What amount of energy is required to change a spherical drop of water with a diameter of 1.60 mm 1.60 mm to six smaller spherical drops of equal size? The surface tension, γ γ , of water at room temperature is 72.0 mJ / m 2 72.0 mJ/m2 .

0.4731 μJ

Explanation:

Surface tension = surface energy/surface area.

Let the radius of the big sphere be R = 0.8 mm = 8 × 10⁻⁴ m

Radius of the small spheres = r

Surface Area of the big sphere = 4πR² = 4π(8 × 10⁻⁴)² = 0.0000080425 m² = 8.0425 × 10⁻⁶ m²

Surface energy of the big sphere = surface tension × surface area = 72 × 10⁻³ × 8.0425 × 10⁻⁶ = 5.791 × 10⁻⁷ J

Volume of big sphere = 4πR³/3 = 4π(8.0 × 10⁻⁴)³/3 = 2.145 × 10⁻⁹ m³

This volume is separated into 6 equal smaller drops,

Volume of one small sphere = (2.145 × 10⁻⁹)/6 = 3.574 × 10⁻¹⁰ m³

Each small sphere would have a radius as calculated from

4πr³/3 = 3.574 × 10⁻¹⁰ = 0.00044 m = 0.44 mm

Surface area of each small sphere = 4πr² = 4π(0.00044)² = 0.0000024356 m² = 2.436 × 10⁻⁶ m²

Surface area of 6 small spheres = 6 × 2.436 × 10⁻⁶ = 0.0000146141 m²

Surface energy of 6 small spheres = surface tension of water × Surface area of 6 small spheres = 72 × 0.001 × 0.0000146141 = 0.0000010522 J = 1.0522 × 10⁻⁶ J

Difference in surface energy of the big sphere and 6 small spheres = (1.0522 × 10⁻⁶) - (5.791 × 10⁻⁷) = 0.0000004731 J = 4.731 × 10⁻⁷ J = 0.4731 μJ

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f=ma newton

f=100x2.5=250n