1. **State the problem:** A mass of 0.25 kg of ice at -32°C is placed in water at 0°C. When thermal equilibrium is reached, the temperature is 0°C. We need to show that about 50 g of water have turned into ice. 2. **Given data:** - Mass of ice, $m_{ice} = 0.25$ kg - Initial temperature of ice, $T_i = -32^\circ C$ - Final temperature, $T_f = 0^\circ C$ - Specific heat capacity of ice, $c_{ice} = 2100$ J/kg/K - Specific latent heat of fusion of ice, $L = 334000$ J/kg 3. **Formula used:** The heat lost by water turning into ice equals the heat gained by ice warming up from -32°C to 0°C plus the heat required to melt some ice: $$m_{water} L = m_{ice} c_{ice} \Delta T + m_{melted} L$$ Since the final temperature is 0°C, the ice warms up and some water freezes. 4. **Set up the energy balance:** Let $m$ be the mass of water that freezes (turns into ice). The heat released by freezing water is $m L$. The heat absorbed by ice warming from -32°C to 0°C is $m_{ice} c_{ice} \Delta T$. Energy balance: $$m L = m_{ice} c_{ice} (0 - (-32))$$ $$m L = m_{ice} c_{ice} \times 32$$ 5. **Calculate $m$:** $$m = \frac{m_{ice} c_{ice} \times 32}{L} = \frac{0.25 \times 2100 \times 32}{334000}$$ 6. **Simplify:** $$m = \frac{16800}{334000} = 0.0503 \text{ kg} = 50.3 \text{ g}$$ 7. **Conclusion:** About 50 g of water have turned into ice when thermal equilibrium is reached.