Final answer:
To warm 15.0 g of water from 75.0 °C to 100.0 °C and convert it to steam, the total energy required is 35.4738 kilojoules. This includes both the energy to heat the water to 100.0 °C and the latent heat of vaporization to convert the water into steam.
Step-by-step explanation:
To calculate the total energy 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 account for two steps: heating the liquid water and converting it to steam (the phase change).
Firstly, we use the specific heat capacity of water (4.184 J/g°C) to calculate the heat needed to raise the temperature from 75.0 °C to 100.0 °C:
q = mcΔT
where:
- q is the heat energy in Joules
- m is the mass of the water in grams
- c is the specific heat capacity in J/g°C
- ΔT is the change in temperature in °C
For the heating step:
q = (15.0 g)(4.184 J/g°C)(100.0 °C - 75.0 °C) = 1573.8 J
Secondly, we calculate the energy required for the phase change from liquid to steam using the latent heat of vaporization for water (2260 J/g):
q = mL
where:
- q is the heat energy in Joules
- m is the mass of the water in grams
- L is the latent heat of vaporization in J/g
For the phase change:
q = (15.0 g)(2260 J/g) = 33900 J
To find the total energy in kilojoules, we add both amounts and convert from Joules to kilojoules (1 kJ = 1000 J):
Total energy = (1573.8 J + 33900 J) / 1000 = 35.4738 kJ
Thus, the total energy required is 35.4738 kilojoules.