A 100.0-mL sample of water is heated to its boiling point. How much heat (in kJ) is required to vaporize it? (Assume a density of 1.00 g/mL.)

Verified step by step guidance

1. Calculate the mass of the water sample using the given density: \( \text{mass} = \text{volume} \times \text{density} \).
2. Use the mass of water to find the number of moles: \( \text{moles} = \frac{\text{mass}}{\text{molar mass of water}} \).
3. Determine the heat of vaporization for water, which is typically given as \( 40.7 \text{ kJ/mol} \).
4. Calculate the total heat required to vaporize the water using the formula: \( \text{heat} = \text{moles} \times \text{heat of vaporization} \).
5. Express the final answer in kilojoules (kJ).

Key Concepts

Heat of Vaporization

The heat of vaporization is the amount of energy required to convert a unit mass of a liquid into vapor without a change in temperature. For water, this value is approximately 40.79 kJ/mol at its boiling point. Understanding this concept is crucial for calculating the total heat needed to vaporize a given mass of water.

Mass and Volume Relationship

The relationship between mass and volume is defined by the density of a substance. In this case, water has a density of 1.00 g/mL, meaning that 100.0 mL of water has a mass of 100.0 grams. This conversion from volume to mass is essential for determining how much water is being vaporized in the heat calculation.

Specific Heat Capacity

Specific heat capacity is the amount of heat required to raise the temperature of a unit mass of a substance by one degree Celsius. While this concept is not directly used in the vaporization calculation, it is important to understand the energy changes involved in heating water to its boiling point before vaporization occurs.