Why is freezing something "fast enough" be approximated as adiabatic? I am reading Scott Shell's Thermodynamics and Statistical Mechanics, and I just saw this statement:

If freezing happens fast enough, to a good approximation the process can be considered adiabatic.

The author has mentioned this statement while talking about a problem regarding the spontaneous freezing of subcooled water at $-10^{\circ}$C.
Why is this the case? I thought freezing is a constant temperature, constant pressure phenomenon. When you freeze something, a certain amount of heat $Q$ is removed from it and it escapes into an environment of some temperature $T$. How can freezing ever be adiabatic?
 A: If the liquid water starts out sub-cooled at -10 C, it is not at thermodynamic equilibrium, and there is a driving force for it to change spontaneously into a mixture of liquid water and ice at 0 C.  Its own heat of freezing provides the energy necessary to bring about this change at constant enthalpy.  There is no need to remove heat from the system by transferring it to the surroundings.
Heat transfer is a process that takes time to occur, and the amount of heat transferred increases with both the temperature difference between the surroundings and the system, and the amount of time available for the heat to be transferred.  In this case, the process takes place very rapidly, and even if there is a temperature difference between the surroundings and the system (assuming, for example, the surroundings are at 0 C), there is not enough time for a significant amount of heat to be transferred.
Response
Let M represent the amount of liquid water at -10 C and let m be the amount of ice resulting at 0 C, and (M-m) the amount of liquid water at 0 C.
Take as a reference state for calculating enthalpy, liquid water at 0 C.  Then, per unit mass relative to the reference state, the enthalpy per unit mass of the liquid water at -10 C is $$h=MC(-10-0)=-10MC$$ where C is the heat capacity of liquid water.
In the final state, the enthalpy per unit mass of the ice is $$h=-m\Delta H_f$$where $\Delta H_f$ is the latent heat of fusion of ice.  And the enthalpy per unit mass of the liquid water is $$h=(M-m)C(0-0)=0$$
So applying the first law of thermodynamics to this system with Q = 0 (adiabatic) yields:  $$-10MC=-m\Delta H_f$$
