Height of Ice Block Below the Water Surface
Consider a large cubic ice block floating in seawater. The specific gravities of ice and seawater are 0.92 and 1.025, respectively. If a 25-cm-high portion of the ice block extends above the surface of the water, determine the height of the ice block below the surface.
The height of the portion of a cubic ice block that extends above the water surface is measured. The height of the ice block below the surface is to be determined.
Assumptions 1 The buoyancy force in air is negligible. 2 The top surface of the ice block is parallel to the surface of the sea.
Properties The specific gravities of ice and seawater are given to be 0.92 and 1.025, respectively, and thus the corresponding densities are 920 kg/m³ and 1025 kg/m³.
Analysis The weight of a body floating in a fluid is equal to the buoyant force acting on it (a consequence of vertical force balance from static equilibrium), as shown in Fig. 3–48. Therefore,
\begin{gathered}W=F_{B} \rightarrow \rho_{\text {body }} g V_{\text {total }}=\rho_{\text {fluid }} g V_{\text {submerged }} \\\frac{V_{\text {submerged }}}{V_{\text {total }}}=\frac{\rho_{\text {body }}}{\rho_{\text {fluid }}}\end{gathered}
The cross-sectional area of a cube is constant, and thus the “volume ratio” can be replaced by “height ratio.” Then,
\frac{h_{\text {submerged }}}{h_{\text {total }}}=\frac{\rho_{\text {body }}}{\rho_{\text {fluid }}} \rightarrow \frac{h}{h+0.25}=\frac{\rho_{\text {ice }}}{\rho_{\text {water }}} \rightarrow \frac{h}{h+0.25 m }=\frac{920 kg / m ^{3}}{1025 kg / m ^{3}}
where h is the height of the ice block below the surface. Solving for h gives
h=\frac{\left(920 kg / m ^{3}\right)(0.25 m )}{(1025-920) kg / m ^{3}}=2.19 m
Discussion Note that 0.92/1.025 = 0.898, so approximately 90% of the volume of an ice block remains under water. For symmetrical ice blocks this also represents the fraction of height that remains under water. This also applies to icebergs; the vast majority of an iceberg is submerged.
