Question

You have a \(1.00\) -mol sample of water at \(-30 .{ }^{\circ} \mathrm{C}\) and you heat it until you have gaseous water at \(140 .^{\circ} \mathrm{C}\). Calculate \(q\) for the entire process. Use the following data. Specific heat capacity of ice \(=2.03 \mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{g}\) Specific heat capacity of water \(=4.18 \mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{g}\) Specific heat capacity of steam \(=2.02 \mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{g}\) \(\begin{array}{lr}\mathrm{H}_{2} \mathrm{O}(s) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l) & \Delta H_{\text {fusion }}=6.02 \mathrm{~kJ} / \mathrm{mol}\left(\text { at } 0^{\circ} \mathrm{C}\right) \\\ \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{2} \mathrm{O}(g) & \Delta H_{\text {vaporization }}=40.7 \mathrm{~kJ} / \mathrm{mol}\left(\text { at } 100 .{ }^{\circ} \mathrm{C}\right)\end{array}\)

Short answer

The heat (q) required for the entire process of heating 1.00 mol of water from -30°C to gaseous water at 140°C is approximately 56794 J.

Step-by-step solution

Heating the ice from -30°C to 0°C

First, we need to calculate the heat required to raise the temperature of the ice from -30°C to 0°C using the specific heat capacity of ice (2.03 J/°C·g). We will use the formula: q = mcΔT where: m = mass (g) c = specific heat capacity (J/°C·g) ΔT = change in temperature (°C) The molar mass of water is 18.015 g/mol, so for a 1.00 mol sample, the mass is: m = 1.00 mol * 18.015 g/mol = 18.015 g Now, calculate the heat (q) required for this step: q1 = (18.015 g)(2.03 J/°C·g)(0°C - (-30°C)) q1 = 1091.56 J

Melting the ice at 0°C (fusion process)

Next, we will calculate the heat required for the fusion process using the enthalpy of fusion value (6.02 kJ/mol) for 1 mol of ice. q2 = 1.00 mol * 6.02 kJ/mol * (1000 J/1 kJ) q2 = 6020 J

Heating the water from 0°C to 100°C

Now, we need to calculate the heat required to raise the temperature of the liquid water from 0°C to 100°C using the specific heat capacity of water (4.18 J/°C·g). q3 = (18.015 g)(4.18 J/°C·g)(100°C - 0°C) q3 = 7531.26 J

Vaporizing the water at 100°C (vaporization process)

Next, we calculate the heat required for the vaporization process using the enthalpy of vaporization value (40.7 kJ/mol) for 1 mol of water. q4 = 1.00 mol * 40.7 kJ/mol * (1000 J/1 kJ) q4 = 40700 J

Heating the steam from 100°C to 140°C

Finally, we need to calculate the heat required to raise the temperature of the steam from 100°C to 140°C using the specific heat capacity of steam (2.02 J/°C·g). q5 = (18.015 g)(2.02 J/°C·g)(140°C - 100°C) q5 = 1450.79 J

Calculating the total heat (q) for the entire process

Now, we can sum up the heat calculated for each step to get the total heat for the entire process: q_total = q1 + q2 + q3 + q4 + q5 q_total = 1091.56 J + 6020 J + 7531.26 J + 40700 J + 1450.79 J q_total ≈ 56794 J So, the heat (q) required for the entire process of heating 1.00 mol of water from -30°C to gaseous water at 140°C is approximately 56794 J.

Key concepts

Specific Heat Capacity

When we talk about specific heat capacity, we're essentially discussing a material's ability to absorb heat. The specific heat capacity (\( c \)) measures how much energy is needed to raise the temperature of 1 gram of a substance by 1 degree Celsius. For water, this number is quite high compared to many other substances, which means it can absorb a lot of heat without drastic changes in temperature.
Understanding the specific heat capacity is crucial when calculating the heat (\( q \)) involved in temperature changes of a substance. The formula used is \( q = mc\Delta T \), where \( m \) is mass, \( c \) is specific heat capacity, and \( \Delta T \) is the change in temperature.
For this exercise, we focus on specific heat capacities of ice, liquid water, and steam:

• Ice: 2.03 J/°C·g
• Water: 4.18 J/°C·g
• Steam: 2.02 J/°C·g

These values help you calculate how much heat is needed to raise the temperature of water through its different states as it goes from ice to steam.

Enthalpy of Fusion

Enthalpy of fusion is a measure of the energy required to change a substance from a solid to a liquid at its melting point. For water, this process happens at \(0^{\circ} C\), and it's when ice melts into liquid water. The energy required for this transition is known as the enthalpy of fusion.
The formula used in this process is \( q = n \Delta H_{fusion} \), where \( n \) is the number of moles and \( \Delta H_{fusion} \) is the enthalpy of fusion. For water, the enthalpy of fusion is 6.02 kJ/mol.
This value means that you need 6.02 kJ of energy to convert one mole of ice at \(0^{\circ} C\) into liquid water, and is a crucial part of understanding the amount of energy involved in phase changes.

Enthalpy of Vaporization

The enthalpy of vaporization refers to the energy required to turn a liquid into a gas at its boiling point. For water, the boiling point is \(100^{\circ} C\), and transitioning from liquid water to steam requires a significant amount of energy. This is quantified by the enthalpy of vaporization.
Using the formula \( q = n \Delta H_{vaporization} \) helps calculate the energy for this phase transition, similar to how we calculate for fusion. For water, the enthalpy of vaporization is 40.7 kJ/mol.
This value indicates that 40.7 kJ of energy is needed to vaporize one mole of water at \(100^{\circ} C\). The greater energy requirement compared to fusion highlights how much more energy is needed to overcome the forces keeping molecules in a liquid state.

Phase Change

Phase changes are those transitions from one state of matter to another, like going from solid ice to liquid water, or from liquid water to steam. They involve energy changes because they rearrange the structure and movement of atoms or molecules without changing temperature.
In our exercise, phase changes occur twice:

• From ice to water (melting): This occurs at \(0^{\circ} C\) using the enthalpy of fusion.
• From water to steam (boiling): Happening at \(100^{\circ} C\) involving the enthalpy of vaporization.

Each phase change requires energy to overcome intermolecular forces without changing temperature. These concepts are crucial for calculating the total energy (\( q \)) required across different states and temperatures as you heat substances through their various phases.