Problem 99 - Chemistry An Atoms First Approach

Question

What quantity of energy does it take to convert 0.500 kg ice at \(-20 .^{\circ} \mathrm{C}\) to steam at \(250 .^{\circ} \mathrm{C} ?\) Specific heat capacities: ice, \(2.03 \mathrm{J} / \mathrm{g} \cdot^{\circ} \mathrm{C} ;\) liquid, \(4.2 \mathrm{J} / \mathrm{g} \cdot^{\circ} \mathrm{C} ;\) steam, \(2.0 \mathrm{J} / \mathrm{g} \cdot^{\circ} \mathrm{C} ; \Delta H_{\mathrm{vap}}=\) \(40.7 \mathrm{kJ} / \mathrm{mol} ; \Delta H_{\mathrm{fus}}=6.02 \mathrm{kJ} / \mathrm{mol}.\)

Step-by-Step Solution

Answer

It takes \(1,744,560 \, J\) of energy to convert \(0.500 \, kg\) of ice at \(-20°C\) to steam at \(250°C\).

Step 1: Convert mass to grams and calculate the number of moles of water in the sample

Since the specific heat capacities and enthalpies are given in terms of grams and moles, we need to convert the given mass of ice from kilograms to grams and calculate the number of moles: \(m = 0.500 \, kg = 500 \, g\) To calculate the number of moles, we will use the molar mass of water, which is 18 g/mol: \(n = \frac{500 \, g}{18 \, g/mol} = 27.78 \, mol\)

Step 2: Calculate the energy required to heat the ice from -20°C to 0°C

Using the specific heat capacity of ice and the given mass and temperature difference, we can calculate the energy required: \(Q_1 = mcΔT\) \(Q_1 = (500 \, g)(2.03 \, J/g°C)(0°C - (-20°C)) = 20300 \, J\)

Step 3: Calculate the energy required to convert the ice at 0°C to water

Using the enthalpy of fusion and the number of moles, we can now calculate the energy needed to turn the ice into water: \(Q_2 = nΔH_{fus}\) \(Q_2 = (27.78 \, mol)(6.02 \, kJ/mol) = 167.20 \, kJ = 167200 \, J\)

Step 4: Calculate the energy required to heat the water from 0°C to 100°C

Now we need to find the energy needed to raise the temperature of the water to its boiling point: \(Q_3 = mcΔT\) \(Q_3 = (500 \, g)(4.2 \, J/g°C)(100°C - 0°C) = 210000 \, J\)

Step 5: Calculate the energy required to convert the water at 100°C to steam

Using the enthalpy of vaporization and the number of moles, calculate the energy needed to turn the water into steam: \(Q_4 = nΔH_{vap}\) \(Q_4 = (27.78 \, mol)(40.7 \, kJ/mol) = 1131.06 \, kJ = 1131060 \, J\)

Step 6: Calculate the energy required to heat the steam from 100°C to 250°C

Finally, we need to find the energy needed to raise the temperature of the steam to 250°C: \(Q_5 = mcΔT\) \(Q_5 = (500 \, g)(2.0 \, J/g°C)(250°C - 100°C) = 150000 \, J\)

Step 7: Calculate the total energy required

Now, simply add up all the energy values calculated in the previous steps: \(Q_{total} = Q_1 + Q_2 + Q_3 + Q_4 + Q_5\) \(Q_{total} = 20300 \, J + 167200 \, J + 210000 \, J + 1131060 \, J + 150000 \, J\) \(Q_{total} = 1744560 \, J\) Thus, it takes 1,744,560 J of energy to convert 0.500 kg of ice at -20°C to steam at 250°C.

Key Concepts

Specific Heat Capacity

Understanding specific heat capacity is crucial when calculating energy changes in substances. It refers to the amount of energy, usually in joules, needed to raise the temperature of 1 gram of a substance by 1 degree Celsius. Each state of matter—solid, liquid, and gas—has its own specific heat capacity, reflecting how much energy is required for changing temperature within that state.
For the problem at hand, the specific heat capacities are:

• Ice: 2.03 J/g°C
• Liquid water: 4.2 J/g°C
• Steam: 2.0 J/g°C

These values represent how readily each state of water can absorb energy. A higher specific heat capacity, like that of liquid water, means more energy is needed to change its temperature. This concept helps explain why it takes more energy to heat water than steam or ice when each is at their respective states.

Enthalpy of Fusion

Enthalpy of fusion is the energy required to change a substance from a solid to a liquid state at its melting point, without changing its temperature. This transition is crucial when ice melts into liquid water.Using the enthalpy of fusion, denoted as \(\Delta H_{fus}\),we can calculate the energy required:

• \(\Delta H_{fus} = 6.02 \mathrm{kJ/mol}\)

This value signifies the energy needed per mole to overcome the forces holding the solid structure together and form a liquid. In our problem, we multiply the enthalpy of fusion by the number of moles of ice, calculated from its mass and the molar mass of water (18 g/mol), to find the total energy needed to completely melt the ice. This concept explains why it takes energy without a temperature change at the melting point.

Enthalpy of Vaporization

When a substance transitions from a liquid to a gaseous state, it undergoes vaporization. The enthalpy of vaporization, \(\Delta H_{vap}\),quantifies the energy involved in this phase change. This process is essential in converting liquid water at its boiling point into steam, also without a change in temperature.For water:

• \(\Delta H_{vap} = 40.7 \mathrm{kJ/mol}\)

By multiplying this value by the number of moles, you can determine the energy required to vaporize the entire liquid sample. This phase change requires more energy than melting, due to the greater separation of molecules needed to form gas from liquid. Knowing the enthalpy of vaporization allows us to accurately calculate the energy to fully transform water into steam, emphasizing the concept of energy exchanges during state changes.

Phase Transitions

Phase transitions refer to the changes a substance undergoes between different states of matter: solid, liquid, and gas. Each transition involves specific energy changes without an accompanying temperature change at the transition point.
Key transitions are:

• Melting (solid to liquid)
• Freezing (liquid to solid)
• Vaporization (liquid to gas)
• Condensation (gas to liquid)
For example, in the discussed problem, several phase transitions are considered:
• Ice melting to water involves the enthalpy of fusion.
• Water boiling into steam involves the enthalpy of vaporization.
These transitions illustrate the energy demands for state changes without altering temperature during the phase shift itself. Understanding these transitions is fundamental to mastering energy calculations in chemistry and thermodynamics.