Problem 76 - Chemistry and Chemical Reactivity

Question

Calculate the quantity of energy required to convert \(60.1 \mathrm{g}\) of \(\mathrm{H}_{2} \mathrm{O}(\mathrm{s})\) at \(0.0^{\circ} \mathrm{C}\) to \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) at \(100.0^{\circ} \mathrm{C} .\) The enthalpy of fusion of ice at \(0^{\circ} \mathrm{C}\) is \(333 \mathrm{J} / \mathrm{g}\) i the enthalpy of vaporization of liquid water at \(100^{\circ} \mathrm{C}\) is \(2256 \mathrm{J} / \mathrm{g}.\)

Step-by-Step Solution

Answer

The energy required is 180904.7 J.

Step 1: Calculate Energy for Melting Ice

To calculate the energy required to melt ice, use the formula: \[ q_1 = m \times \Delta H_{\text{fusion}} \]where \( m = 60.1 \text{ g} \) is the mass of the ice and \( \Delta H_{\text{fusion}} = 333 \text{ J/g} \) is the enthalpy of fusion. Substituting in the values gives:\[ q_1 = 60.1 \times 333 = 20013.3 \text{ J} \] This is the energy required to convert ice at \(0^{\circ}C\) to water at \(0^{\circ}C\).

Step 2: Calculate Energy to Heat Water

Next, calculate the energy required to heat the water from \(0^{\circ}C\) to \(100^{\circ}C\) using the formula:\[ q_2 = m \times c \times \Delta T \]where \( c = 4.18 \text{ J/g°C} \) is the specific heat of water, and \( \Delta T = 100^{\circ}C - 0^{\circ}C = 100^{\circ}C \). Substituting the values gives:\[ q_2 = 60.1 \times 4.18 \times 100 = 25185.8 \text{ J} \] This is the energy required to heat water from \(0^{\circ}C\) to \(100^{\circ}C\).

Step 3: Calculate Energy for Vaporization

Now, calculate the energy required to vaporize the water at \(100^{\circ}C\), using:\[ q_3 = m \times \Delta H_{\text{vaporization}} \]where \( \Delta H_{\text{vaporization}} = 2256 \text{ J/g} \). Substituting the values gives:\[ q_3 = 60.1 \times 2256 = 135705.6 \text{ J} \]This is the energy required to convert water at \(100^{\circ}C\) to steam at \(100^{\circ}C\).

Step 4: Calculate Total Energy Required

The total energy required is the sum of the energies calculated in Steps 1, 2, and 3:\[ \text{Total Energy} = q_1 + q_2 + q_3 = 20013.3 + 25185.8 + 135705.6 = 180904.7 \text{ J} \]Hence, the total amount of energy required to convert the ice to steam is \( 180904.7 \text{ J} \).

Key Concepts

Understanding Enthalpy of Fusion

Enthalpy of fusion is an important concept when studying phase changes, particularly the transition of a substance from solid to liquid. It is the amount of energy needed to change 1 gram of a solid into a liquid at its melting point without changing temperature. This energy is absorbed by the substance, causing the molecules to break free from their solid lattice structure, which is why it's an endothermic process. In our example, we have water in its solid form, commonly known as ice. To melt 60.1 grams of ice at 0°C, we calculated the energy needed using the enthalpy of fusion for water, which is 333 J/g. The formula utilized here is:\[ q_1 = m \times \Delta H_{\text{fusion}} \]where \( m \) is the mass of the ice and \( \Delta H_{\text{fusion}} \) is the enthalpy of fusion. By substituting, we discover that 20013.3 Joules are necessary to turn the ice into liquid water at the same temperature.

Exploring Enthalpy of Vaporization

The enthalpy of vaporization is a critical factor in understanding the energy required for converting a liquid into a gas at its boiling point. This transformation, like fusion, involves an endothermic process where energy is absorbed by the substance, allowing molecules to escape into the gaseous phase. The energy that we calculated here is significant because water has a high enthalpy of vaporization, attributed to the strong hydrogen bonds between water molecules.To determine how much energy is needed to vaporize 60.1 grams of liquid water at 100°C, we apply:\[ q_3 = m \times \Delta H_{\text{vaporization}} \]where \( \Delta H_{\text{vaporization}} \) is 2256 J/g. Inserting the mass of water, we find that approximately 135705.6 Joules are required to convert the liquid water at boiling point into steam.

Understanding Specific Heat of Water

The specific heat of water is a unique property that quantifies how much energy it takes to raise the temperature of 1 gram of water by 1°C. Water has a specific heat of 4.18 J/g°C, which is relatively high compared to many other substances. This high specific heat capacity is due to the hydrogen bonding in water, which makes it a great of heat storage and a stabilizer of temperature in various environments.In our calculation, we used this property to find the amount of energy required to raise 60.1 grams of water from 0°C to 100°C. The formula used is:\[ q_2 = m \times c \times \Delta T \]Here, \( c \) stands for the specific heat, and \( \Delta T \) is the change in temperature. By applying these values, the energy calculated to heat the water was 25185.8 Joules. This understanding helps explain why large amounts of heat are necessary to warm water, making it vital in numerous natural and industrial processes.