Question

Under constant-volume conditions, the heat of combustion of glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) is \(15.57 \mathrm{~kJ} / \mathrm{g}\). A 3.500 -g sample of glucose is burned in a bomb calorimeter. The temperature of the calorimeter increased from \(20.94^{\circ} \mathrm{C}\) to \(24.72^{\circ} \mathrm{C}\). (a) What is the total heat capacity of the calorimeter? (b) If the size of the glucose sample had been exactly twice as large, what would the temperature change of the calorimeter have been?

Short answer

The total heat capacity of the calorimeter is \(14,403.17 \; \frac{\text{J}}{\text{°C}}\). If the size of the glucose sample had been exactly twice as large, the temperature change of the calorimeter would have been 7.56°C.

Step-by-step solution

Calculate the heat released

To find the total heat released, multiply the heat of combustion of glucose by the mass of the glucose sample: \[q = m(\text{glucose}) \cdot \Delta H_c(\text{glucose})\] Plug in the given values for the mass of glucose and heat of combustion: \[q = (3.500 \; \text{g})\left(15.57 \; \frac{\text{kJ}}{\text{g}}\right)\] Now, convert the heat released to J by multiplying by 1,000 J/kJ: \[q = (3.500 \cdot 15.57 \cdot 1,000) \; \text{J}\] Calculate the value of q: \[q = 54,495 \; \text{J}\]

Determine the heat capacity of the calorimeter

Using the formula: \[q = C\Delta T\] We can solve for the heat capacity of the calorimeter (C), where q is the heat released, and ΔT is the change in temperature: \[C = \frac{q}{\Delta T}\] Calculate the change in temperature using the given initial and final temperatures: \[\Delta T = T_{final} - T_{initial} = 24.72 - 20.94 = 3.78 \; \text{°C}\] Now plug in the values of q and ΔT: \[C = \frac{54,495}{3.78}\] Calculate the heat capacity C: \[C = 14,403.17 \; \frac{\text{J}}{\text{°C}}\]

Calculate the temperature change for twice the mass of glucose

Now, let's determine the temperature change for twice the mass of the glucose sample. First, find the heat released by twice the mass of glucose: \[q' = 2 \cdot (3.500 \; \text{g}) \cdot (15.57\; \frac{\text{kJ}}{\text{g}}) \cdot 1,000\; J\] Calculate q': \[q' = 108,990 \; \text{J}\] With q' and the heat capacity of the calorimeter known, we can find the new temperature change for the calorimeter: \[\Delta T' = \frac{q'}{C}\] Plug in the values of q' and C: \[\Delta T' = \frac{108,990}{14,403.17}\] Calculate the new temperature change ΔT': \[\Delta T' = 7.56 \; \text{°C}\] So, if the size of the glucose sample had been exactly twice as large, the temperature change of the calorimeter would have been 7.56°C.

Key concepts

Heat of Combustion

The heat of combustion refers to the amount of energy released when a substance is burned completely in the presence of oxygen. For glucose, this value is provided as 15.57 kJ/g. This means that when 1 gram of glucose burns, 15.57 kJ of energy is liberated.
Understanding the heat of combustion is crucial in calorimetry, as it provides insight into the energy content of a fuel. Here's why it's important:

• It helps in comparing the energy yields of different fuels. The higher the heat of combustion, the more energy a fuel can release.
• It's used to calculate the total energy released in combustion reactions, which is essential for energy management and efficiency evaluation.

In the exercise above, the calculation involves multiplying the mass of the glucose sample by its heat of combustion to determine the total energy released. This calculation gives insight into how much heat is involved in the reaction, setting the stage for subsequent calorimetric analysis.

Bomb Calorimeter

A bomb calorimeter is a device designed to measure the heat of combustion of a substance under constant-volume conditions. When glucose is burned, it releases heat, which is absorbed by the water in the calorimeter, leading to a measurable change in temperature.
Here are some key features of a bomb calorimeter:

• It's a highly insulated apparatus to ensure that no heat is lost to the environment, allowing for precise measurements of energy changes.
• It operates at constant volume, meaning the volume inside the calorimeter doesn't change even as the reaction progresses. This simplifies calculations of heat exchange.
• The device measures changes in temperature that arise from the burning process. These measurements help in determining the heat transferred to the calorimeter's surroundings.

In this exercise, the bomb calorimeter played a crucial role in capturing the heat exchange from the combustion of glucose, allowing us to calculate the heat capacity based on the observed temperature increase.

Heat Capacity

Heat capacity is a measure of the amount of heat energy needed to raise the temperature of an object by one degree Celsius. In the context of a calorimeter, it represents how much heat the calorimeter can absorb before its temperature changes by a specific amount.
Here’s why it’s a pivotal concept:

• Heat capacity helps in quantifying the calorimeter's response to heat input, indicating how much heat is stored within it during an experiment.
• It depends on the mass and specific heat of the materials composing the calorimeter, affecting how quickly the temperature rises.

In this particular problem, the heat capacity was calculated using the formula \[ C = \frac{q}{\Delta T} \]where \( q \) is the heat absorbed, and \( \Delta T \) is the temperature change. Understanding the relationship between heat energy, heat capacity, and temperature change is key to interpreting calorimetry results and analyzing how different sample sizes affect temperature shifts, as evidenced when they hypothesized burning a larger mass of glucose.