Final answer:
To heat and vaporize 15.0 g of water from 75.0 °C to 100.0 °C, a total of 35.470 kilojoules are needed. This includes 1.570 kJ to increase the temperature to the boiling point and 33.9 kJ for the phase transition to steam.
Step-by-step explanation:
To answer how many kilojoules are required to warm 15.0 g of water from 75.0 °C to 100.0 °C and convert it to steam at 100.0 °C, we need to calculate two parts of heat energy: the energy needed to raise the temperature of the water to boiling point and the energy needed for the phase change from liquid to vapor (this is where the heat of vaporization comes into play).
To raise the temperature, we use the specific heat of water, which is 4.184 J/g °C. The calculation is:
Q1 = mass x specific heat x ΔT
= 15.0 g x 4.184 J/g °C x (100 °C – 75 °C)
= 15.0 g x 4.184 J/g °C x 25 °C
= 1,570 J or 1.570 kJ
Next, we calculate the energy required to convert the water at 100.0 °C to steam, using the heat of vaporization of water which is approximately 2260 J/g. The calculation is:
Q2 = mass x heat of vaporization
= 15.0 g x 2260 J
= 33,900 J or 33.9 kJ
Finally, the total energy required is the sum of both steps:
Total energy = Q1 + Q2
= 1.570 kJ + 33.9 kJ
= 35.470 kJ
Therefore, 35.470 kilojoules are required to warm 15.0 g of water from 75.0 °C to 100.0 °C and convert it to steam at 100.0 °C.