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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

Respuesta :

Answer:

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

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