THERMAL BYU 6

1. Problem Setup

We need to freeze $m_w = 1.0 \text{ kg}$ of supercooled water initially at $-10^{\circ}\text{C}$ by adding ice of mass $m_{ice}$ at $-20^{\circ}\text{C}$.
Final State: All water freezes to ice. Equilibrium temperature assumed to be $0^{\circ}\text{C}$ (maximum temp for ice).

2. Heat Balance

Heat Released by Water:
1. Water at $-10^{\circ}\text{C}$ warms to $0^{\circ}\text{C}$: $Q_1 = m_w s_w \Delta T = 1.0 \times 4200 \times 10 = 42,000 \text{ J}$.
2. Water at $0^{\circ}\text{C}$ freezes to Ice at $0^{\circ}\text{C}$: $Q_2 = m_w L = 1.0 \times 336,000 = 336,000 \text{ J}$.
Note: Since the water is supercooled, it is in a metastable state. The net energy released to the surroundings (the added ice) to turn it into stable ice at 0°C is:

$$ Q_{released} = m_w L – m_w s_w (0 – (-10)) $$ $$ Q_{released} = 336,000 – 42,000 = 294,000 \text{ J} $$

Heat Absorbed by Added Ice:
Ice warms from $-20^{\circ}\text{C}$ to $0^{\circ}\text{C}$.

$$ Q_{absorbed} = m_{ice} s_i \Delta T = m_{ice} \times 2100 \times 20 $$ $$ Q_{absorbed} = 42,000 m_{ice} $$

3. Calculation

Equating Heat Released and Heat Absorbed:

$$ 294,000 = 42,000 m_{ice} $$ $$ m_{ice} = \frac{294,000}{42,000} $$ $$ m_{ice} = 7.0 \text{ kg} $$