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The African bombardier beetle (Stenaptinus insignis) can emit a jet of defensive spray from the movable tip of its abdomen (Fig. P17.91). The beetle's body has reservoirs containing two chemicals; when the beetle is disturbed, these chemicals combine in a reaction chamber, producing a compound that is warmed from 20\(^\circ\)C to 100\(^\circ\)C by the heat of reaction. The high pressure produced allows the compound to be sprayed out at speeds up to 19 m/s 168 km/h2, scaring away predators of all kinds. (The beetle shown in Fig. P17.91 is 2 cm long.) Calculate the heat of reaction of the two chemicals (in J/kg). Assume that the specific heat of the chemicals and of the spray is the same as that of water, \(4.19 \times 10{^3} J/kg \cdot K\), and that the initial temperature of the chemicals is 20\(^\circ\)C.

Short Answer

Expert verified
The heat of reaction is 335,200 J/kg.

Step by step solution

01

Identify the Given Data

We begin by noting the given information: initial temperature \(T_i = 20^{\circ}\)C, final temperature \(T_f = 100^{\circ}\)C, and the specific heat \(c = 4.19 \times 10^3 \, \text{J/kg} \cdot \text{K}\).
02

Determine the Temperature Change

Calculate the change in temperature \(\Delta T\) by subtracting the initial temperature \(T_i\) from the final temperature \(T_f\):\[\Delta T = T_f - T_i = 100^{\circ}C - 20^{\circ}C = 80\,^{\circ}C.\]
03

Use Specific Heat Capacity Formula

The formula relating heat change \(Q\), mass \(m\), specific heat \(c\), and temperature change \(\Delta T\) is:\[Q = mc\Delta T.\]Since we need \(Q\) per unit mass (J/kg), we divide by mass \(m\) to focus on the specific heat of reaction \(\frac{Q}{m}\):\[\frac{Q}{m} = c\Delta T.\]
04

Calculate the Heat of Reaction per Kilogram

Substitute the known values \(c = 4.19 \times 10^3\, \text{J/kg} \cdot \text{K}\) and \(\Delta T = 80\,^{\circ}C\):\[\frac{Q}{m} = (4.19 \times 10^3 \, \text{J/kg} \cdot \text{K})(80 \, \text{K}) = 335,200 \, \text{J/kg}.\]

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Key Concepts

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

Specific Heat Capacity
Specific heat capacity is a fundamental concept in thermodynamics, describing the amount of heat required to change a substance's temperature by a certain amount. It is usually expressed in units of joules per kilogram per degree Celsius (\( J/kg \, \cdot \, ^\circ \text{C} \)) or (\( J/kg \, \cdot \, \text{K} \)). For water, a common reference point, the specific heat capacity is (\( 4.19 \times 10^3 \text{J/kg} \, \cdot \, \text{K}\)).
This property depends on the material itself, as different substances accept and retain heat differently due to molecular structure. When the bombardier beetle mixes chemicals in its reaction chamber, these have a specific heat similar to water. This indicates that the heat needed to raise their temperature is worked out the same way as for water.
  • Specific heat capacity helps predict how a substance behaves under heating.
  • The beetle's known specific heat lets us calculate the reaction heat during the temperature rise from 20°C to 100°C.
Temperature Change
Temperature change is a key variable when calculating the heat absorbed or released in chemical reactions and physical changes. In our example, the bombardier beetle's chemical reaction causes a temperature increase from 20°C to 100°C.
The temperature change (\( \Delta T \)) is calculated simply by subtracting the initial temperature (\( T_i \)) from the final temperature (\( T_f \)). Therefore, (\( \Delta T = T_f - T_i = 100\,^{\circ}\text{C} - 20\,^{\circ}\text{C} = 80\,^{\circ}\text{C} \)).
  • A larger temperature change generally means more heat exchange.
  • In the beetle's case, this involves the release of heat during the chemical reaction that leads to the ejection response.
Calculating this change is crucial for determining how much energy was involved in warming the substance.
Heat of Reaction
The heat of reaction is the heat exchanged when a chemical reaction occurs. In thermodynamic terms, it is crucial for understanding how energy is absorbed or emitted during reactions.
In this specific context, the bombardier beetle chemically reacts substances in its body to produce a heat that raises the temperature significantly. This is measured in joules per kilogram (\( \text{J/kg} \)), making it easier to understand how much energy per kilogram is involved in the reaction, irrespective of the mass.
The heat of reaction can be calculated using the formula:\[\frac{Q}{m} = c\Delta T\]Where:
  • \( Q \) is the total heat.
  • \( m \) is the mass of the substance.
  • \( c \) is the specific heat capacity.
  • \( \Delta T \) is the temperature change.
For the beetle, substituting the given specific heat and temperature change provides 335,200 J/kg, indicating a very high energy release necessary to defend itself effectively against predators.

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Most popular questions from this chapter

Like the Kelvin scale, the Rankine scale is an absolute temperature scale: Absolute zero is zero degrees Rankine (0\(^\circ\)R). However, the units of this scale are the same size as those of the Fahrenheit scale rather than the Celsius scale. What is the numerical value of the triple-point temperature of water on the Rankine scale?

(a) Calculate the one temperature at which Fahrenheit and Celsius thermometers agree with each other. (b) Calculate the one temperature at which Fahrenheit and Kelvin thermometers agree with each other.

Careful measurements show that the specific heat of the solid phase depends on temperature (Fig. P17.117). How will the actual time needed for this cryoprotectant to come to equilibrium with the cold plate compare with the time predicted by using the values in the table? Assume that all values other than the specific heat (solid) are correct. The actual time (a) will be shorter; (b) will be longer; (c) will be the same; (d) depends on the density of the cryoprotectant.

CP A person of mass 70.0 kg is sitting in the bathtub. The bathtub is 190.0 cm by 80.0 cm; before the person got in, the water was 24.0 cm deep. The water is at 37.0\(^\circ\)C. Suppose that the water were to cool down spontaneously to form ice at 0.0\(^\circ\)C, and that all the energy released was used to launch the hapless bather vertically into the air. How high would the bather go? (As you will see in Chapter 20, this event is allowed by energy conservation but is prohibited by the second law of thermodynamics.)

A 25,000-kg subway train initially traveling at 15.5 m/s slows to a stop in a station and then stays there long enough for its brakes to cool. The station's dimensions are 65.0 m long by 20.0 m wide by 12.0 m high. Assuming all the work done by the brakes in stopping the train is transferred as heat uniformly to all the air in the station, by how much does the air temperature in the station rise? Take the density of the air to be 1.20 kg/m\(^3\) and its specific heat to be 1020 J /kg \(\cdot\) K.

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