Final answer:
To convert 14.0 g of water at 32.0°C to steam at 100.0°C, the total energy required is calculated in two steps: heating the water to the boiling point and vaporizing it into steam. The sum of energies for these processes is 35627.36 joules.
Step-by-step explanation:
To calculate the amount of energy needed to convert 14.0 g of water at 32.0 °C to steam at 100.0°C, you need to consider two steps: heating the water from 32.0°C to 100.0°C, and then vaporizing the water into steam at 100.0°C.
First, use the specific heat capacity of water to calculate the energy required to raise the temperature of the water to the boiling point. The specific heat of water is 4.184 J/g°C. Therefore, the energy required to heat the water from 32.0°C to 100.0°C can be calculated with the equation:
E = m × C × ΔT
Where E is the energy, m is the mass of the water, C is the specific heat capacity, and ΔT is the change in temperature. For the first step:
E = 14.0 g × 4.184 J/g°C × (100.0°C - 32.0°C) = 3987.36 J
Next, you need to calculate the energy required to vaporize the water at 100.0°C. The heat of vaporization of water is typically around 2260 J/g.
E = m × ΔHvap
For the second step, the energy to vaporize water:
E = 14.0 g × 2260 J/g = 31640 J
The total energy required is the sum of the energy required to heat the water and the energy to vaporize it:
Total Energy = 3987.36 J + 31640 J = 35627.36 J
Therefore, 35627.36 joules of energy is needed to convert 14.0 g of water at 32.0°C to steam at 100.0°C.