Subtopic: Preservation of food by cooling and freezing
Question Statement:
A food product has an initial moisture content of $80\%\text{ (w/w)}$ ($m_0 = 0.80\text{ kg water/kg food}$) and an initial freezing point of $T_f = -1.50^\circ\text{C}$. When frozen to a storage temperature of $T = -15.0^\circ\text{C}$, the mass fraction of unfrozen water remaining in the food (calculated using the Raoult-derived inverse temperature freezing model: $m_{\text{uf}} = m_0 \frac{T_f}{T}$) in $\text{kg water/kg food}$ is ________ (round off to three decimal places).
Answer Key:
0.080
Step-by-Step Solution & Explanation
By Raoult's law of freezing point depression, the equilibrium concentration of solute is inversely proportional to temperature below freezing:
$m_{\text{uf}} = m_0 \times \frac{T_f}{T} = 0.80 \times \frac{-1.50}{-15.0} = 0.80 \times 0.10 = 0.080\text{ kg water/kg food}$
Thus, $90\%$ of the water is frozen into ice and $10\%$ remains in the unfrozen concentrated phase.