Water has a molar heat capacity of 75.38 J/(mol.°C)and its vaporization enthalpy at 100°C is 40.7 kJ/mol.How much energy is needed to convert 124 g liquid H₂O at 20.0°C to steam at 100°C?

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The correct answer is A.Explanation:The energy needed to convert 124 g of liquid H2O at 20.0°C to steam at 100°C can be calculated as follows:1. Calculate the moles of H2O:n = m/M = 124 g / 18.015 g/mol = 6.88 mol2. Calculate the energy needed to raise the temperature of the liquid water from 20.0°C to 100°C:Q1 = n * Cp * ΔT = 6.88 mol * 75.38 J/(mol °C) * (100°C - 20°C) = 39.6 kJ3. Calculate the energy needed to vaporize the water at 100°C:Q2 = n * ΔHvap = 6.88 mol * 40.7 kJ/mol = 282 kJ4. The total energy needed is the sum of Q1 and Q2:Qtotal = Q1 + Q2 = 39.6 kJ + 282 kJ = 322 kJTherefore, the energy needed to convert 124 g of liquid H2O at 20.0°C to steam at 100°C is 322 kJ.

Water Has a Molar Heat Capacity of 75