Answer:
5.9×10^6 J
Step-by-step explanation:
We can use the specific heat capacity and the heat of vaporization of water to solve this problem. The specific heat capacity of water is 4.18 J/g·°C and the heat of vaporization of water is 2.26×10^6 J/kg.
First, we need to calculate the amount of heat required to raise the temperature of the water from its initial temperature to its boiling point. This can be calculated using the formula:
Q1 = m × c × ΔT
where Q1 is the heat required, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the temperature change. Since we want to know the initial temperature, we can rearrange the formula to get:
ΔT = Q1 / (m × c)
Substituting the values, we get:
ΔT = Q1 / (m × c) = (4.91×10^5 J) / (2390 g × 4.18 J/g·°C) ≈ 49.14 °C
This means that the initial temperature of the water was 100°C - 49.14°C = 50.85 °C (since water boils at 100°C at standard pressure).
Next, we need to calculate the amount of heat required to vaporize the water at its boiling point. This can be calculated using the formula:
Q2 = m × Lv
where Q2 is the heat required, m is the mass of the water, and Lv is the heat of vaporization of water. Substituting the values, we get:
Q2 = m × Lv = (2390 g) × (2.26×10^6 J/kg) / 1000 = 5401.4 kJ
Finally, we can add up the heat required for temperature change and vaporization to get the total heat absorbed by the water:
Q = Q1 + Q2 = 4.91×10^5 J + 5401.4 kJ = 5.9×10^6 J
This is the total heat absorbed by the water during the heating process.