THERMAL CYU 5

Step 1: Condition for Initial Sinking

Initially, the ice ball (mass $M$) contains a lead pellet (mass $m$). For the system to sink, the total weight must exceed the buoyant force.

$$ (M + m)g > F_B $$

$$ (M + m)g > (V_{ice} + V_{lead}) \rho_w g $$

Substituting volumes $V = \text{mass}/\text{density}$:

$$ M + m > \left( \frac{M}{\rho_i} + \frac{m}{\rho_p} \right) \rho_w $$

Rearranging to solve for $m$:

$$ m \left( 1 – \frac{\rho_w}{\rho_p} \right) > M \left( \frac{\rho_w}{\rho_i} – 1 \right) $$

Substituting values: $\rho_w = 1, \rho_i = 0.9, \rho_p = 11, M = 0.9 \text{ kg}$.

$$ m \left( 1 – \frac{1}{11} \right) > 0.9 \left( \frac{1}{0.9} – 1 \right) $$

$$ m \left( \frac{10}{11} \right) > 0.9 \left( \frac{0.1}{0.9} \right) \Rightarrow m \left( \frac{10}{11} \right) > 0.1 $$

$$ m > 0.11 \text{ kg} \Rightarrow m > 110 \text{ g} $$

Step 2: Thermal Equilibrium and Ice Accretion

The ice is initially at $\theta = -42^\circ\text{C}$. The water is at $0^\circ\text{C}$. The ice ball absorbs heat from the surrounding water until it reaches $0^\circ\text{C}$. This heat removal causes some surrounding water to freeze onto the ball.

$$ \text{Heat absorbed by ice ball} = \text{Heat released by freezing water} $$

$$ M s_i (0 – \theta) = m_{new} L $$

$$ m_{new} = \frac{M s_i |\theta|}{L} $$

Substituting values ($s_i = 2.4 \text{ J/g}^\circ\text{C}, L = 336 \text{ J/g}$):

$$ m_{new} = \frac{900 \times 2.4 \times 42}{336} = 270 \text{ g} $$

Total mass of ice is now $M’ = M + m_{new} = 900 + 270 = 1170 \text{ g} = 1.17 \text{ kg}$.

Step 3: Condition for Floating

After the new ice forms, the ball floats. This means the buoyant force must be at least equal to the weight.

$$ (V’_{ice} + V_{lead}) \rho_w g \ge (M’ + m)g $$

$$ \left( \frac{M’}{\rho_i} + \frac{m}{\rho_p} \right) \rho_w \ge M’ + m $$

Rearranging for $m$:

$$ M’ \left( \frac{\rho_w}{\rho_i} – 1 \right) \ge m \left( 1 – \frac{\rho_w}{\rho_p} \right) $$

$$ 1.17 \left( \frac{1}{0.9} – 1 \right) \ge m \left( \frac{10}{11} \right) $$

$$ 1.17 \left( \frac{1}{9} \right) \ge m \left( \frac{10}{11} \right) $$

$$ 0.13 \ge m \left( \frac{10}{11} \right) \Rightarrow m \le \frac{0.13 \times 11}{10} $$

$$ m \le 0.143 \text{ kg} \Rightarrow m \le 143 \text{ g} $$