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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

The hot glowing surfaces of stars emit energy in the form of electromagnetic radiation. It is a good approximation to assume e = 1 for these surfaces. Find the radii of the following stars (assumed to be spherical): (a) Rigel, the bright blue star in the constellation Orion, which radiates energy at a rate of \(2.7 \times 10{^3}{^2} W\) and has surface temperature 11,000 K; (b) Procyon B (visible only using a telescope), which radiates energy at a rate of \(2.1 \times 10{^2}{^3} W\) and has surface temperature 10,000 K. (c) Compare your answers to the radius of the earth, the radius of the sun, and the distance between the earth and the sun. (Rigel is an example of a supergiant star, and Procyon B is an example of a white dwarf star.)

A typical doughnut contains 2.0 g of protein, 17.0 g of carbohydrates, and 7.0 g of fat. Average food energy values are 4.0 kcal/g for protein and carbohydrates and 9.0 kcal/g for fat. (a) During heavy exercise, an average person uses energy at a rate of 510 kcal/h. How long would you have to exercise to 'work off' one doughnut? (b) If the energy in the doughnut could somehow be converted into the kinetic energy of your body as a whole, how fast could you move after eating the doughnut? Take your mass to be 60 kg, and express your answer in m/s and in km/h.

BIO Before going in for his annual physical, a 70.0-kg man whose body temperature is 37.0\(^\circ\)C consumes an entire 0.355-L can of a soft drink (mostly water) at 12.0\(^\circ\)C. (a) What will his body temperature be after equilibrium is attained? Ignore any heating by the man’s metabolism. The specific heat of the man’s body is 3480 J/kg \(\cdot\) K. (b) Is the change in his body temperature great enough to be measured by a medical thermometer?

A copper pot with a mass of 0.500 kg contains 0.170 kg of water, and both are at 20.0\(^\circ\)C. A 0.250-kg block of iron at 85.0\(^\circ\)C is dropped into the pot. Find the final temperature of the system, assuming no heat loss to the surroundings.

In very cold weather a significant mechanism for heat loss by the human body is energy expended in warming the air taken into the lungs with each breath. (a) On a cold winter day when the temperature is -20\(^\circ\)C, what amount of heat is needed to warm to body temperature (37\(^\circ\)C) the 0.50 L of air exchanged with each breath? Assume that the specific heat of air is 1020 J / kg \(\cdot\) K and that 1.0 L of air has mass \(1.3 \times 10{^-}{^3} kg\). (b) How much heat is lost per hour if the respiration rate is 20 breaths per minute?

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