Problem 57 - University Physics with Modern Physics

Question

A Thermodynamic Process in an Insect. The African bombardier beetle (Stenaptinus insignis) can emit a jet of defensive spray from the movable tip of its abdomen (Fig. Pl9.57). The beetle's body has reservoirs of two different chemicals; when the beetle is disturbed, these chemicals are combined in a reaction chamber, producing a compound that is warmed from \(20^{\circ} \mathrm{C}\) to \(100^{\circ} \mathrm{C}\) by the heat of reaction. The high pressure produced allows the compound to be sprayed out at speeds up to 19 \(\mathrm{m} / \mathrm{s}(68 \mathrm{km} / \mathrm{h})\) , scaring away predators of all kinds. (The beetle shown in the figure is 2 \(\mathrm{cm}\) long.) Calculate the heat of reaction of the two chemicals (in \(\mathrm{J} / \mathrm{kg} ) .\) Assume that the specific heat of the two chemicals and the spray is the same as that of water, \(4.19 \times 10^{3} \mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\) , and that the initial temperature of the chemicals is \(20^{\circ} \mathrm{C}\) .

Step-by-Step Solution

Answer

The heat of reaction is \(3.352 \times 10^5 \mathrm{J/kg}\).

Step 1: Understanding the Given Problem

We need to calculate the heat of reaction for chemicals in a beetle, which involves a temperature change from \(20^{\circ}\mathrm{C}\) to \(100^{\circ}\mathrm{C}\). Given data includes specific heat capacity \(c = 4.19 \times 10^{3} \mathrm{J/kg} \cdot \mathrm{K}\) for the chemicals, similar to water, and an initial temperature \(T_i = 20^{\circ}\mathrm{C}\).

Step 2: Identifying the Final Temperature

The compound is warmed to \(100^{\circ}\mathrm{C}\), making this the final temperature (\(T_f\)). The problem involves heating the chemicals from \(20^{\circ}\mathrm{C}\) to \(100^{\circ}\mathrm{C}\).

Step 3: Calculating the Temperature Change

The temperature change (\(\Delta T\)) is the difference between the final and initial temperatures:\[ \Delta T = T_f - T_i = 100^{\circ}C - 20^{\circ}C = 80^{\circ}C \]Convert this into Kelvin since the specific heat is in \( \mathrm{J/kg} \cdot \mathrm{K}\):\[ \Delta T = 80\,\mathrm{K} \] (since the size of the degree is the same for Celsius and Kelvin).

Step 4: Using the Specific Heat Formula to Find Heat of Reaction

Use the formula for heat transfer: \( Q = m \cdot c \cdot \Delta T \), where:\- \(m\) is mass,- \(c\) is the specific heat capacity,- \(\Delta T\) is the temperature change.Since we're asked to find heat per unit mass (\(Q/m\)), we rearrange to find \(q = c \cdot \Delta T \).Plug in the known values:\[ q = 4.19 \times 10^3 \mathrm{J/kg} \cdot \mathrm{K} \times 80\, \mathrm{K} = 3.352 \times 10^5 \mathrm{J/kg} \]

Step 5: Concluding the Calculation

The heat of reaction for the chemicals combined in the beetle is \(3.352 \times 10^5 \mathrm{J/kg}\). This value represents the energy needed to increase the temperature to \(100^{\circ}\mathrm{C}\) per kilogram of the chemical mix.

Key Concepts

Specific Heat Capacity

Specific heat capacity is a critical concept when discussing thermodynamic processes. It tells us how much energy is needed to change the temperature of one kilogram of a substance by one Kelvin (or degree Celsius). This property is specific to each material. For water and the chemicals in the beetle's reaction, this value is given as \(4.19 \times 10^3 \text{ J/kg} \cdot \text{K}\).
This means that for every kilogram of the beetle's chemical mix, 4190 Joules are required to raise its temperature by 1 degree Celsius. Knowing the specific heat capacity helps us understand how much energy is needed when the beetle uses its unique defense mechanism.

• If a substance has a high specific heat capacity, it can absorb a lot of energy without a large temperature change.
• This property is why water, and these specific chemicals, are effective at storing and releasing heat during the reaction.

The specific heat capacity allows us to solve for the energy needed in the beetle example through the formula \( Q = m \cdot c \cdot \Delta T \), where we isolate \( c \) to use per unit mass.

Temperature Change Calculation

Temperature change calculation is vital for understanding how much energy is transferred in the beetle's thermodynamic process. When calculating temperature change, you find the difference between the final and initial temperatures.
In this exercise, we start at \(20^{\circ}\text{C}\) and raise it to \(100^{\circ}\text{C}\). The temperature change is calculated as:

• \(\Delta T = T_f - T_i = 100^{\circ}\text{C} - 20^{\circ}\text{C}\)
• This gives us \(\Delta T = 80^{\circ}\text{C}\), which converts directly to 80 Kelvin.

The conversion between Celsius and Kelvin is straightforward because the temperature increment size is the same. The temperature change calculation is necessary to apply the specific heat capacity result and find the energy change \( q \). This helps understand how efficiently the beetle can achieve the high-energy spray to deter predators.

Heat of Reaction

The heat of reaction refers to the energy change that occurs during a chemical process. For the African bombardier beetle, the chemicals react to produce heat, raising the compound's temperature from \(20^{\circ}\text{C}\) to \(100^{\circ}\text{C}\).
The formula used to calculate this is: \[ q = c \cdot \Delta T \] Where:

• \( q \) is the heat per kilogram \((\text{J/kg})\).
• \( c \) is the specific heat capacity (\(4.19 \times 10^3 \text{ J/kg} \cdot \text{K}\)).
• \( \Delta T \) is the temperature change (80 K).

When these values are plugged in, we get: \[ q = 4.19 \times 10^3 \cdot 80 = 3.352 \times 10^5 \text{ J/kg} \] This computation shows us the energy needed per kilogram to reach the desired temperature for this reactive spray. Understanding the heat of reaction provides insight into the impressive energy release managed by such a small insect.