Energy required to change a spherical drop

Question

What amount of energy is required to change a spherical drop of water with a diameter of 1.40 mm to two smaller spherical drops of equal size? The surface tension, 𝛾 , of water at room temperature is 72.0 mJ/m2 .

Answer 1

First, we need to determine the initial and final surface areas of the water droplets.

1. Calculate the initial surface area (A1):

The initial drop has a diameter of 1.40 mm, so its radius (r1) is half of that:

r1 = 1.40 mm / 2 = 0.70 mm = 0.70 × 10^(-3) m

The surface area of a sphere is given by the formula:

A = 4πr^2

A1 = 4π(r1)^2 = 4π(0.70 × 10^(-3) m)^2 ≈ 6.154 × 10^(-6) m²

1. Calculate the final surface area (A2):

Since the initial drop is split into two equal-sized smaller drops, the volume of each smaller drop will be half of the initial drop's volume. The volume of a sphere is given by:

V = (4/3)πr^3

Let V1 be the initial drop's volume and V2 be the volume of each smaller drop.

Then:

V2 = V1 / 2

Since both the initial drop and the smaller drops have a spherical shape, we can write their volumes as:

(4/3)π(r1)^3 = 2 × (4/3)π(r2)^3

Where r2 is the radius of the smaller drops. Canceling out common terms, we get:

(r1)^3 = 2 × (r2)^3

Now, we can solve for r2:

r2 = (r1 / 2^(1/3))

r2 ≈ 0.605 × 10^(-3) m

Now, we can find the surface area of one smaller drop (A2_single) and then multiply by 2 to find the total

surface area of both smaller drops (A2):

A2_single = 4π(r2)^2 ≈ 4.598 × 10^(-6) m²

A2 = 2 × A2_single ≈ 9.196 × 10^(-6) m²

1. Calculate the change in surface area (ΔA):

ΔA = A2 - A1 ≈ 9.196 × 10^(-6) m² - 6.154 × 10^(-6) m² ≈ 3.042 × 10^(-6) m²

1. Calculate the energy required (ΔE):

Now, we can use the surface tension of water (γ = 72.0 mJ/m² = 72.0 × 10^(-3) J/m²) and the change in surface area (ΔA) to find the energy required:

ΔE = γ × ΔA ≈ (72.0 × 10^(-3) J/m²) × (3.042 × 10^(-6) m²) ≈ 2.190 × 10^(-7) J

Therefore, the amount of energy required to change the 1.40 mm spherical drop of water into two smaller equal-sized drops is approximately 2.19 × 10^(-7) J.