How much heat is absorbed when 52.3 g H2O(l) at 100°C and 101.3 kPa is converted to steam at 100°C? (The molar heat of vaporization of water is 40.7 kJ/mol.)

Answer 1

To find the answer to this question, we can use the following equation:

heat absorbed = (mass of water)(heat of vaporization of water)
heat absorbed = (52.3 g)(2260 J/g)
heat absorbed = 118198 J
heat absorbed = 118.198 kJ or 1.18 × 10^2 kJ
Answer: About 1.18 × 10^2 kJ of heat is absorbed when 52.3 g of H2O at 100°C and 101.3 kPa is converted to steam at 100°C. The answer is choice "B".