Seawater

Seawater has a density of 1025 kg/m³, ice 920 kg/m³. 8 liters of seawater froze and created a cube. Calculate the size of the cube edge.

Correct result:

x₂ =  2.0734 dm

Solution:

$$\begin{gathered} V_1=8\text{ l}\to {\text{dm}}^3=8\cdot 1{\text{dm}}^3=8{\text{dm}}^3 \\ {x}_1=\sqrt[3]{V_1}=\sqrt[3]{8}=2\text{ dm } \\ {V}_2={\displaystyle \frac{1025}{920}}\cdot V_1={\displaystyle \frac{1025}{920}}\cdot 8={\displaystyle \frac{205}{23}}\doteq 8.913{\text{dm}}^3 \\ {x}_2=\sqrt[3]{V_2}=\sqrt[3]{8.913}=2.0734\text{ dm} \end{gathered}$$

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