Question 11.1: Let us model the dissolution of sugar in a cup of tea by car...
Let us model the dissolution of sugar in a cup of tea by carefully stratifying a large quantity of 80°C temperature water onto a saturated solution of sucrose at the bottom of a tall cylinder. Let us assume that at t = 0, the total amount of sucrose, n_{0} = 0.01 mol (3.423 g), is at the plane of the bottom of the cup in a negligibly thin layer. Let us consider that the cylinder is infinitely tall compared to the distance sugar would diffuse during the experiment and ignore any convective flow inside the cylinder. Estimate the time needed for the sucrose concentration to attain c = 0.001 mol dm^{-3} (roughly 0.34 g L^{-1}) 1 cm away from the bottom.
The diffusion coefficient of sucrose is 6.73 × 10^{–6} cm² s^{–1} and the cross-sectional area of the cup is 25 cm².
The concentration profile of sucrose along the vertical direction z can be described by Eq. (11.18).
c=\frac{n_{0}}{A / 2} \frac{1}{\sqrt{4 \pi D t}} \mathrm{e}^{-\frac{z^{2}}{4 D t}} (11.18)
By substituting the given quantities and z = 1 cm, we get the equation
10^{-6} \mathrm{~mol} \mathrm{~cm}^{-3}=\frac{0.01 \mathrm{~mol}}{25 \mathrm{~cm}^{2} \sqrt{6.73 \times 10^{-6} \mathrm{~cm}^{2} \mathrm{~s}^{-1} \cdot \pi t}} e^{-\frac{(\mathrm{lcm})^{2}}{4 t\cdot 6.37 \times 10^{-6} \mathrm{~cm}^{2}s^{-1}}}It is obvious that we can eliminate units by “simplification” using mol, cm, and s throughout, thus having only dimensionless numbers left in the exponential equation. Solution of this equation can be found by taking the logarithm of both sides, providing 5238.1 s, i.e., 87.3 min. (Obviously, it is worth stirring if we want sugar to be dissolved in our cup before the tea gets cold.)