Question

A liquid has a density at 289K of 955kg/m^3. Its(volume) thermal expansion coefficient is 1.05E-3/K. What is its density at 316K? Answer 929 kg/m^3 I'd appreciate explanation ! Thank you :)

Answer 1

Volume thermal expansion coefficient = = 1.05 x 10^-3 K^-1

Initial temperature of the liquid = T₁ = 289 K

Initial density of the liquid = ₁ = 955 kg/m³

Initial volume of the liquid = V₁

New temperature of the liquid = T₂ = 316 K

New density of the liquid =  ₂

New volume of the liquid = V₂

Change in volume of the liquid = V

V = V₁(T₂ - T₁)

V₂ = V₁ + V

V₂ = V₁ + V₁(T₂ - T₁)

V₂ = V₁(1 + (T₂ - T₁))

The mass of the liquid remains same

Mass of the liquid = M

M = ₁V₁ = ₂V₂

₁V₁ = ₂V₂

₁V₁ = ₂V₁(1 + (T₂ - T₁))

₁ = ₂(1 + (T₂ - T₁))

955 = ₂(1 + (1.05x10^-3)(316 - 289))

₂ = 928.67 kg/m³

Density of the liquid at 316 K = 928.67 kg/m³