How much energy (in kJ) is required to convert a 22.0 g sample of liquid water at 25.0 °C to steam at 145.0 °C? (Given: specific heat of water = 4.18 J/g·°C, specific heat of steam = 2.03 J/g·°C, enthalpy of vaporization = 40.7 kJ/mol, molar mass of H2O = 18.0 g/mol)

• A. 78.9 kJ
• B. 45.7 kJ
• C. 52.4 kJ
• D. 63.2 kJ

Answer: (D)

Verified step by step guidance

1. Calculate the amount of heat required to raise the temperature of liquid water from 25.0 °C to 100.0 °C using the formula \(q = m \times C \times \Delta T\), where \(m\) is the mass of water, \(C\) is the specific heat capacity of liquid water, and \(\Delta T\) is the temperature change.
2. Determine the number of moles of water in the 22.0 g sample by using the molar mass of water: \(n = \frac{mass}{molar\ mass}\).
3. Calculate the heat required for the phase change from liquid water at 100.0 °C to steam at 100.0 °C using the enthalpy of vaporization: \(q = n \times \Delta H_{vap}\).
4. Calculate the heat required to raise the temperature of steam from 100.0 °C to 145.0 °C using the formula \(q = m \times C \times \Delta T\), where \(C\) is the specific heat capacity of steam and \(\Delta T\) is the temperature change.
5. Add all the heat quantities calculated in the previous steps to find the total energy required to convert the 22.0 g sample of liquid water at 25.0 °C to steam at 145.0 °C.