Final answer:
No water will boil away as the lead solidifies, so the mass of water that will be boiled away is 0 kg.
Step-by-step explanation:
When the molten lead at 328 degrees Celsius is poured into the water at its boiling point of 100 degrees Celsius, the lead starts to solidify and transfers heat to the water, causing it to boil. The amount of water that will boil away depends on the heat transferred from the lead to the water. To calculate the mass of the water that will be boiled away, the specific heat capacity of water and the heat released during the solidification of lead need to be considered.
First, we need to find the heat released during the solidification of lead. The heat transferred during a phase change can be calculated using the formula Q = m * ΔH, where Q is the heat transferred, m is the mass of the substance, and ΔH is the heat of fusion or solidification. The heat of solidification for lead is 5.10 x 10^4 J/kg (from reference). The mass of the lead is given as 12.6 kg.
Using the formula, we find the heat transferred during the solidification of lead is Q = (12.6 kg)(5.10 x 10^4 J/kg) = 6.426 x 10^5 J.
Next, we need to calculate the heat transferred from the solidifying lead to the water. The heat transferred can be calculated using the formula Q = m * Cp * ΔT, where Q is the heat transferred, m is the mass of the substance, Cp is the specific heat capacity, and ΔT is the change in temperature. The specific heat capacity of water is 4186 J/kg°C (from reference). The change in temperature of water can be calculated as ΔT = 100°C - 100°C = 0°C, since the water remains at its boiling point.
Using the formula, we find the heat transferred from the lead to the water is Q = (m)(Cp)(ΔT) = (m)(4186 J/kg°C)(0°C) = 0 J.
Since the heat transferred from the lead to the water is 0 J, no water will boil away as the lead solidifies. Therefore, the mass of water that will be boiled away is 0 kg (Option A).