Question

(a) When \(0.350 \mathrm{~g}\) of biodiesel \(\left(\mathrm{C}_{19} \mathrm{H}_{38} \mathrm{O}_{2}\right)\) is burned in a bomb calorimeter, the temperature of both the water and the calorimeter rise in temperature by \(6.85^{\circ} \mathrm{C}\). Assuming that the bath contains \(300.0 \mathrm{~g}\) of water and that the heat capacity for the calorimeter is \(675 \mathrm{~J} /{ }^{\circ} \mathrm{C}\), calculate combustion en\(\operatorname{ergy}(\Delta E)\) for biodiesel in units of \(\mathrm{k} \mathrm{J} / \mathrm{g}\) (b) Calculate combustion energy \((\Delta E)\) for biodiesel in units of \(\mathrm{kJ} / \mathrm{mol}\) (c) If the density of biodiesel is \(0.880 \mathrm{~g} / \mathrm{mL}\), calculate combustion energy \((\Delta E)\) for biodiesel in units of \(\mathrm{kJ} / \mathrm{mL}\).

Short answer

(a) 37.757 kJ/g. (b) 11258 kJ/mol. (c) 33.226 kJ/mL.

Step-by-step solution

Calculate Heat Absorbed by Water

First, determine the heat absorbed by the water using the formula: \[ q = m \cdot c \cdot \Delta T \]where: - \( m = 300.0 \text{ g} \) (mass of the water), - \( c = 4.18 \text{ J/g}^\circ\text{C} \) (specific heat capacity of water), and - \( \Delta T = 6.85^\circ\text{C} \) (temperature change).Calculate:\[ q_{\text{water}} = 300.0 \text{ g} \times 4.18 \text{ J/g}^\circ\text{C} \times 6.85^\circ\text{C} \]\[ q_{\text{water}} = 8591 \text{ J} \approx 8.591 \text{ kJ} \]

Calculate Heat Absorbed by Calorimeter

Next, calculate the heat absorbed by the calorimeter using the formula: \[ q = C \cdot \Delta T \]where:- \( C = 675 \text{ J/}^\circ\text{C} \) (heat capacity of the calorimeter) and - \( \Delta T = 6.85^\circ\text{C} \) (temperature change).Calculate:\[ q_{\text{calorimeter}} = 675 \text{ J/}^\circ\text{C} \times 6.85^\circ\text{C} \]\[ q_{\text{calorimeter}} = 4624 \text{ J} \approx 4.624 \text{ kJ} \]

Total Heat Released by Combustion

Add the heat absorbed by the water and the calorimeter to find the total heat released:\[ q_{\text{total}} = q_{\text{water}} + q_{\text{calorimeter}} \]\[ q_{\text{total}} = 8.591 \text{ kJ} + 4.624 \text{ kJ} = 13.215 \text{ kJ} \]

Calculate Combustion Energy per Gram

Now, calculate the combustion energy \( \Delta E \) per gram of biodiesel using the formula:\[ \Delta E = \frac{q_{\text{total}}}{\text{mass of biodiesel}} \]where:- mass of biodiesel = 0.350 g.Calculate:\[ \Delta E = \frac{13.215 \text{ kJ}}{0.350 \text{ g}} \approx 37.757 \text{ kJ/g} \]

Calculate Combustion Energy per Mole

The molar mass of biodiesel \(\text{C}_{19} \text{H}_{38} \text{O}_{2} \) is approximately 298 g/mol.Calculate combustion energy \( \Delta E \) per mole using:\[ \Delta E = 37.757 \text{ kJ/g} \times 298 \text{ g/mol} \]\[ \Delta E = 11258.186 \text{ kJ/mol} \approx 11258 \text{ kJ/mol} \]

Calculate Combustion Energy per mL

Given the density of biodiesel is 0.880 g/mL, find the combustion energy \( \Delta E \) per mL using:\[ \Delta E = 37.757 \text{ kJ/g} \times 0.880 \text{ g/mL} \]\[ \Delta E = 33.226 \text{ kJ/mL} \]

Key concepts

bomb calorimeter

A bomb calorimeter is an essential device in the study of combustion reactions. It measures the energy released when a substance burns. This tool is especially useful for determining the energy content of fuels like biodesel.

The bomb calorimeter consists of a sealed container (the "bomb") where the fuel is placed and burned. It's immersed in a water bath to capture the heat released by the combustion. This setup ensures no heat escapes into the surroundings, providing a closed system where all energy changes are accounted for.

By measuring the temperature change in the water bath, we can infer the heat generated by the combustion. Since the calorimeter's heat capacity is known, this information allows us to determine the energy produced by the burning fuel.

• This makes bomb calorimeters crucial for energy content studies in various substances.
• It allows researchers to calculate the heat of reaction directly.
• Ensures precise measurement due to its closed nature, minimizing energy loss.

Understanding how a bomb calorimeter works is fundamental in energy measurement and essential in assessing the efficiency of fuels.

calorimetry calculations

Calorimetry calculations are at the heart of using a bomb calorimeter. These calculations reveal the energy release during combustion by assessing changes in temperature.

The primary calculation involves measuring the heat absorbed by both the water and the calorimeter itself. This is done using the equation: \[ q = m \cdot c \cdot \Delta T \] for water, where \( q \) is the heat energy, \( m \) is the mass, \( c \) is the specific heat capacity, and \( \Delta T \) is the temperature change.

For the calorimeter, the formula used is:\[ q = C \cdot \Delta T \] where \( C \) is the calorimeter's heat capacity.

After calculating the heat absorbed by both the water and the calorimeter, these values are summed to find the total heat released by the combustion process. This total gives a direct measure of the energy content of the substance tested (like biodiesel).

• Step-by-step calculations ensure precision.
• Each variable represents a specific part of the system—mass, heat capacity, and temperature change—all crucial for accurate results.
• These calculations help in understanding heat transfer during chemical reactions.

With these detailed steps, students can accurately determine combustion energy, a key concept in chemistry and physics.

specific heat capacity

Specific heat capacity is a critical factor in understanding why some substances heat up quickly while others need more energy. It represents the amount of heat required to change the temperature of one gram of a substance by one degree Celsius.

Different materials have different specific heat capacities. For instance, water has a high specific heat capacity of \( 4.18 \, \text{J/g}^\circ\text{C} \), meaning it can absorb a lot of heat before its temperature rises significantly. This property makes water an excellent substance for capturing and measuring heat in calorimetry experiments.

When applied to a bomb calorimeter experiment, understanding specific heat capacity allows us to calculate the amount of heat absorbed by the water, and consequently the total energy released in the combustion process.

• Specific heat capacity helps in designing systems like heating and cooling that rely on heat transfer.
• It is a characteristic property that varies from one substance to another.
• With a known specific heat capacity, we can predict how different materials will respond to heat energy.

This concept not only aids in energy calculations but also gives insight into the thermal properties of substances, an important area of study in materials science and engineering.