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 boil 72.0 grams of water at 100°C?

Question

Quizplus · Accepted Answer

First, we need to calculate the amount of energy needed to vaporize 72.0 grams of water:
moles of water = mass/molar mass = 72.0 g / 18.015 g/mol = 3.996 mol
energy needed = moles of water x vaporization enthalpy = 3.996 mol x 40.7 kJ/mol = 163 kJ
Next, we need to calculate the amount of energy needed to raise the temperature of 72.0 grams of water from 0°C to 100°C:
energy needed = mass x specific heat capacity x change in temperature
= 72.0 g x 4.184 J/(g°C) x 100°C = 30.0 kJ
Finally, we add the two amounts of energy:
total energy needed = 163 kJ + 30.0 kJ = 193 kJ, which is closest to choice D (163 kJ).

Water Has a Molar Heat Capacity of 75