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?

1) Steel, Ethyl alcohol

2) Lead, Water

3) Lead, Ethyl alcohol

4) Steel, Water

correct answer 2

Explanation:

For this exercise we must calculate the change in volume of the container and the liquid, the education that describes the change in volume is

ΔV = V₀ β ΔT

Let's write this equation for each material

Steel

ΔV₁ = V₀ 36 10⁻⁶ ΔT

Lead

ΔV₂ = V₀ 87 10⁻⁶ ΔT

We write the same equations for liquids

Water

ΔV₃ = V₀ 207 10⁻⁶ ΔT

Alcohol

ΔV₄ = V₀ 1120 10⁻⁶ ΔT

So that the spill is minimal, we can find, the volume of liquid minus the change in volume of content

Steel-alcohol

ΔV₁₄ = ΔV₄ - ΔV₁

ΔV₁₄ = V₀ ΔT 10⁻⁶ (1120-36) =

ΔV₁₄ = (V₀ ΔT 10⁻⁶) 1084

Lead - water

ΔV₂₃ = ΔVw - ΔVpb

ΔV₂₃ = V₀ ΔT 10⁻⁶ (207-87)

ΔV₂₃ = V₀ ΔT 10⁻⁶ 120

Lead - Alcohol

ΔV₄₃ = ΔV₄ - ΔV₂

ΔV₄₃ = V₀ ΔT 10⁻⁶ (1120-87)

ΔV₄₃ = V₀ ΔT 10⁻⁶ 1033

Steel - Water

ΔV₃₁ = ΔV₃ - ΔV₁

ΔV₃₁ = V₀ ΔT 10⁻⁶ (207 - 36)

ΔV₃₁ = V₀ ΔT 10⁻⁶ 171

We can see that there is the smallest spill for the combination lead filled with water

correct answer 2