Question

A diver 45 mm deep in 10∘C∘C fresh water exhales a 0.50 cmcm diameter bubble.

What is the bubble's diameter just as it reaches the surface of the lake, where the water temperature is 20∘C∘C? Assume that the air bubble is always in thermal equilibrium with the surrounding water.

Answer 1

The diameter of the bubble just as it reaches the surface of the lake is 5.06 mm.

The diameter of the bubble just as it reaches the surface of the lake is calculated as follows;

V₁ / T₁ = V₂/T₂

where;

V₁ is the initial volume of the bubble
V₂ is the final volume of the bubble
T₁ is the initial temperature of the bubble
T₂ is the final temperature of the bubble
The volume of the bubble is calculated using the formula for volume of a sphere;

V = ⁴/₃πr³

r³₁ / T₁ = r³₂/T₂

r³₂ = (T₂/T₁) x r³₁

r³₂ = [ (20 + 273) / (10 + 273) ] x (0.25 x 10⁻²)³

r³₂ = 1.62 x 10⁻⁸

r₂ = ∛ (1.62 x 10⁻⁸)

r₂ = 2.53 x 10⁻³ m

The diameter = 2 x r₂

diameter = 5.06 x 10⁻³ m

diameter = 5.06 mm

Answer 2

The diameter of the bubble just as it reaches the surface of the lake is 5.06 mm.

How to calculate the diameter of the bubble?

The diameter of the bubble just as it reaches the surface of the lake is calculated as follows;

V₁ / T₁ = V₂/T₂

where;

• V₁ is the initial volume of the bubble
• V₂ is the final volume of the bubble
• T₁ is the initial temperature of the bubble
• T₂ is the final temperature of the bubble

The volume of the bubble is calculated using the formula for volume of a sphere;

V = ⁴/₃πr³

r³₁ / T₁ = r³₂/T₂

r³₂ = (T₂/T₁) x r³₁

r³₂ = [ (20 + 273) / (10 + 273) ] x (0.25 x 10⁻²)³

r³₂ = 1.62 x 10⁻⁸

r₂ = ∛ (1.62 x 10⁻⁸)

r₂ = 2.53 x 10⁻³ m

The diameter = 2 x r₂

diameter = 5.06 x 10⁻³ m

diameter = 5.06 mm