A sample of water with a mass of 100.0 g and a density of 1.00 g/mL is heated from 25.0 °C to its boiling point (100.0 °C) and then completely vaporized. How much heat (in kJ) is required for this process? (Specific heat of water = 4.18 J/g·°C; enthalpy of vaporization = 40.7 kJ/mol)

• A. 4.07 kJ
• B. 226 kJ
• C. 41.8 kJ
• D. 18.7 kJ

Answer: (B)

Verified step by step guidance

1. Identify the two stages of heating: first, heating the liquid water from 25.0 °C to 100.0 °C, and second, vaporizing the water at 100.0 °C.
2. Calculate the heat required to raise the temperature of the water using the formula: \(q_1 = m \times C \times \Delta T\), where \(m\) is the mass of water (100.0 g), \(C\) is the specific heat capacity (4.18 J/g·°C), and \(\Delta T\) is the temperature change (100.0 °C - 25.0 °C).
3. Convert the heat from joules to kilojoules by dividing the result by 1000, since 1 kJ = 1000 J.
4. Calculate the number of moles of water using the molar mass of water (approximately 18.015 g/mol) with the formula: \(n = \frac{m}{M}\), where \(m\) is the mass and \(M\) is the molar mass.
5. Calculate the heat required for vaporization using the formula: \(q_2 = n \times \Delta H_{vap}\), where \(n\) is the number of moles and \(\Delta H_{vap}\) is the enthalpy of vaporization (40.7 kJ/mol). Finally, add \(q_1\) and \(q_2\) to find the total heat required.