Properties of Matter Part 3

Concept: The energy released when multiple small drops coalesce into one large drop is equal to the decrease in the total surface energy of the system.Formula/Principle:1. Volume of a sphere:$V = \;\frac{4}{3} \pi r^3$2. Surface Area of a sphere:$A = 4 \pi r^2$3. Surface Energy:$U = A \times S$4. Energy Released:$\Delta U = U_{\text{initial}} - U_{\text{final}}$Solution/Analysis:1. Volume Conservation: Since the eight small drops coalesce into one large drop, the total volume remains constant. Let$r$be the radius of the small drops and$R$be the radius of the final large drop.$8 \times \left( \;\frac{4}{3} \pi r^3 \right) = \;\frac{4}{3} \pi R^3$2. Finding the Final Radius ($R$): Canceling the common terms$\;\frac{4}{3} \pi$from both sides:$8 r^3 = R^3$$R = \sqrt[3]{8 r^3} = 2r$3. Calculating Initial Surface Energy ( $U_{\text{initial}}$ ): The initial surface energy is the sum of the surface energies of the eight individual drops.$U_{\text{initial}} = 8 \times (\text{Surface Area of one drop}) \times S$$U_{\text{initial}} = 8 \times (4 \pi r^2) \times S = 32 \pi r^2 S$4. Calculating Final Surface Energy ( $U_{\text{final}}$ ): The final surface energy is the surface energy of the single large drop of radius $R$.$U_{\text{final}} = (\text{Surface Area of large drop}) \times S$$U_{\text{final}} = 4 \pi R^2 S$Substituting$R = 2r$:$U_{\text{final}} = 4 \pi (2r)^2 S = 4 \pi (4r^2) S = 16 \pi r^2 S$5. Calculating Energy Released ( $\Delta U$ ): The energy released is the difference between the initial and final surface energies. Energy Released = $U_{\text{initial}} - U_{\text{final}}$ Energy Released =$32 \pi r^2 S - 16 \pi r^2 S$ Energy Released =$16 \pi r^2 S$