Answer:
To calculate the heat change for the water, we need to know how much energy is required to raise the temperature of the water from 27.2°C to 0°C, how much energy is required to melt the ice, how much energy is required to raise the temperature of the resulting liquid water from 0°C to 100°C, how much energy is required to vaporize the water, and how much energy is required to raise the temperature of the resulting water vapor from 100°C to 74.36°C. The total heat change is the sum of these five contributions.
The energy required to raise the temperature of the water from 27.2°C to 0°C is:
Q1 = mcΔT = (395 g)(4.18 J/(g·°C))(27.2°C - 0°C) = 46,324.4 J
The energy required to melt the ice is:
Q2 = mL = (395 g)(333.55 J/g) = 131,977.25 J
The energy required to raise the temperature of the resulting liquid water from 0°C to 100°C is:
Q3 = mcΔT = (395 g)(4.18 J/(g·°C))(100°C - 0°C) = 165,371 J
The energy required to vaporize the water is:
Q4 = mL = (395 g)(2257 J/g) = 892,615 J
The energy required to raise the temperature of the resulting water vapor from 100°C to 74.36°C is:
Q5 = mcΔT = (395 g)(2.08 J/(g·°C))(74.36°C - 100°C) = -34,439.76 J
Note that the negative sign in front of Q5 indicates that the water vapor is losing heat.
Therefore, the total heat change for the water is:
Q = Q1 + Q2 + Q3 + Q4 + Q5 = 1,181,849.89 J
Rounding to three decimal places, the answer is 1,181,849.890 J.