will take for all of it to transfer to the atmosphere.

Water Evaporation Process

The water evaporation process is the transformation of water from liquid to gas. This process occurs when the water molecules acquire enough energy to break free from the liquid's surface and become vapor. Several factors influence this process, including temperature, air pressure, surface area, and air humidity. In the case provided, we have a controlled temperature and a given room pressure, which affects the pace of evaporation.

The concept of evaporation rate is critical here as it represents how fast the water is expected to evaporate, given these conditions. It is usually determined by the difference between the saturated pressure at the saturation temperature and the actual room pressure.
The exercise demonstrates that a lacking constant value (K) inhibits the determination of an exact timeframe for the water to fully evaporate. Moreover, the principle of mass transfer is implicitly used when relating the mass of water to the evaporation rate in order to find out the required time for complete evaporation.