Archimedes' Principle — Question 5

Question 5

Melting floating ice depends on the receiving liquid

A $2.0\,\mathrm{kg}$ piece of pure ice floats in a uniform liquid of density $1200\,\mathrm{kg/m^3}$. On melting it becomes water of density $1000\,\mathrm{kg/m^3}$. Assume volume additivity on mixing, no evaporation or thermal expansion, and negligible air buoyancy. The container has constant horizontal area $0.10\,\mathrm{m^2}$.

1. Find the level change after melting.

2. Explain why the familiar no-change result for ice floating in fresh water does not apply here, and recover it as a limiting case.

Question 5 - Solution

Melting floating ice depends on the receiving liquid

(a) Derivation and calculation

$$V_i=m/\rho_\ell=2.0/1200=.00166667\,\mathrm{m^3}.$$$$V_f=m/\rho_w=2.0/1000=.00200000\,\mathrm{m^3}.$$$$\boxed{\Delta h=\frac{V_f-V_i}{A}=\frac{.000333333}{.100}=.00333333\,\mathrm m\ \text{upward}}.$$

(b) Result and interpretation

In general,$$\boxed{\Delta h=\frac mA\left(\frac1{\rho_w}-\frac1{\rho_\ell}\right)}.$$$$\rho_\ell=\rho_w\Rightarrow\Delta h=0.$$Here the denser receiving liquid requires less displacement than the final meltwater volume. Mixing contraction and thermal volume changes are excluded.

Physical check

The no-change result depends on equality of receiving-liquid and meltwater densities, not on ice density.