Question

Given the following data for water:

Heat of fusion = 334 J/g
Heat of vaporization = 2,256 J/g
Specific heat of solid = 2.09 J/g °C)
Specific heat of liquid = 4.184 J/g °C)
Specific heat of gas = 1.84 J/g °C)
Calculate how much energy is needed to change 100.0 grams of liquid water at 15.0 °C to vapor at 125.0 °C. (3 points)
Oa
O
b
44,000 J
89,400 J
104,000 J
266,000 J

Answer 1

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.