A container can be made from steel [β = 36 × 10-6 (C°)-1] or lead [β = 87 × 10-6 (C°)-1]. A liquid is poured into the container, filling it to the brim. The liquid is either water [β = 207 × 10-6 (C°)-1] or ethyl alcohol [β = 1120 × 10-6 (C°)-1]. When the full container is heated, some liquid spills out. To keep the overflow to a minimum, from what should the container be made and what should the liquid be

The container should be made of lead and the liquid should be water.

Explanation:

Since the volume of the container of liquid after expansion is V = V₀(1 + βΔθ) where V = initial volume, β = coefficient of volume expansion, Δθ = temperature change.

So, the volume change V₂ - V₁ where V₁ = volume of liquid and V₂ = volume of container

For steel, V₂ = V₀(1 + β₂Δθ) and V₁ = V₀(1 + β₁Δθ)

So, ΔV = V₀(1 + β₂Δθ) - V₀(1 + β₁Δθ) = V₀[1 + β₂Δθ - 1 - β₁Δθ] = V₀[β₂Δθ - β₁Δθ]

Since we want a minimum value for ΔV and V₀ and Δθ are the same, we need ΔV/V₀Δθ = β₂ - β₁ to be a minimum

where β₂ = coefficient of volume expansion of liquid and β₁ = coefficient of volume expansion of container.

So, trying each combination, with  β₂ = 207 × 10⁻⁶ (C°)⁻¹] and β₁ = 36 × 10⁻⁶ (C°)⁻¹

β₂ - β₁ = 207 × 10⁻⁶ (C°)⁻¹ - 36 × 10⁻⁶ (C°)⁻¹ = 171 × 10⁻⁶ (C°)⁻¹

With  β₂ = 207  × 10⁻⁶ (C°)⁻¹] and β₁ = 87  × 10⁻⁶ (C°)⁻¹

β₂ - β₁ = 207 × 10⁻⁶ (C°)⁻¹ - 87 × 10⁻⁶ (C°)⁻¹ = 120 × 10⁻⁶ (C°)⁻¹

With  β₂ = 1120 × 10⁻⁶ (C°)⁻¹] and β₁ = 36  × 10⁻⁶ (C°)⁻¹

β₂ - β₁ = 1120 × 10⁻⁶ (C°)⁻¹ - 36 × 10⁻⁶ (C°)⁻¹ = 1084 × 10⁻⁶ (C°)⁻¹

With  β₂ = 1120 × 10⁻⁶ (C°)⁻¹] and β₁ = 87 ×  10⁻⁶ (C°)⁻¹

β₂ - β₁ = 207 × 10⁻⁶ (C°)⁻¹ - 87 × 10⁻⁶ (C°)⁻¹ = 1033 × 10⁻⁶ (C°)⁻¹

The combination that gives the lowest value for β₂ - β₁ is β₂ = 207 × 10⁻⁶ (C°)⁻¹] and β₁ = 87 × 10⁻⁶ (C°)⁻¹

Since  β₁ = 87 × 10⁻⁶ (C°)⁻¹ = coefficient of expansion for lead β₂ = 207 × 10⁻⁶ (C°)⁻¹]  = coefficient of expansion for water, the container should be made of lead and the liquid should be water.