Question

The average moisture binding energy of a textured protein product (TPP) at 8% moisture content (dry basis) is 3200 cal·mol$^{-1}$. If the water activity of the TPP at the above moisture content is 0.30 at $30^{\circ}$C, the water activity of the sample at $45^{\circ}$C is ................ (rounded off to 2 decimal places). Gas constant $R=1.987$ cal·mol$^{-1}$·K$^{-1}$.

Hint:

At fixed moisture, increasing temperature usually {increases} water activity. Use the van ’t Hoff form with the moisture-binding energy to translate $a_w$ across temperatures.

Answer 1

Step 1: Temperature dependence of $a_w$. For a fixed moisture content, water activity follows a van ’t Hoff–type relation \[ \ln a_{w,2} = \ln a_{w,1} - \frac{\Delta H}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right), \] where $\Delta H$ is the (average) moisture binding energy. 
Step 2: Substitute the numbers. $T_1=30^{\circ}\text{C}=303\ \text{K}$, $T_2=45^{\circ}\text{C}=318\ \text{K}$, $a_{w,1}=0.30$, $\Delta H=3200\ \text{cal mol}^{-1}$. \[ \frac{1}{T_2}-\frac{1}{T_1}=\frac{1}{318}-\frac{1}{303}=-1.5568\times 10^{-4}\ \text{K}^{-1}, \] \[ \frac{\Delta H}{R}= \frac{3200}{1.987}=1610.8. \] Hence \[ \ln a_{w,2}=\ln(0.30) - 1610.8\times(-1.5568\times10^{-4}) = -1.20397 + 0.251 \approx -0.953. \] Step 3: Exponentiate. \[ a_{w,2}=e^{-0.953}\approx 0.38, \] which lies within 0.32–0.41.