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The energy content of a certain food is to be determined in a bomb calorimeter that contains \(3 \mathrm{kg}\) of water by burning a \(2-\mathrm{g}\) sample of it in the presence of \(100 \mathrm{g}\) of air in the reaction chamber. If the water temperature rises by \(3.2^{\circ} \mathrm{C}\) when equilibrium is established, determine the energy content of the food, in \(\mathrm{kJ} / \mathrm{kg}\), by neglecting the thermal energy stored in the reaction chamber and the energy supplied by the mixer. What is a rough estimate of the error involved in neglecting the thermal energy stored in the reaction chamber?

Short Answer

Expert verified
Answer: The energy content of the food sample is approximately 20,088 kJ/kg.

Step by step solution

01

Determine the energy absorbed by water

To find the total energy absorbed by the water, we will use the formula: Energy absorbed = mass of water * specific heat capacity of water * change in temperature where the specific heat capacity of water is \(4.18\ \mathrm{kJ/(kg\cdot ^\circ C)}\). Given the mass of water \(m_{water} = 3\ \mathrm{kg}\) and the temperature rise \(\Delta T = 3.2\ ^\circ\mathrm{C}\), we can calculate the energy absorbed.
02

Calculate the energy absorbed by water

Now, let's plug in the values: Energy absorbed = \(3\ \mathrm{kg} * 4.18\ \mathrm{kJ/(kg\cdot ^\circ C)} * 3.2\ ^\circ\mathrm{C} = 40.176\ \mathrm{kJ}\)
03

Determine the energy content of the food

To find the energy content per kg of food, we will first determine the energy content of the \(2\ \mathrm{g}\) sample and then convert it to per kg value. Since all the energy absorbed by the water is released by the food, the energy content of the \(2\ \mathrm{g}\) sample equals the energy absorbed by the water. Therefore, the energy content of the sample is \(40.176\ \mathrm{kJ}\). Now, let's convert this value to per kg: Energy content per kg = \(\frac{40.176\ \mathrm{kJ}}{2\ \mathrm{g}} * \frac{1000\ \mathrm{g}}{1\ \mathrm{kg}} = 20,\!088\ \mathrm{kJ/kg}\)
04

Discuss the error involved in neglecting the thermal energy stored in the reaction chamber

The main source of error in this calculation is due to the neglect of thermal energy stored in the reaction chamber. In a real experiment, some energy would have been absorbed by the reaction chamber, and the actual energy content of the food might be different from our calculated value. However, without more information about the reaction chamber's properties, it is difficult to estimate the actual error magnitude. Thus, this calculated energy content per kg of food should be regarded as an approximation.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Energy Content Determination
Understanding the energy content of food is crucial for various applications such as nutrition science and energy efficiency studies. In our exercise, we utilize a bomb calorimeter to determine this energy content. A calibrated device, the bomb calorimeter is designed to measure the amount of heat produced by a substance when it undergoes complete combustion.

In the experiment, a small sample of food is burned inside the calorimeter's reaction chamber. The released heat is absorbed by the surrounding water, and the temperature increase provides us with data to compute the energy content. By reassessing the water's temperature change and knowing its specific heat capacity, we can accurately calculate the energy released by the sample during combustion. Afterward, using the mass of the food sample, we convert it to a standardized 'per kilogram' basis which lets us compare energy contents of different foods to one another.
Thermal Energy Calculation
The thermal energy calculation is essential in a bomb calorimetry experiment as it allows us to quantify the heat absorbed by the water. This process involves applying the formula Energy absorbed = mass of water × specific heat capacity of water × change in temperature.

The specific heat capacity, a substance's inherent property, signifies the amount of heat required to raise the temperature of one kilogram of the substance by one-degree Celsius. By plugging the known values into the formula, we obtained the energy absorbed by the water when the food sample was combusted. This energy quantity, expressed in kilojoules (kJ), represents the energy that was previously stored in the food, which we then used to determine the energy content per unit mass of the food sample.
Specific Heat Capacity
Specific heat capacity is a pivotal factor in thermal physics, defining how a substance reacts to the addition of heat. It is a measure of the energy required to raise the temperature of a unit mass of a material by one degree Celsius (or one Kelvin). In the context of our exercise, we focus on the water's specific heat capacity, which is typically around 4.18 kJ/(kg·°C).

What is unique about water is its high specific heat capacity, implying it can absorb a lot of heat before its temperature rises significantly. This makes it an ideal medium for calorimetry experiments, as it stably captures the heat released by the tested sample, facilitating an accurate temperature-based energy calculation.
Error Estimation in Measurements
Error estimation is a fundamental component of experimental science, as it helps assess the reliability and precision of results. In our exercise, we recognize that ignoring the calorimeter's reaction chamber's capacity to store thermal energy introduces an approximation. While the chamber's thermal properties can significantly affect the final result, without specific details, quantifying this error becomes challenging.

To minimize such errors, calorimeters are typically constructed with materials that have lower heat capacities, allowing for negligible heat storage compared to the sample and water. Accurate error estimation often relies on knowledge about all involved materials' specific heat capacities, their masses, and the setup's heat exchange efficiency with its surroundings. This information helps scientists correct their measurements, ensuring their results are as close to reality as possible.

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Most popular questions from this chapter

Nitrogen gas to 20 psia and \(100^{\circ} \mathrm{F}\) initially occupies a volume of \(1 \mathrm{ft}^{3}\) in a rigid container equipped with a stirring paddle wheel. After 5000 lbf.ft of paddle wheel work is done on nitrogen, what is its final temperature?

An ideal gas at a given state expands to a fixed final volume first at constant pressure and then at constant temperature. For which case is the work done greater?

In order to cool 1 ton of water at \(20^{\circ} \mathrm{C}\) in an insulated tank, a person pours \(80 \mathrm{kg}\) of ice at \(-5^{\circ} \mathrm{C}\) into the water. Determine the final equilibrium temperature in the tank. The melting temperature and the heat of fusion of ice at atmospheric pressure are \(0^{\circ} \mathrm{C}\) and \(333.7 \mathrm{kJ} / \mathrm{kg},\) respectively.

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1-m^{3}\( of saturated liquid water at \)200^{\circ} \mathrm{C}$ is expanded isothermally in a closed system until its quality is 80 percent. Determine the total work produced by this expansion, in kJ.

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