Consider now curve S for asolutionof one compound, e.g., a sugar, in
water. The heat capacity will generally differ somewhat from that of pure
water, leading to a different slope. Cooling will proceed to a point B^0 ,
generally at a lower temperature than B, since the initial freezing
temperatureTfis lower than that of water. Again, nucleation has to occur,
and after that ice is formed (at B^0 ) and heat is given up until the equilibrium
freezing curve is reached at point C^0. Notice that this is at a temperature
below Tf, because the solution is already somewhat concentrated. Ice
formation, and hence the concentrating of the remaining solution, and
temperature decrease go on until a point D^0. The system is then undercooled
with respect to solute crystallization, again because nucleation has to occur
first. After that, heat of crystallization is given up, the temperature rises to
E^0 , after which equilibrium crystallization of both solute and water occur,
until everything is frozen (point F^0 ). Finally, the mass is further reduced in
temperature.
Uponslow heatingof the frozen system, about the same curve results
in the opposite direction, but now the equilibrium lines will be followed.
Hence, the ‘‘dips’’ near D^0 and B^0 will not occur. Neither will nucleation
cause delay of melting, since initiation of solid–liquid transitions generally
needs negligible overheating (Sections 14.1 and 14.2.2).
FIGURE15.15 Examples of cooling curves for water (W) and an aqueous solution
(S). The rate of heat removalðJ?kg^1 ?s^1 Þis constant.Tis temperature,tis time.Te
is eutectic andTfinitial freezing temperature of the solution. Broken lines denote
hypothetical relations if no undercooling were to occur.