Question

The method of Exercise 97 is used in some bomb calorimetry experiments. A 1.148 g sample of benzoic acid is burned in excess \(\mathrm{O}_{2}(\mathrm{g})\) in a bomb immersed in 1181 g of water. The temperature of the water rises from 24.96 to \(30.25^{\circ} \mathrm{C}\). The heat of combustion of benzoic acid is \(-26.42 \mathrm{kJ} / \mathrm{g} .\) In a second experiment, a \(0.895 \mathrm{g}\) powdered coal sample is burned in the same calorimeter assembly. The temperature of \(1162 \mathrm{g}\) of water rises from 24.98 to \(29.81^{\circ} \mathrm{C}\). How many metric tons (1 metric ton \(=1000 \mathrm{kg}\) ) of this coal would have to be burned to release \(2.15 \times 10^{9} \mathrm{kJ}\) of heat?

Short answer

The mass of coal required to release \(2.15 \times 10^9 kJ\) of energy is \(m_{ton}\) metric tons.

Step-by-step solution

Calculate the heat released by benzoic acid

First, calculate the heat (q) released by the combustion of benzoic acid by using the given mass of benzoic acid and its heat of combustion: \(q_{benzoic} = (1.148g) * (-26.42 kJ/g)\)

Calculate the specific heat capacity of the calorimeter

The heat from the combustion is absorbed by water, causing an increase in temperature. Use the calorimetry equation \(q = mc\Delta T\) to find the specific heat capacity (c) of calorimeter: \[q = (mass_{water})*(c_{calorimeter})*(T_{final}-T_{initial})\] Solve this equation for \(c_{calorimeter}\) to get \(c_{calorimeter} = q_{benzoic} /(mass_{water}*(T_{final}-T_{initial}))\] Insert the given values into this equation to calculate \(c_{calorimeter}\)

Calculate the heat generated by the coal sample

Use the calorimetry equation and the specific heat capacity which we just calculated to find the heat (q) generated by the coal sample. Use the same formula as in the previous step, but this time solve for q for the coal sample: \(q_{coal} = (mass_{water})*(c_{calorimeter})*(T_{final}-T_{initial})\) Input the related data from the exercise to calculate \(q_{coal}\)

Calculate the heat of combustion per gram of coal

After calculating the total energy released by coal, divide it by the mass of the coal sample to find the energy per gram: \((q_{coal}/mass_{coal})\)

Determine the mass of coal required to release the desired amount of heat

Use the energy per gram of coal calculated in the previous step to calculate the mass (m) of coal required to release \(2.15 * 10^{9} kJ\) of heat. Divide the total energy required by the energy per gram value to get the mass of coal in grams \(m = (2.15*10^{9} kJ / energyPerGramCoal)\), and then convert this mass from grams to metric tons: \(m_{ton} = m_{gram} / 10^{6}\)

Key concepts

Heat of Combustion

The heat of combustion is a crucial concept in understanding how substances release energy when burned. It represents the energy produced as heat when a compound completely combusts in the presence of oxygen. In bomb calorimetry, this measurement is essential for determining the energy content of fuels or foods.
To calculate the heat of combustion, you use the mass of the substance and its known heat of combustion value, expressed in terms of energy per mass (like kJ/g). For instance, benzoic acid's heat of combustion is given as \(-26.42 \text{kJ/g}\). This means that when 1 gram of benzoic acid is burned, it releases 26.42 kJ of energy.

• The sign of the heat of combustion is negative. This indicates that energy is released during combustion, making it an exothermic process.
• In practical applications, knowing the heat of combustion allows industries to choose suitable fuels based on energy output and efficiency.

Understanding this concept is pivotal for energy calculations, as it directly connects to the quantity of energy released from a specific amount of fuel.

Calorimetry Equation

The calorimetry equation, \(q = mc\Delta T\), forms the backbone of calculating energy changes in bomb calorimetry. In this formula, \(q\) represents the heat transferred, \(m\) is the mass of the substance undergoing the temperature change (like water in a calorimetry experiment), \(c\) is the specific heat capacity, and \(\Delta T\) is the change in temperature.
This equation helps determine the heat absorbed or released by the system being studied. In the given calorimetry experiment, when substances like benzoic acid or coal are burned, their heat is transferred to the surrounding water, causing its temperature to rise.

• To use the equation effectively, ensure you have the accurate values for all components, particularly the mass of the water and the temperature change.
• Solve for the unknown, which could be the heat released or absorbed, a specific heat capacity, or a temperature change.
• This precise method allows scientists to measure energy release and better understand reactions' energetics.

Mastering the calorimetry equation is key to interpreting calorimetric experiments and making energy-related calculations.

Specific Heat Capacity

Specific heat capacity is a fundamental property of materials, indicating how much energy is required to change the temperature of a given mass by a degree. This value is usually expressed in units of \(\text{J/g°C}\).
In calorimetry, the specific heat capacity helps us understand how substances react to heat addition or removal. Water, often used as a medium in calorimeters, has a specific heat capacity of \(4.18 \text{J/g°C}\), meaning it requires 4.18 joules to raise one gram of water by one degree Celsius.

• Specific heat capacity is crucial for calculating temperature changes in calorimetric experiments, as it links energy change with temperature change.
• Knowing the specific heat capacity of substances helps in adjusting compositions in thermal processes to ensure desired temperature changes and energy efficiencies.
• Using the calorimetry equation, we can isolate the specific heat capacity \(c\) to solve problems related to energy transfer.

Understanding specific heat capacity is vital for interpreting how different substances absorb and release heat, impacting both the design of calorimetry experiments and practical applications in energy management.