Seawater

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

Correct answer:

x₂ =  2.0734 dm

Step-by-step explanation:

$$\begin{gathered} V_1=8\text{l}\to {\text{dm}}^3=8\cdot 1{\text{dm}}^3=8{\text{dm}}^3 \\ V=x^3 \\ {x}_1=\sqrt[3]{V_1}=\sqrt[3]{8}=2\text{dm} \\ {\rho }_1=1025{\text{kg/m}}^3 \\ {\rho }_2=920{\text{kg/m}}^3 \\ {V}_2=\frac{{\rho }_1}{{\rho }_2}\cdot V_1=\frac{1025}{920}\cdot 8=\frac{1025\cdot 8}{920}=\frac{8200}{920}=\frac{205}{23}\doteq 8.913{\text{dm}}^3 \\ {x}_2=\sqrt[3]{V_2}=\sqrt[3]{8.913}=2.0734\text{dm} \end{gathered}$$

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