Final answer:
To convert 17.0 g of water at 29.0 °C to steam at 100.0°C, one must first calculate the energy required to heat the water to boiling point and then the energy for the phase change to vapor. The energy needed to heat the water is 5012.28 J, and for vaporization, it is 38420 J, totalling 43432.28 J.
Step-by-step explanation:
Calculating the Energy Required to Convert Water to Steam
To calculate the amount of energy needed to convert 17.0 g of water at 29.0 °C to steam at 100.0°C, we need to account for the energy required to heat the water to its boiling point and the energy required for the phase change from liquid to gas. The specific heat of water is 4.184 J/g°C, which is the amount of energy needed to raise the temperature of 1 gram of water by 1°C.
First, calculate the energy required to heat the water from 29.0 °C to 100.0 °C:
- Energy to heat water (E) = mass (m) × specific heat (C) × change in temperature (ΔT)
- E = 17.0 g × 4.184 J/g°C × (100.0°C - 29.0°C)
- E = 17.0 g × 4.184 J/g°C × 71.0°C
- E = 17.0 g × 4.184 J/g°C × 71.0°C
- E = 5012.28 J
Next, to convert the water at 100.0 °C to steam, we use the latent heat of vaporization, which for water is 2260 J/g. The energy (E) for the phase change is:
- E = mass (m) × latent heat of vaporization (L)
- E = 17.0 g × 2260 J/g
- E = 38420 J
Therefore, the total energy required is the sum of the energy to heat the water to boiling point and the energy for the phase change:
- Total energy = energy to heat water + energy for phase change
- Total energy = 5012.28 J + 38420 J
- Total energy = 43432.28 J
The total energy required to convert 17.0 g of water at 29.0 °C to steam at 100.0°C is 43432.28 J.