GSO002 Problem 10

Let the volume of the original water in the tank be $V_0$ . The total mass of the floating system (air mass neglected) is:

M_{total} = M + m_1 + m_2

By Archimedes' principle, the submerged volume $V_{sub}$ satisfies:

\rho_w V_{sub} g = M_{total} g \implies V_{sub} = \frac{M + m_1 + m_2}{\rho_w}

The initial water level:

h_{init} = \frac{1}{S_0} \left( V_0 + \frac{M}{\rho_w} + \frac{m_1}{\rho_w} + \frac{m_2}{\rho_w} \right)

Each substance contributes to the total volume as follows:

h_{final} = \frac{1}{S_0} \left( V_0 + \frac{M}{\rho_w} + \frac{m_1}{\rho_1} + \frac{m_2}{\rho_2} \right)

\Delta V = \left( V_0 + \frac{M}{\rho_w} + \frac{m_1}{\rho_w} + \frac{m_2}{\rho_w} \right) - \left( V_0 + \frac{M}{\rho_w} + \frac{m_1}{\rho_1} + \frac{m_2}{\rho_2} \right)

\Delta V = m_1 \left( \frac{1}{\rho_w} - \frac{1}{\rho_1} \right) + m_2 \left( \frac{1}{\rho_w} - \frac{1}{\rho_2} \right)

(Note: the air volume $V_a$ and ice mass $M$ cancel — they do not affect the result.)

Metal lump term:

m_1 \left( \frac{1}{\rho_w} - \frac{1}{\rho_1} \right) = 0.27 \times \left( \frac{1}{1000} - \frac{1}{7500} \right) = 0.27 \times \frac{13}{15000} = \frac{3.51}{15000} = 0.000234 \ \mathrm{m^3}

Oil term:

m_2 \left( \frac{1}{\rho_w} - \frac{1}{\rho_2} \right) = 0.17 \times \left( \frac{1}{1000} - \frac{1}{850} \right) = 0.17 \times \frac{-3}{17000} = -0.000030 \ \mathrm{m^3}

\Delta V = 0.000234 - 0.000030 = 0.000204 \ \mathrm{m^3}

\Delta h = \frac{0.000204}{0.012} = 0.017 \ \mathrm{m}

Multiplied by $10^5$ :

0.017 \times 10^5 = 1700

Answer: 1700

Melting of Ice with Inclusions and Liquid Level Change