Problem 94 - Chemistry The Molecular Science

Question

A \(25.0-\mathrm{mL}\) sample of benzene at \(19.9{ }^{\circ} \mathrm{C}\) was cooled to its melting point, \(5.5^{\circ} \mathrm{C}\), and then frozen. Calculate how much heat transfer to the surroundings occurred in this process. (The density of benzene is \(0.880 \mathrm{~g} / \mathrm{mL} ;\) its specific heat capacity is \(1.74 \mathrm{~J} \mathrm{~g}^{-1} \mathrm{~K}^{-1}\), and its enthalpy of fusion is \(127 \mathrm{~J} / \mathrm{g} .)\)

Step-by-Step Solution

Answer

The total heat transfer is 3344.15 J.

Step 1: Calculate Mass of Benzene

To find the mass of the benzene sample, we use the volume and density with the formula: \[ m = V \times \text{density} = 25.0 \, \text{mL} \times 0.880 \, \text{g/mL} \].This gives us:\[ m = 22.0 \, \text{g} \].

Step 2: Calculate Heat Lost During Cooling

We apply the formula for calculating heat lost during cooling: \[ q_1 = m \times c \times \Delta T \],where \( q_1 \) is the heat lost, \( m = 22.0 \, \text{g} \) is the mass, \( c = 1.74 \, \text{J/g K} \) is the specific heat capacity, and \( \Delta T = 19.9 - 5.5 \, \text{K} = 14.4 \, \text{K} \).Thus, \[ q_1 = 22.0 \, \text{g} \times 1.74 \, \text{J/g K} \times 14.4 \, \text{K} = 550.15 \, \text{J} \].

Step 3: Calculate Heat Lost During Freezing

The heat lost during the phase change (freezing) is calculated using the heat of fusion:\[ q_2 = m \times \Delta H_f = 22.0 \, \text{g} \times 127 \, \text{J/g} \].This results in:\[ q_2 = 2794.0 \, \text{J} \].

Step 4: Calculate Total Heat Transfer

The total heat transfer to the surroundings is the sum of the heat lost during cooling and freezing:\[ q_{\text{total}} = q_1 + q_2 = 550.15 \, \text{J} + 2794.0 \, \text{J} \].Thus, \[ q_{\text{total}} = 3344.15 \, \text{J} \].

Key Concepts

Heat Transfer

In thermodynamics, heat transfer is the process where heat energy moves from one body or substance to another, often due to a temperature difference. When you cool substances, like benzene, the heat is transferred away from it to the surroundings. This is what happened when the benzene sample in the exercise was cooled from 19.9°C to its freezing point, 5.5°C.
This flow of heat can be calculated using the specific heat capacity of the substance, the mass, and the temperature change. For benzene, the specific heat capacity is given as 1.74 J/g K, allowing us to compute how much energy was lost as it cooled down.

• The formula used is: \[ q = m \times c \times \Delta T \], where \( q \) is the heat transferred, \( m \) is the mass, \( c \) is the specific heat capacity, and \( \Delta T \) is the change in temperature.
• This calculation is crucial to determine the first stage of energy loss before the substance undergoes any phase change.

Phase Change

A phase change is a transition of matter from one state to another, like from liquid to solid. During this process, the temperature of the substance remains constant because the energy is used to change the phase of the substance rather than to change its temperature.
In the case of benzene, once it reached its freezing point at 5.5°C, it underwent a phase change from a liquid to a solid.

• Even though the temperature stays consistent, the heat energy used in this transformation is significant.
• This energy, known as latent heat, is necessary to overcome molecular attractions without altering the temperature.

Phase changes are part of thermodynamics and are calculated separately because they only depend on the mass of the substance and its enthalpy of fusion.

Enthalpy of Fusion

The enthalpy of fusion is the heat required to melt (or freeze) a certain mass of a substance without changing its temperature. For benzene, as given in the problem, this value is 127 J/g. It represents the energy needed to disrupt intermolecular forces and change the state from solid to liquid or vice versa.
This concept is critical when calculating the phase change portion of the heat transfer.

• The formula used for calculating this heat is: \[ q = m \times \Delta H_f \], where \( q \) is the heat, \( m \) is the mass, and \( \Delta H_f \) is the enthalpy of fusion.
• Even though the temperature of the benzene does not change, a substantial amount of energy is still transferred during its phase change.
• This energy helps in changing the rigid molecular arrangement of a solid into the more fluid structure of a liquid or vice versa.

This is why phase changes involve a significant amount of energy relative to the mass of the substance, demonstrating the importance of the enthalpy of fusion in thermodynamics.