Drying and Dehydration Questions and Answers

Answer: B

The most efficient solution is 20% NaCl solution.

Water moves out of the potato only if the solution’s water activity is below the potato’s 0.942, and the lower it is, the stronger the drive. By Raoult’s law, water activity is the mole fraction of water. Take 100 g of each solution; the 20% solutions contain 80 g of water, which is 80/18 = 4.44 mol:

\[ \begin{aligned} \text{NaCl} &: \tfrac{20}{58.5} = 0.342\ \text{mol},\ a_{w} = 0.929 \\ \text{sucrose} &: \tfrac{20}{342} = 0.058\ \text{mol},\ a_{w} = 0.987 \end{aligned} \]

The mixtures give 0.200 mol of solute (10% + 10%), \(a_{w} = 0.957\), and 0.129 mol (15% + 5%), \(a_{w} = 0.972\).

Only the 20% NaCl solution falls below 0.942, because salt, with its far smaller molar mass, supplies many more moles of solute per gram than sucrose. (Counting the dissociation of NaCl into two ions would lower its value further and strengthen the same conclusion.)

Answer: A

The partial pressure is equal to the partial pressure of water vapour in air.

Water activity is defined as \(a_{w} = p/p_{0}\), the vapour pressure of water over the food divided by that of pure water at the same temperature. The relative humidity of air is \(p_{v}/p_{0}\) at that temperature. If the food is in equilibrium with the air, \(a_{w}\) equals the equilibrium relative humidity, and since both ratios share \(p_{0}\), the vapour pressures themselves are equal.

Equilibrium also means no net moisture transfer: if the food's vapour pressure were lower it would take up moisture from the air, and if higher it would dry. So 'lesser' and 'greater' describe a food that is not in equilibrium, and 'not dependent' contradicts the definition itself.

Answer: D

The rate of drying is 0.0485 kg/h.

Dry solids = 650 - 405 = 245 g. A final moisture of 6.8% on a dry basis means water left = 0.068 x 245 g:

\[ \begin{aligned} \text{water left} &= 0.068 \times 245 = 16.66\ \text{g} \\ \text{removed} &= 405 - 16.66 = 388.34\ \text{g} \\ \text{rate} &= \frac{0.38834\ \text{kg}}{8\ \text{h}} = 0.0485\ \text{kg/h} \end{aligned} \]

A sample of 650 g cannot lose anywhere near 78 to 129 kg in an hour, so the large options cannot be right whatever the working. Because the whole drying is in the constant-rate period, the average rate equals the constant rate.

Answer: D

The relative humidity is about 81.80%.

First the saturation pressure at 30°C from the given equation:

\[ \begin{aligned} \ln p_{w}^{0} &= 18.6556 - \frac{5217.635}{303} \\ &= 18.6556 - 17.2199 = 1.4357 \\ p_{w}^{0} &= 4.20\ \text{kPa} \end{aligned} \]

Then the actual vapour pressure from the absolute humidity, \(H = \dfrac{18}{29}\dfrac{p_{w}}{P - p_{w}}\):

\[ \begin{aligned} p_{w} &= \frac{HP}{0.6207 + H} = \frac{0.0218 \times 101.325}{0.6425} \\ &= 3.438\ \text{kPa} \\ RH &= \frac{3.438}{4.2025} \times 100 = 81.8\% \end{aligned} \]

Using 0.622 for the molar mass ratio gives 81.6%, still closest to 81.80. 62.82, 68.22 and 86.62 do not follow from these numbers.