To calculate the amount of heat required to convert a given mass of water from a certain initial temperature to steam at 100°C, we need to consider two main steps: heating the water from 20°C to its boiling point (100°C) and then converting it from liquid to steam at the same temperature. We'll use the specific heat capacity and latent heat of vaporization for water in these calculations.
- Heating the water from 20°C to 100°C: The specific heat capacity of water is approximately 4.18 J/g°C. To raise the temperature of 600g of water by 80°C (from 20°C to 100°C), we can use the formula:
Q1 = m * c * ΔT
Where: Q1 is the heat energy required, m is the mass of water (600g), c is the specific heat capacity of water (4.18 J/g°C), and ΔT is the change in temperature (100°C - 20°C = 80°C).
Q1 = 600g * 4.18 J/g°C * 80°C Q1 ≈ 200,640 J (or 200.64 kJ)
- Converting the water to steam at 100°C: The latent heat of vaporization for water is approximately 2260 J/g. To convert 600g of water into steam at 100°C, we can use the formula:
Q2 = m * Lv
Where: Q2 is the heat energy required for vaporization, m is the mass of water (600g), and Lv is the latent heat of vaporization (2260 J/g).
Q2 = 600g * 2260 J/g Q2 ≈ 1,356,000 J (or 1,356 kJ)
Therefore, the total heat energy required to convert 600g of water from 20°C to steam at 100°C is the sum of Q1 and Q2:
Total heat energy = Q1 + Q2 Total heat energy ≈ 200,640 J + 1,356,000 J Total heat energy ≈ 1,556,640 J (or 1,556.64 kJ)