Water at 293K flows through a naphthalene tube with an inner diameter of $20 \mathrm{mm}$ at a main flow velocity of 1.2m/s. Given that the Schmidt number when naphthalene is dissolved in water is $S_c = 2330$, use the Reynolds, Prandtl-Taylor, von Kármán, and Colburn analogy equations to calculate the convective mass transfer coefficient after full development.

Options:
A. Reynolds analogy $k_{c}^{\circ}=3.815 \times 10^{-3} \, \mathrm{m/s}$; Prandtl-Taylor analogy $k_{c}^{\circ}=5.9 \times 10^{-6} \, \mathrm{m/s}$; von Kármán analogy $k_{c}^{\circ}=5.78 \times 10^{-6} \, \mathrm{m/s}$; Colburn analogy $k_{c}^{\circ}=2.17 \times 10^{-5} \, \mathrm{m/s}$
B. Reynolds analogy $k_{c}^{\circ}=3.815 \times 10^{-4} \, \mathrm{m/s}$; Prandtl-Taylor analogy $k_{c}^{\circ}=5.8 \times 10^{-6} \, \mathrm{m/s}$; von Kármán analogy $k_{c}^{\circ}=5.78 \times 10^{-6} \, \mathrm{m/s}$; Colburn analogy $k_{c}^{\circ}=2.17 \times 10^{-5} \, \mathrm{m/s}$
C. Reynolds analogy $k_{c}^{\circ}=3.815 \times 10^{-3} \, \mathrm{m/s}$; Prandtl-Taylor analogy $k_{c}^{\circ}=5.8 \times 10^{-6} \, \mathrm{m/s}$; von Kármán analogy $k_{c}^{\circ}=5.78 \times 10^{-6} \, \mathrm{m/s}$; Colburn analogy $k_{c}^{\circ}=2.17 \times 10^{-5} \, \mathrm{m/s}$
D. Reynolds analogy $k_{c}^{\circ}=3.815 \times 10^{-3} \, \mathrm{m/s}$; Prandtl-Taylor analogy $k_{c}^{\circ}=5.8 \times 10^{-6} \, \mathrm{m/s}$; von Kármán analogy $k_{c}^{\circ}=5.78 \times 10^{-6} \, \mathrm{m/s}$; Colburn analogy $k_{c}^{\circ}=2.16 \times 10^{-5} \, \mathrm{m/s}$
E. Reynolds analogy $k_{c}^{\circ}=3.815 \times 10^{-3} \, \mathrm{m/s}$; Prandtl-Taylor analogy $k_{c}^{\circ}=5.7 \times 10^{-6} \, \mathrm{m/s}$; von Kármán analogy $k_{c}^{\circ}=5.78 \times 10^{-6} \, \mathrm{m/s}$; Colburn analogy $k_{c}^{\circ}=2.17 \times 10^{-5} \, \mathrm{m/s}$
F. Reynolds analogy $k_{c}^{\circ}=3.815 \times 10^{-3} \, \mathrm{m/s}$; Prandtl-Taylor analogy $k_{c}^{\circ}=5.8 \times 10^{-6} \, \mathrm{m/s}$; von Kármán analogy $k_{c}^{\circ}=5.78 \times 10^{-6} \, \mathrm{m/s}$; Colburn analogy $k_{c}^{\circ}=2.17 \times 10^{-6} \, \mathrm{m/s}$
G. Reynolds analogy $k_{c}^{\circ}=3.815 \times 10^{-3} \, \mathrm{m/s}$; Prandtl-Taylor analogy $k_{c}^{\circ}=5.8 \times 10^{-6} \, \mathrm{m/s}$; von Kármán analogy $k_{c}^{\circ}=5.78 \times 10^{-6} \, \mathrm{m/s}$; Colburn analogy $k_{c}^{\circ}=3.17 \times 10^{-5} \, \mathrm{m/s}$
H. Reynolds analogy $k_{c}^{\circ}=3.815 \times 10^{-3} \, \mathrm{m/s}$; Prandtl-Taylor analogy $k_{c}^{\circ}=5.8 \times 10^{-6} \, \mathrm{m/s}$; von Kármán analogy $k_{c}^{\circ}=6.78 \times 10^{-6} \, \mathrm{m/s}$; Colburn analogy $k_{c}^{\circ}=2.17 \times 10^{-5} \, \mathrm{m/s}$
I. Reynolds analogy $k_{c}^{\circ}=3.815 \times 10^{-2} \, \mathrm{m/s}$; Prandtl-Taylor analogy $k_{c}^{\circ}=5.8 \times 10^{-6} \, \mathrm{m/s}$; von Kármán analogy $k_{c}^{\circ}=5.78 \times 10^{-6} \, \mathrm{m/s}$; Colburn analogy $k_{c}^{\circ}=2.17 \times 10^{-5} \, \mathrm{m/s}$
J. Reynolds analogy $k_{c}^{\circ}=3.810 \times 10^{-3} \, \mathrm{m/s}$; Prandtl-Taylor analogy $k_{c}^{\circ}=5.8 \times 10^{-6} \, \mathrm{m/s}$; von Kármán analogy $k_{c}^{\circ}=5.78 \times 10^{-6} \, \mathrm{m/s}$; Colburn analogy $k_{c}^{\circ}=2.17 \times 10^{-5} \, \mathrm{m/s}$
Answer: C