Given,
Initial volume of the water, V₁=3.00 m³
The initial temperature of the water, T₁=20.0 °C=293.15 K
The final temperature of the water, T₂=60 °C=333.15 K
From Charle's law, we have,
[tex]\frac{V_1}{T_1_{}}=\frac{V_2}{T_2}[/tex]On rearranging the above equation,
[tex]V_2=\frac{V_1T_2}{T_1}[/tex]On substituting the known values in the above equation,
[tex]V_2=\frac{3.00\times333.15}{293.15}=3.41m^3[/tex]Therefore the change in the volume is,
[tex]\Delta V=V_2-V_1[/tex]i.e.,
[tex]\Delta V=3.41-3.00=0.41m^3[/tex]Therefore, the volume of the water will expand by 0.41 m³
