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

37.1 J of energy is released.

Step 1: Calculate the mass of mercury

Determine the mass using the volume and density of mercury. Since the density is given in \( \mathrm{g/cm}^3 \), and 1 mL is equivalent to 1 cm³, calculate the mass: \[ \text{mass} = \text{density} \times \text{volume} = 13.6 \ \mathrm{g/cm}^3 \times 1 \ \mathrm{cm}^3 = 13.6 \ \mathrm{g} \]

Step 2: Calculate energy released during cooling

Use the specific heat capacity to calculate the energy released while cooling from \( 23.0^{\circ} \mathrm{C} \) to \(-38.8^{\circ} \mathrm{C} \). The formula is:\[ q = mc\Delta T \]where \( m = 13.6 \ \mathrm{g} \), \( c = 0.140 \ \mathrm{J/g\cdot K} \), and \( \Delta T = -38.8^{\circ} \mathrm{C} - 23.0^{\circ} \mathrm{C} = -61.8^{\circ} \mathrm{C} \).\[ q = 13.6 \ \mathrm{g} \times 0.140 \ \mathrm{J/g\cdot K} \times (-61.8 \ \mathrm{K}) = -117.9 \ \mathrm{J} \]

Step 3: Calculate energy released during solidification

Use the heat of fusion to determine the energy released when the mercury freezes.\[ q = m \times \text{heat of fusion} \]where \( m = 13.6 \ \mathrm{g} \) and the heat of fusion is \( 11.4 \ \mathrm{J/g} \).\[ q = 13.6 \ \mathrm{g} \times 11.4 \ \mathrm{J/g} = 155.0 \ \mathrm{J} \]

Step 4: Calculate total energy released

Add the energy calculated from cooling and solidification to find the total energy released:\[ q_{\text{total}} = q_{\text{cooling}} + q_{\text{fusion}} = -117.9 \ \mathrm{J} + 155.0 \ \mathrm{J} = 37.1 \ \mathrm{J} \]

Key Concepts

Heat Energy

Heat energy is the energy transferred between substances or systems due to a temperature difference. It flows from a warmer object to a cooler one until thermal equilibrium is reached. In this exercise, we're looking at how much heat energy is lost by mercury as it cools down and undergoes phase change. This involves cooling mercury from 23.0°C to its freezing point at -38.8°C and then freezing it.
Heat energy can be released or absorbed during such processes, depending on the direction of the temperature change. In this case, as mercury cools and freezes, it releases heat.

• During cooling: This involves temperature change within the same phase (liquid to colder liquid).
• During freezing: This involves phase change (liquid to solid) without temperature change.

Understanding how heat energy is transferred is essential in solving thermodynamics problems.

Phase Change

A phase change is when a substance changes from one state of matter to another, such as from liquid to solid. This process occurs when energy is added or removed. In our exercise, mercury freezes at -38.8°C, changing from a liquid to a solid.
When a substance undergoes a phase change, the temperature remains constant despite the ongoing energy transfer. This is because the energy is used to change the material's structure rather than altering its temperature. Mercury, as it freezes, requires energy removal called heat of fusion.

• Freezing: Energy is released as mercury changes to a solid.
• Constant temperature: Even as energy is released, the temperature stays at the freezing point until the phase change is complete.

Specific Heat Capacity

The specific heat capacity is a property that describes how much energy is needed to change the temperature of a certain mass of a substance by 1°C (or 1 K). For mercury, this value is 0.140 J/g·K.
To calculate the heat energy exchanged during the cooling of mercury, the specific heat capacity is used in the formula:
\[ q = mc\Delta T \]
where:

• q is the heat energy exchanged,
• m is the mass,
• c is the specific heat capacity,
• ΔT is the change in temperature.

This formula helps in figuring out the energy released or absorbed during temperature changes without phase change.

Heat of Fusion

The heat of fusion is the energy required to change a substance from solid to liquid or vice versa at its melting/freezing point. For mercury, the heat of fusion is 11.4 J/g. This means 11.4 joules of energy must be removed per gram of mercury to freeze it.
Calculating the energy during a phase change involves using the formula:
\[ q = m \times \text{heat of fusion} \]
where:

• q is the total heat energy exchanged during the phase change,
• m is the mass of the substance.

The heat of fusion comes into play after the substance reaches its freezing point and begins to solidify. The steady release of energy ensures that mercury transitions properly to a solid phase.