The process of changing 100.0 grams of liquid water at 15.0 °C to vapor at 125.0 °C involves several steps, and we need to calculate the energy required for each step and then add them up:
1. Heating the liquid water from 15.0 °C to 100.0 °C:
q = m * Cp * ΔT
= 100.0 g * 4.184 J/g °C * (100.0 °C - 15.0 °C)
= 34,972 J
2. Vaporizing the liquid water at 100.0 °C:
q = m * Hvap
= 100.0 g * 2,256 J/g
= 225,600 J
3. Heating the water vapor from 100.0 °C to 125.0 °C:
q = m * Cp * ΔT
= 100.0 g * 1.84 J/g °C * (125.0 °C - 100.0 °C)
= 4,600 J
The total energy required is the sum of the three steps:
Q = q1 + q2 + q3
= 34,972 J + 225,600 J + 4,600 J
= 265,172 J
Therefore, the energy needed to change 100.0 grams of liquid water at 15.0 °C to vapor at 125.0 °C is approximately 265,172 J, which is closest to option (d) 266,000 J.