Question

The average moisture binding energy of a plant protein-based snack at 8% moisture content (dry basis) is 3200 cal.mol$^{-1}$. If the water activity of the snack at the above moisture content is 0.30 at 30°C, the water activity of the sample at 45°C is ............. (rounded off to 2 decimal places). Use $R = 1.987$ cal.mol$^{-1}$.K$^{-1}$.

Hint:

Use the Clausius–Clapeyron relation for estimating water activity changes with temperature.

Answer 1

Correct Answer: 0.32

Step 1: Use the Arrhenius-type relation for water activity.
\[ \ln \left( \frac{a_{w2}}{a_{w1}} \right) = - \frac{\Delta H}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \] where, $\Delta H =$ moisture binding energy = 3200 cal/mol, $T_1 = 30^\circ C = 303 \, K$, $T_2 = 45^\circ C = 318 \, K$, $a_{w1} = 0.30$.
Step 2: Calculate constants.
\[ \frac{\Delta H}{R} = \frac{3200}{1.987} = 1611.63 \] \[ \left( \frac{1}{T_2} - \frac{1}{T_1} \right) = \left( \frac{1}{318} - \frac{1}{303} \right) = 0.003145 - 0.003300 = -0.000155 \] \[ -\frac{\Delta H}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) = -1611.63 \times (-0.000155) = 0.249 \]
Step 3: Calculate new $a_{w2$.}
\[ \ln \left( \frac{a_{w2}}{0.30} \right) = 0.249 \] \[ \frac{a_{w2}}{0.30} = e^{0.249} = 1.283 \] \[ a_{w2} = 0.30 \times 1.283 = 0.385 \] Final Answer: \[ \boxed{0.39} \]