At room temperature, the surface tension of water is 72.0 mJ·m^–2. What is the energy required to change a spherical drop of water with a diameter of 1.80 mm to three smaller spherical drops of equal size? The surface area of a sphere of radius r is 4πr^2 and the volume is 4πr^3/3.

To calculate the energy required to change a spherical drop of water into three smaller spherical drops of equal size, we need to determine the surface area and volume of the original drop and the three smaller drops.

1. Start by calculating the radius of the original drop:
- The diameter is given as 1.80 mm, so the radius is half of the diameter: r = 0.90 mm = 0.90 × 10^(-3) m.

2. Calculate the surface area of the original drop:
- Use the formula for the surface area of a sphere: Surface area = 4πr^2.
- Substitute the radius into the formula: Surface area = 4π(0.90 × 10^(-3))^2.

3. Calculate the volume of the original drop:
- Use the formula for the volume of a sphere: Volume = 4πr^3/3.
- Substitute the radius into the formula: Volume = 4π(0.90 × 10^(-3))^3/3.

4. Calculate the surface area and volume of each smaller drop:
- Since we have three smaller drops, we divide the surface area and volume of the original drop by 3.

5. Calculate the energy required to change the original drop into three smaller drops:
- The energy is equal to the surface tension multiplied by the change in surface area: Energy = surface tension × (3 × surface area of smaller drop - surface area of original drop).

Now let's calculate step by step:

Step 1:
Radius (r) = 0.90 × 10^(-3) m

Step 2:
Surface area of the original drop = 4π(0.90 × 10^(-3))^2

Step 3:
Volume of the original drop = 4π(0.90 × 10^(-3))^3/3

Step 4:
Surface area of each smaller drop = (4π(0.90 × 10^(-3))^2) / 3
Volume of each smaller drop = (4π(0.90 × 10^(-3))^3/3) / 3

Step 5:
Energy = 72.0 × 10^(-3) × (3 × (4π(0.90 × 10^(-3))^2 / 3) - 4π(0.90 × 10^(-3))^2)

Evaluate these steps to find the final answer.

To find the energy required to change a spherical drop of water into three smaller spherical drops, we need to calculate the change in surface area.

First, let's calculate the original surface area of the larger drop.
We are given the diameter of the larger drop, which is 1.80 mm. So, the radius (r) of the larger drop is half of the diameter, which is 0.9 mm or 0.9 x 10^(-3) m.

The formula for the surface area of a sphere is given by A = 4πr^2, where A is the surface area and r is the radius.
Substituting the given radius, we have:
A_larger = 4π(0.9 x 10^(-3))^2 = 4π(0.81 x 10^(-6)) = 3.24π x 10^(-6) m^2

Next, we need to find the surface area of each smaller drop.
Since we're splitting the larger drop into three equal-sized drops, the radius of the smaller drops will be one-third of the original radius.

The radius of each smaller drop is (1/3) x 0.9 x 10^(-3) = 0.3 x 10^(-3) m.

Using the formula for the surface area, we can calculate the surface area of each smaller drop:
A_smaller = 4π(0.3 x 10^(-3))^2 = 4π(0.09 x 10^(-6)) = 0.36π x 10^(-6) m^2

Now, we can find the total surface area of the three smaller drops by multiplying the individual surface area of one smaller drop by 3:
A_total_smaller = 3 x A_smaller = 3 x (0.36π x 10^(-6)) = 1.08π x 10^(-6) m^2

Finally, we can calculate the change in surface area by subtracting the total surface area of the smaller drops from the surface area of the larger drop:
ΔA = A_larger - A_total_smaller = (3.24π x 10^(-6)) - (1.08π x 10^(-6))
= 2.16π x 10^(-6) m^2

The energy required to change the drop can be found using the formula:
Energy = surface tension x change in surface area

We are given that the surface tension of water is 72.0 mJ·m^–2, which can be converted to Joules per square meter (J/m^2) by dividing by 1000:
Surface tension = 72.0 mJ·m^–2 = 72.0 x 10^(-3) J/m^2

Substituting the values, we can now calculate the energy required:
Energy = (72.0 x 10^(-3)) x (2.16π x 10^(-6))
= 0.15552π x 10^(-9) J

Therefore, the energy required to change the spherical drop of water into three smaller spherical drops is approximately 0.15552π x 10^(-9) J.