the sorption isotherm is thus of importance for deciding under what
humidity of the air a food material should be stored.
In many foods, the situation is more complicated. Often, the material,
or an important component, may occur either in crystalline or in amorphous
form. Figure 8.6a gives examples for sucrose: crystalline, amorphous, and in
solution. Crystalline sucrose, if pure, cannot contain water and any water
present is adsorbed on the crystal faces, and this is a very small amount.
Amorphous sucrose can take up water, because this means in fact dilution of
an extremely concentrated solution. If sufficient water has been taken up,
the diffusion coefficient of sucrose has become large enough for it to
crystallize. This then leads to the release of water. Theawand the rate at
which these changes occur greatly depend on temperature.
Figure 8.6b shows sorption isotherms for dried skim milk, of which
about 50%is lactose. The lactose can remain amorphous when the milk is
spray-dried. It can also be made with crystalline lactose, and the normal
crystalline form is a monohydrate. The water of crystallization is not
available as a solvent, which means thatawis smaller than for amorphous
sugar at the same water content, provided the latter is fairly small. At higher
water content, the curves cross, and amorphous sugar may crystallize,
taking up part of the water.
It may further be noted, that water content, and therebyaw, may
greatly vary from place to place in a food during water uptake (as well as
during drying).
Question 1
Consider a powder recently obtained by spray drying a liquid food. At room
temperature a water activity of 0.3 is observed. How doesawalter if the following
changes are applied?
(a) Increase the temperature to 80 8 C. (b) Keep at that temperature for some
days in completely dry air, then cool again to room temperature. (c) Then put the
powder in moist air at room temperature until it has the same weight (hence, the
same water content) as originally. (d) Keep this powder in a closed container at room
temperature for a month.
Answer
(a) According to the relation of Clausius–Clapeyron [Eq. (8.7)]awwill increase; see
also Figure 8.5b. (b) This treatment implies that water has been lost, henceaw<0.3.
(c) In the original powder, the water activity would correspond to that on a
desorption isotherm, in this case on an adsorption isotherm. According to
Figure 8.4a, nowawwill be>0.3. (d) The treatment described in (c) cannot have