A hollow glass sphere has a density of 1.30g/cm cubic at 20°C. Glycerine has a density of 1.26g/cm cubic at 20°C. At what temperature would the sphere begin to float in glycerine?

Given coefficient of volume expansion of glycerine is 53 * 10^-5 / °C.

-38°C

Explanation:

When glycerine is heated, it expands from volume V₁ to volume V₂:

V₂ - V₁ = V₁ β (T₂ - T₁)

Since the mass of glycerine is the same before and after:

m = m

ρ₁ V₁ = ρ₂ V₂

V₂ = (ρ₁ / ρ₂) V₁

Substituting:

(ρ₁ / ρ₂) V₁ - V₁ = V₁ β (T₂ - T₁)

(ρ₁ / ρ₂) - 1 = β (T₂ - T₁)

(ρ₁ - ρ₂) / ρ₂ = β (T₂ - T₁)

(ρ₁ - ρ₂) / (ρ₂ β) = T₂ - T₁

T₂ = T₁ + (ρ₁ - ρ₂) / (ρ₂ β)

The glass sphere will float when it has the same density as the glycerine.

Given:

T₁ = 20°C

ρ₁ = 1.26 g/cm³

ρ₂ = 1.30 g/cm³

β = 53*10⁻⁵/°C

T₂ = 20 + (1.26 - 1.30) / (1.30 * 53*10⁻⁵)

T₂ = - 38

The glycerine must be cooled to - 38°C for the hollow glass sphere to float.