The human body obtains 915 kJ of energy from a candy bar. If this energy were used to vaporize water at 100.0 °C, how much water (in liters) could be vaporized? (Assume the density of water is 1.00 g/mL.)

Verified step by step guidance

1. Identify the energy required to vaporize water, which is the heat of vaporization (\(\Delta H_{vap}\)) for water at 100.0 °C. This value is typically 40.7 kJ/mol.
2. Convert the energy obtained from the candy bar from kilojoules to joules. Since 1 kJ = 1000 J, multiply 915 kJ by 1000 to get the energy in joules.
3. Calculate the number of moles of water that can be vaporized using the energy from the candy bar. Use the formula: \(\text{moles of water} = \frac{\text{energy in joules}}{\Delta H_{vap} \times 1000}\), where \(\Delta H_{vap}\) is in kJ/mol.
4. Convert the moles of water to grams using the molar mass of water (18.02 g/mol). Use the formula: \(\text{mass of water (g)} = \text{moles of water} \times 18.02\, \text{g/mol}\).
5. Convert the mass of water in grams to volume in liters. Since the density of water is 1.00 g/mL, the volume in mL is equal to the mass in grams. Convert mL to liters by dividing by 1000.