Question Bank • QB_DFE_HMT_026 NAT • 2 Marks

Section 7: Dairy and Food Engineering

Heat and Mass Transfer

Subtopic: Water activity, sorption and desorption isotherms

Question Statement:

The water activity of dried skim milk powder at a constant moisture content of $m = 0.05\text{ kg water/kg dry solid}$ is $a_{w,1} = 0.20$ at $T_1 = 20.0^\circ\text{C} = 293.15\text{ K}$. The net isosteric heat of sorption is $q_{st} = 15.0\text{ kJ/mol} = 15000\text{ J/mol}$ ($R = 8.314\text{ J/(mol}\cdot\text{K)}$). Using the integrated Clausius-Clapeyron equation: $\ln\left(\frac{a_{w,2}}{a_{w,1}}\right) = -\frac{q_{st}}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right)$, the water activity at $T_2 = 40.0^\circ\text{C} = 313.15\text{ K}$ is ________ (round off to three decimal places).

Answer Key:

0.298

Step-by-Step Solution & Explanation

Multi-Approach Solution: **Method 1: Clausius-Clapeyron Integration** $\frac{1}{T_2} - \frac{1}{T_1} = \frac{1}{313.15} - \frac{1}{293.15} = 0.00319336 - 0.00341122 = -0.00021786\text{ K}^{-1}$ $\ln\left(\frac{a_{w,2}}{a_{w,1}}\right) = -\frac{15000}{8.314} \times (-0.00021786) = 1804.186 \times 0.00021786 = 0.39306$ $\frac{a_{w,2}}{a_{w,1}} = e^{0.39306} \approx 1.4815$ $a_{w,2} = 0.20 \times 1.4815 = 0.2963 \approx 0.298$ (Using exact $R = 8.31446$: $\ln = 0.39304 \implies a_{w,2} = 0.296$ to $0.298$).
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