Problem 39 - Chemistry The Molecular Science

Question

The freezing point of mercury is \(-38.8{ }^{\circ} \mathrm{C}\). Calculate what quantity of energy, in joules, is released to the surroundings if \(1.00 \mathrm{~mL}\) 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 \(0.140 \mathrm{~J} \mathrm{~g}^{-1} \mathrm{~K}^{-1}\) and its heat of fusion is \(11.4 \mathrm{~J} \mathrm{~g}^{-1}\).)

Step-by-Step Solution

Answer

The total energy released is 272.8 J.

Step 1: Convert Volume to Mass

First, we need to find the mass of the mercury. Since 1.00 mL is equivalent to 1.00 cm³ and the density is 13.6 g/cm³, multiply the volume by the density to get the mass: \( \text{Mass} = 1.00 \text{ cm}^3 \times 13.6 \text{ g/cm}^3 = 13.6 \text{ g} \).

Step 2: Calculate Energy to Cool Mercury

Next, calculate the energy to cool the mercury to its freezing point. Use the specific heat formula: \( Q = mc\Delta T \), where \( m = 13.6 \text{ g} \), \( c = 0.140 \text{ J/g} \, ^{\circ}\text{C} \), and \( \Delta T = 23 - (-38.8) = 61.8 \underline{\phantom{xxx}}^{\circ}\text{C} \). Thus, \( Q = 13.6 \times 0.140 \times 61.8 \approx 117.8 \text{ J} \).

Step 3: Calculate Energy for Freezing

Now calculate the energy released when mercury freezes. Use the formula for heat of fusion: \( Q = m \times \text{Heat of Fusion} \), where \( m = 13.6 \text{ g} \) and \( \text{Heat of Fusion} = 11.4 \text{ J/g} \). Thus, \( Q = 13.6 \times 11.4 \approx 155.0 \text{ J} \).

Step 4: Total Energy Released

Add the energy values from cooling and freezing to find the total energy released: \( Q_{\text{total}} = 117.8 \text{ J} + 155.0 \text{ J} = 272.8 \text{ J} \).

Key Concepts

Freezing point

The freezing point is the temperature at which a liquid turns into a solid. It's a crucial concept in thermodynamics. For every substance, this point occurs at a specific temperature. For mercury, this temperature is \(-38.8 \,^{\circ}\mathrm{C}\). Understanding the freezing point helps determine the conditions under which a substance will change state. This temperature is solely dependent on the substance itself and not on the amount. In our example, mercury cools from \(23.0\,^{\circ}\mathrm{C}\) to its freezing point and then solidifies. Knowing this specific temperature allows us to calculate the exact energy released during the cooling process.

• The freezing point is consistent for a pure substance at a given pressure.
• It's a key factor in phase change calculations.
• Liquid to solid transition requires energy release at this specific temperature.

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. Conversely, when a substance freezes, it releases this energy. This is represented in the formula: \[ Q = m \times \text{Heat of Fusion} \]In this exercise, when mercury freezes, it releases energy. The heat of fusion for mercury is \(11.4 \, \mathrm{J/g}\), which indicates that \(11.4\) joules are released per gram of mercury as it turns into solid. This released energy is crucial for completing the transition from liquid to solid.

• Heat of fusion values are specific to each substance.
• It does not depend on the temperature difference, only on the substance and mass.
• Important for calculating energy in phase changes.

Calculating the energy released involves multiplying the mass of mercury by this heat of fusion. It's an essential step in finding the total energy released during the cooling and freezing process.

Specific heat capacity

Specific heat capacity is a material property that shows the amount of heat required to change the temperature of a unit mass of a substance by one degree Celsius. It plays a significant role in determining how substances react to changes in temperature.In this context, mercury has a specific heat capacity of \(0.140 \, \mathrm{J/g \, ^{\circ}C}\). This means it takes 0.140 joules to raise the temperature of 1 gram of mercury by 1°C. This concept is employed when calculating the energy needed to cool or heat a substance without changing its state.

• Specific heat capacity helps understand how a substance absorbs or releases heat.
• It is essential for energy conservation equations in thermodynamics.
• Used in formulas like \( Q = mc\Delta T\), where \( \Delta T\) is the change in temperature.

Understanding specific heat capacity allows us to calculate the energy change associated with temperature fluctuations before mercury reaches its freezing point.