Final answer:
To determine the mass of ice that can be melted and heated to 15.7 °C with 5.85×103 kJ of heat, calculate the cumulative energy required to warm the ice to 0 °C, melt it, and then heat the resulting water to the final temperature.
Step-by-step explanation:
The student is asking how many grams of ice at -26.2 °C can be completely converted to liquid at 15.7 °C with an available heat of 5.85×103 kJ. To solve this, we first calculate the heat required to raise the temperature of the ice from -26.2 °C to 0 °C using the specific heat capacity of ice. Next, we calculate the heat of fusion to melt the ice. Finally, we determine the heat required to raise the temperature of the liquid water from 0 °C to 15.7 °C using the specific heat capacity of water.
To begin, we use the formula Q = mcΔT, where Q is the heat energy, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. The specific heat capacity of ice is approximately 2.09 J/(g·°C) and that of water is 4.18 J/(g·°C). The heat of fusion for ice is 334 J/g. Thus, the total heat Q needed involves three parts: warming the ice to 0 °C, melting the ice, and then warming the water to 15.7 °C.
The equation to calculate the mass of ice that can be melted and warmed by Q heat energy is Q = (mcΔT)ice + (mLf) + (mcΔT)water. Given the total available heat, and knowing c and ΔT for both the ice and water, along with Lf, we can re-arrange this equation to solve for mass m.