Question

Two Identical drop of water with radiu 0. 2cm Join together to form a big drop. Calculate the lo or gain In urface energy during the proce. (urface tenion of water = 0. 75 Nm)

Answer 1

Step-by-step explanation:

Given:

r₁ = 0.2 cm = 0.002 m

σ = 75·10⁻³ N·m

______________

ΔE - ?

The volume of one small drop:

V₁ = (4/3)·π·r³ = (4/3)·3.14·0.002³ ≈ 3.35·10⁻⁸ m³

Large drop volume:

V = 2·V₁ = 2·3.35·10⁻⁸ = 6.70·10⁻⁸ m³

V = (4/3)·π·R³

Big drop radius:

R = ∛ (3·V / (4·π)) = ∛ (3·6.70·10⁻⁸ / (4·3.14) ) ≈ 0.0025 m

Surface area of a large drop:

S = 4·π·R² = 4·3.14·0.0025² ≈ 78.5·10⁻⁶ m²

Surface area of a small drop:

S₁ = 4·π·r₁² = 4·3.14·0.002² = 50.2·10⁻⁶ m²

S₂ = 2·S₁ = 100.4·10⁻⁶ m²

ΔS = S₂ - S₁ = (100.4 - 78.5)·10⁻⁶ m² = 21.9·10⁻⁶ m²

Energy change:

ΔE = σ·ΔS = 75·10⁻³·21.9·10⁻⁶ ≈ 1.64·10⁻⁶ J