Question

How much energy (in kilojoules) is absorbed when 10.0 g of liquid water at 75.0 °C is converted to water vapor at 150.0 °C? (The ΔH vap of water is 40.67 kJ/mol, and the molar heat capacity is 75.3 J/K·mol for the liquid and 33.6 J/K·mol for the vapor.)

A) 15.2 kJ
B) 18.6 kJ
C) 22.4 kJ
D) 27.8 kJ

Answer 1

The amount of energy absorbed when 10.0 g of liquid water at 75.0 °C is converted to water vapor at 150.0 °C is 122.8 kJ.

Step-by-step explanation:

To determine the amount of energy absorbed when 10.0 g of liquid water at 75.0 °C is converted to water vapor at 150.0 °C, we need to calculate the energy required to raise the temperature of the liquid water to its boiling point, then add the energy required to vaporize the water at its boiling point.

First, calculate the energy required to raise the temperature of the liquid water using the formula: Q = m * C * ΔT, where Q is the energy, m is the mass, C is the specific heat capacity, and ΔT is the temperature change. In this case, Q = 10.0 g * 75.3 J/K·mol * (100.0 °C - 75.0 °C) / 18.015 g/mol = 100.3 kJ.

Next, calculate the energy required to vaporize the water using the formula: Q = n * ΔHvap, where Q is the energy, n is the number of moles, and ΔHvap is the molar heat of vaporization. In this case, n = m / M, where M is the molar mass of water. n = 10.0 g / 18.015 g/mol = 0.554 mol. Q = 0.554 mol * 40.67 kJ/mol = 22.5 kJ.

The total energy absorbed is the sum of the two calculated energies: 100.3 kJ + 22.5 kJ = 122.8 kJ. Therefore, the correct answer is D) 122.8 kJ.