Conservation of salt during freezing ($S_1=S_0/(1-f_i)$)

If a fraction $f_i$ of initial liquid mass freezes into salt-free ice, all original salt remains in a liquid mass fraction $1-f_i$. Writing salinity as salt concentration per mass gives $S_1(1-f_i)=S_0$, hence $\Delta S=S_0{f}_i/(1-f_i)$. For equal initial cooling and subsequent freezing heat extraction, $f_i=c_{p,w}\Delta T/L_f$. The liquid specific heat capacity, rather than the ice value, determines the initial heat removal. If ice retains salt, the mass balance instead reads $S_1(1-f_i)+S_i{f}_i=S_0$.

Let liquid mass $M$ freeze incrementally, and let each new ice increment retain salinity $rS$ with constant $0\le r\le 1$, where $S$ is the instantaneous well-mixed liquid salinity. Conservation gives $d(MS)=rSdM$, hence $dS/S=(r-1)dM/M$. Integrating yields $S/S_0=(M/M_0{)}^{r-1}$. For frozen mass fraction $f$, $\Delta S=S_0[(1-f{)}^{-(1-r)}-1]\simeq (1-r)S_{0}f$. Specifying a fixed bulk ice salinity instead produces a different, though first-order compatible, finite-mass balance.

For equal heat extraction, the density increase caused by brine rejection can exceed that caused by cooling near the freezing point because the thermal density response is weak. Using conservation of salt during freezing, the ratio is $R=(\partial \rho /\partial S)S_0{f}_i/[(1-f_i)|\partial \rho /\partial T|\Delta T]$, with $f_i=c_{p,w}\Delta T/L_f$. This is a parcel comparison, not a guarantee that freezing overcomes any ambient density stratification.