Problem 21 - Chemistry and Chemical Reactivity

Question

The freezing point of mercury is \(-38.8^{\circ} \mathrm{C} .\) What quantity of energy, in joules, is released to the surroundings if \(1.00 \mathrm{mL}\) of mercury is cooled from \(23.0^{\circ} \mathrm{C}\) to \(-38.8^{\circ} \mathrm{C}\) and then frozen to a solid? (The density of liquid mercury is \(13.6 \mathrm{g} / \mathrm{cm}^{3}\). Its specific heat capacity is 0.140 J/g \cdot K and its heat of fusion is \(11.4 \mathrm{J} / \mathrm{g} .\) )

Step-by-Step Solution

Answer

273.13 J of energy is released.

Step 1: Calculate Mass of Mercury

First, convert the volume of mercury to grams using its density. The volume is given as \(1.00 \text{ mL}\), which is the same as \(1.00 \text{ cm}^3\). Using the density formula \( \text{density} = \frac{\text{mass}}{\text{volume}} \), rearrange to find the mass: \(\text{mass} = \text{density} \times \text{volume}\). With \(\text{density} = 13.6 \text{ g/cm}^3\), the mass is \(13.6 \text{ g}\).

Step 2: Calculate Energy to Cool Mercury

Use the specific heat capacity to calculate the energy needed to cool mercury from \(23.0^{\circ} \text{C}\) to \(-38.8^{\circ} \text{C}\). The formula to use is \(q = m \cdot c \cdot \Delta T\), where \(q\) is the energy, \(m\) is the mass, \(c\) is the specific heat capacity, and \(\Delta T\) is the change in temperature. Here, \(c = 0.140 \text{ J/g} \cdot \text{K}\), \(\Delta T = -38.8 - 23.0 = -61.8\) K. Therefore, \( q = 13.6 \times 0.140 \times -61.8 = -118.0928 \text{ J}\).

Step 3: Calculate Energy for Phase Change

Now calculate the energy needed to freeze mercury using its heat of fusion. Use the formula \( q = m \cdot \Delta H_f\), where \(m\) is the mass and \(\Delta H_f\) is the heat of fusion (\(11.4 \text{ J/g}\)). Thus, \(q = 13.6 \times 11.4 = 155.04 \text{ J}\). Note that this energy will be released as the mercury freezes.

Step 4: Calculate Total Energy Released

Add the energy released from cooling the mercury and the energy released from the phase change to find the total energy released. Both quantities are positive when considered in terms of energy released to the surroundings:\[ q_{\text{total}} = |q_{\text{cooling}}| + |q_{\text{freezing}}| = 118.0928 \text{ J} + 155.04 \text{ J} = 273.1328 \text{ J} \].

Key Concepts

Density of Mercury

To find the mass of mercury from its volume, we start off with density, a fundamental concept in physics and chemistry. Density (\( \rho \)) is the mass of a substance per unit volume and is expressed in units like g/cm³. It tells us how much matter is squeezed into a given space. To calculate the mass of mercury, you use the formula \( \text{mass} = \text{density} \times \text{volume} \). For mercury, with a density of 13.6 g/cm³ and a volume of 1 mL (which is the same as 1 cm³), this calculation gives us:

• Mass of Mercury = 13.6 g.

Understanding density helps not just with calculations like these, but also in comprehending how substances behave under different conditions, which can be crucial in advanced science topics.

Specific Heat Capacity of Mercury

The specific heat capacity is the amount of energy needed to raise the temperature of 1 gram of a substance by 1°C (or 1 K). For mercury, this value is 0.140 J/g·K, which indicates it doesn't take much energy to change its temperature compared to substances with a higher specific heat capacity. To calculate the energy released when cooling mercury, we'd use the formula:

• \( q = m \cdot c \cdot \Delta T \)

In this scenario, \( m \) is the mass (13.6 g), \( c \) is the specific heat capacity (0.140 J/g·K), and \( \Delta T \) is the temperature change (-61.8 K, obtained by subtracting the initial and final temperatures: 23.0 - (-38.8)).Through this calculation:

• Energy required to cool mercury: \(-118.0928 \text{ J}\).

The concept of specific heat capacity helps in understanding how different materials respond to changes in heat, which is critical in areas such as thermodynamics.

Heat of Fusion of Mercury

Heat of fusion is the amount of energy needed to change 1 gram of a substance from solid to liquid or vice versa at constant temperature. For mercury, the heat of fusion is 11.4 J/g. This implies that when 1 gram of mercury freezes, it releases 11.4 J of energy.To determine how much energy is released during the phase change:

• \( q = m \cdot \Delta H_f \)

With \( m \) as 13.6 g and \( \Delta H_f \) as 11.4 J/g, the energy released in this phase change is:

• \( 155.04 \text{ J} \).

By understanding this concept, you grasp why energy is released or absorbed during phase changes, important for fields like materials science. Combining these calculations, the total energy released when cooling and freezing mercury is the sum of energies from cooling and freezing, giving us a total release of 273.1328 J.