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Contests/Gnit Sunday Open 002 (GSO002)/Problem 10 Melting of Ice with Inclusions and Liquid Level Change
Problem 10

Melting of Ice with Inclusions and Liquid Level Change

Finished
300 ptsLv.5 ElementaryThermodynamics
2026/03/22 19:30〜2026/03/22 21:00
Author: admin

Problem Statement

A cylindrical tank of cross-sectional area S0S_0S0​ contains a sufficient amount of water with density ρw\rho_wρw​.

A single large block of ice floats in this water. The ice block consists primarily of pure ice (mass MMM, density ρi\rho_iρi​) and completely contains the following three foreign objects:

  1. A metal lump of mass m1m_1m1​ and density ρ1\rho_1ρ1​ (ρ1>ρw\rho_1 > \rho_wρ1​>ρw​)
  2. Oil of mass m2m_2m2​ and density ρ2\rho_2ρ2​ (ρ2<ρw\rho_2 < \rho_wρ2​<ρw​)
  3. Air of volume VaV_aVa​

The ice block floats at rest on the water surface without touching the walls or bottom of the tank. Let the water level at this point be hinith_{init}hinit​.

After sufficient time passes and the ice fully melts, the metal lump sinks to the bottom, the oil separates and forms a uniform layer on top of the water, and all the air escapes into the atmosphere. Let the height of the topmost liquid surface (top of the oil layer) when the system is again at rest be hfinalh_{final}hfinal​.

Find the change in liquid level Δh=hinit−hfinal\Delta h = h_{init} - h_{final}Δh=hinit​−hfinal​.

Evaporation of water, changes in density due to temperature, and dissolution of oil in water can all be neglected. The mass of air is negligible.

Constraints

  • S0=1.2×10−2 m2S_0 = 1.2 \times 10^{-2} \ \mathrm{m^2}S0​=1.2×10−2 m2
  • M=3.0 kgM = 3.0 \ \mathrm{kg}M=3.0 kg
  • ρi=9.2×102 kg/m3\rho_i = 9.2 \times 10^2 \ \mathrm{kg/m^3}ρi​=9.2×102 kg/m3
  • ρw=1.0×103 kg/m3\rho_w = 1.0 \times 10^3 \ \mathrm{kg/m^3}ρw​=1.0×103 kg/m3
  • m1=2.7×10−1 kgm_1 = 2.7 \times 10^{-1} \ \mathrm{kg}m1​=2.7×10−1 kg
  • ρ1=7.5×103 kg/m3\rho_1 = 7.5 \times 10^3 \ \mathrm{kg/m^3}ρ1​=7.5×103 kg/m3
  • m2=1.7×10−1 kgm_2 = 1.7 \times 10^{-1} \ \mathrm{kg}m2​=1.7×10−1 kg
  • ρ2=8.5×102 kg/m3\rho_2 = 8.5 \times 10^2 \ \mathrm{kg/m^3}ρ2​=8.5×102 kg/m3
  • Va=8.0×10−4 m3V_a = 8.0 \times 10^{-4} \ \mathrm{m^3}Va​=8.0×10−4 m3

Input Format

Find the numerical value of Δh\Delta hΔh in m\mathrm{m}m and give the answer as a positive integer equal to the value multiplied by 10510^5105.

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