Using the values for the heat of fusion, specific heat of water, and/or heat of vaporization, calculate the amount of heat energy in each of the following:c. kilojoules needed to melt 24.0 g of ice at 0 °C, warm the liquid to 100 °C, and change it to steam at 100 °C

Verified step by step guidance

1. Identify the three stages of the process: melting ice, heating water, and vaporizing water.
2. Calculate the heat required to melt 24.0 g of ice at 0 °C using the heat of fusion. Use the formula: \( q = m \times \Delta H_{fus} \), where \( m \) is the mass and \( \Delta H_{fus} \) is the heat of fusion.
3. Calculate the heat required to warm the liquid water from 0 °C to 100 °C using the specific heat capacity of water. Use the formula: \( q = m \times c \times \Delta T \), where \( c \) is the specific heat capacity and \( \Delta T \) is the temperature change.
4. Calculate the heat required to vaporize the water at 100 °C using the heat of vaporization. Use the formula: \( q = m \times \Delta H_{vap} \), where \( \Delta H_{vap} \) is the heat of vaporization.
5. Add the heat values from each step to find the total heat energy required for the entire process.

Key Concepts

Heat of Fusion

The heat of fusion is the amount of energy required to change a substance from solid to liquid at its melting point without changing its temperature. For water, this value is approximately 334 joules per gram. In the context of the question, it is essential for calculating the energy needed to melt ice into liquid water.

Specific Heat Capacity

Specific heat capacity is the amount of heat energy required to raise the temperature of one gram of a substance by one degree Celsius. For water, this value is about 4.18 joules per gram per degree Celsius. This concept is crucial for determining the energy needed to warm the liquid water from 0 °C to 100 °C in the given problem.

Heat of Vaporization

The heat of vaporization is the amount of energy needed to convert a unit mass of a liquid into vapor at its boiling point without changing its temperature. For water, this value is approximately 2260 joules per gram. This concept is necessary for calculating the energy required to convert the liquid water at 100 °C into steam.