Let's solve each part **step by step**. Given Data: - Surface salt concentration, \( C_s = 380 \, \text{kg/m}^3 \) - Initial water concentration, \( C_ = \) - Diffusivity, \( D = 1.2 \times 10^{-9} \, \text{m}^2/\text{s} \) - Density of salt, \( \rho = 2165 \, \text{kg/m}^3 \) - Time, \( t = 24 \, \text{h} = 86400 \, \text{s} \) - Distance from interface, \( x = 10.2 \, \text{mm} = .0102 \, \text{m} \) --- ## (a) **Salt Concentration at 10.2 mm after 24 h** The solution for diffusion into a semi-infinite medium is: \[ C(x, t) = C_s \, \text{erfc} \left( \frac{x}{2\sqrt{Dt}} \right) \] Calculate \( \frac{x}{2\sqrt{Dt}} \): \[ 2\sqrt{Dt} = 2\sqrt{1.2 \times 10^{-9} \times 86400} \] \[ = 2\sqrt{1.0368 \times 10^{-4}} \] \[ = 2 \times .01018 = .02036 \, \text{m} \] \[ \frac{x}{2\sqrt{Dt}} = \frac{.0102}{.02036} = .501 \] From the table, for \( x = .5 \): - \( \text{erfc}(.5) = .4795 \) \[ C(x, t) = 380 \times .4795 = 182 \, \text{kg/m}^3 \] **Final Answer:** > **Salt concentration at 10.2 mm after 24 hours = \(\boxed{182 \, \text{kg/m}^3}\)** --- ## (b) **Thickness of Salt Dissolved in 24 hours** The amount of salt dissolved per unit area (assuming semi-infinite diffusion): \[ m = 2 C_s \sqrt{\frac{D t}{\pi}} \] The thickness (\( \delta \)) is: \[ \delta = \frac{m}{\rho} \] Calculate \( m \): \[ m = 2 \times 380 \times \sqrt{\frac{1.2 \times 10^{-9} \times 86400}{\pi}} \] \[ = 760 \times \sqrt{\frac{1.0368 \times 10^{-4}}{3.1416}} \] \[ = 760 \times \sqrt{3.302 \times 10^{-5}} \] \[ = 760 \times .005745 = 4.367 \, \text{kg/m}^2 \] \[ \delta = \frac{4.367}{2165} = .00202 \, \text{m} = 2.02 \, \text{mm} \] **Final Answer:** > **Thickness of salt dissolved in 24 hours = \(\boxed{2.02 \, \text{mm}}\)** --- ## (c) **Flux of Diffusion at the Slab-Water Interface at \( t = 24 \) h** Flux (\( J \)) at the interface: \[ J = -D \left( \frac{\partial C}{\partial x} \right)_{x=} = \frac{C_s}{\sqrt{\pi D t}} \] \[ J = \frac{380}{\sqrt{3.1416 \times 1.2 \times 10^{-9} \times 86400}} \] \[ = \frac{380}{\sqrt{3.255 \times 10^{-4}}} \] \[ = \frac{380}{.01805} = 21058 \, \text{kg/m}^2\text{/s} \] **Final Answer:** > **Flux at the interface after 24 h = \(\boxed{.02106 \, \text{kg/m}^2\text{/s}}\)** --- ## (d) **Draw Diagrams** ### (i) **Instantaneous Diffusion Rate vs Time at Interface** - The rate \( J \) decreases with time as \( 1/\sqrt{t} \). - Sketch a curve starting high at \( t= \), decreasing with \( t \). ### (ii) **Total Amount of Salt Diffused vs Time** - The total amount increases as \( \propto \sqrt{t} \). - Sketch a curve starting at zero and increasing with diminishing slope. --- ## (e) **Penetration Depth at \( t = 24 \) h and Concentration Profile** Penetration depth (\( \delta \)) is usually taken as the distance where \( C(x,t) = .01 C_s \): \[ .01 = \text{erfc}\left( \frac{x}{2\sqrt{Dt}} \right) \] From Table Q2, - \( \text{erfc}(2.3) \approx .01 \) So, \[ \frac{x}{2\sqrt{Dt}} = 2.3 \] \[ x = 2.3 \times .02036 = .0468 \, \text{m} = 46.8 \, \text{mm} \] **Diagram:** Plot \( C(x, t) \) vs \( x \), starting at \( 380 \) at \( x = \), dropping to nearly zero at \( x \approx 47 \) mm. **Final Answer:** > **Penetration depth = \(\boxed{46.8 \, \text{mm}}\)** --- ## **Summary Table** | Part | Final Answer | |------|----------------------------------------------| | (a) | \( C = 182 \, \text{kg/m}^3 \) | | (b) | \( \delta = 2.02 \, \text{mm} \) | | (c) | \( J = .02106 \, \text{kg/m}^2\text{/s} \) | | (e) | Penetration depth = \( 46.8 \, \text{mm} \) | *Let me know if you want the diagrams sketched!*