/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 91 The African bombardier beetle (S... [FREE SOLUTION] | 91影视

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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. \(\mathbf{P 1 7 . 9 1}\) ). 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 \(20^{\circ} \mathrm{C}\) 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 \(19 \mathrm{~m} / \mathrm{s}(68 \mathrm{~km} / \mathrm{h}),\) scar- ing away predators of all kinds. (The beetle shown in Fig. \(\mathrm{P} 17.91\) 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 chemicals and of 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 \(20^{\circ} \mathrm{C}\).

Short Answer

Expert verified
The heat of reaction of the two chemicals is \(3.35 \times 10^{5} \, J/kg \)

Step by step solution

01

Identify Given Information

In the exercise, the following information are given: \n\n The initial temperature, \(T_{i} = 20^{\circ} \mathrm{C}\) or in Kelvin, \(T_{i} = 20 + 273 = 293 \, K\)\n The final temperature, \(T_{f} = 100^{\circ} \mathrm{C}\) or in Kelvin, \(T_{f} = 100 + 273 = 373 \, K\)\n The specific heat capacity, \(c = 4.19 \times 10^{3} \, J / kg \cdot K\)
02

Calculate Temperature Change

The change in temperature can be calculated by subtracting the initial temperature from the final temperature. \n\n That is, \n\n \(\Delta{T} = T_{f} - T_{i} = 373 \, K - 293 \, K = 80 \, K\)
03

Calculate The Heat of Reaction

Assuming the mass of the chemicals to be one kilogram, one can apply the heat equation \(Q = mc\Delta{T}\) to find the heat of the reaction.\n\n Here, \n\n\(Q = (1 \, kg)(4.19 \times 10^{3} \, J / kg \cdot K)(80 \, K) = 3.35 \times 10^{5} \, J\)\n\n The heat of reaction of the two chemicals per kg is \(3.35 \times 10^{5} J/kg \)

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Key Concepts

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

Bombardier Beetle
The African bombardier beetle, known scientifically as Stenaptinus insignis, is a fascinating creature. This small beetle possesses an impressive defense mechanism that involves the ejection of a hot, noxious spray. When threatened, the beetle releases this spray from the tip of its abdomen. The spray consists of chemicals stored in separate reservoirs within its body. These chemicals mix in a reaction chamber, which leads to an explosive release of heat and pressure.
The high-pressure spray travels at speeds up to 19 meters per second (or 68 kilometers per hour). This startling burst can effectively deter predators and ensure the beetle's safety. The beetle is only 2 centimeters long, yet it wields this powerful defense mechanism. Studying the bombardier beetle provides insights into natural chemical processes and has even inspired technological innovations in human industries.
Chemical Reactions
Chemical reactions are processes where substances, known as reactants, transform into new substances called products. The bombardier beetle's defense spray is a perfect example of a rapid and exothermic chemical reaction, which means it releases heat.
In the beetle, specific chemicals are stored separately to prevent premature reactions. Upon mixing in the reaction chamber, these chemicals undergo a series of reactions that generate heat and pressure, leading to the ejection of the spray.
Understanding chemical reactions is crucial as they form the basis of countless processes in nature and human activities. Features of a reaction include:
  • Change of temperature
  • Emission or absorption of heat
  • Formation of new products
  • Change of color
Learning about chemical reactions helps students grasp not only biological processes but also applications in chemistry and other sciences.
Specific Heat Capacity
Heating involves energy transfer, related to the concept of specific heat capacity. The specific heat capacity is the amount of heat required to raise the temperature of one kilogram of a substance by one Kelvin (or one degree Celsius).
In the bombardier beetle exercise, we assume that the chemicals in the reaction have a specific heat capacity equivalent to that of water, which is 4.19 x 10鲁 J/kg路K. This means that for every kilogram, raising the temperature by one Kelvin requires 4.19 x 10鲁 joules of energy.
This concept is vital in thermochemistry as it helps to understand how substances respond to heat. Different substances have different specific heat capacities, which influence how they undergo temperature changes during chemical reactions. Specific heat capacity applies to various scientific fields,
Heat of Reaction
The heat of reaction, also known as enthalpy change, is the heat energy absorbed or released during a chemical reaction. For the bombardier beetle, the reaction heats the spray from 20掳C to 100掳C.
In studies, the heat of reaction is calculated by using the formula: \[ Q = mc\Delta{T} \]where \( Q \) is the heat energy, \( m \) is the mass, and \( \Delta{T} \) is the temperature change. For this beetle, this amounts to 3.35 x 10鈦 J/kg, indicating a significant energy release per unit mass of chemical reacting.
Heat of reaction is an essential part of thermochemistry, helping to predict whether a reaction will be exothermic (releasing heat) or endothermic (absorbing heat). It provides insights into energy transfer processes and is crucial for applications in various scientific fields, such as energy production and material synthesis.

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Most popular questions from this chapter

What is the amount of heat input to your skin when it receives the heat released (a) by \(25.0 \mathrm{~g}\) of steam initially at \(100.0^{\circ} \mathrm{C},\) when it is cooled to skin temperature \(\left(34.0^{\circ} \mathrm{C}\right) ?\) (b) By \(25.0 \mathrm{~g}\) of water initially at \(100.0^{\circ} \mathrm{C},\) when it is cooled to \(34.0^{\circ} \mathrm{C}\) ? (c) What does this tell you about the relative severity of burns from steam versus burns from hot water?

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

You have probably seen people jogging in extremely hot weather. There are good reasons not to do this! When jogging strenuously, an average runner of mass \(68 \mathrm{~kg}\) and surface area \(1.85 \mathrm{~m}^{2}\) produces energy at a rate of up to \(1300 \mathrm{~W}, 80 \%\) of which is converted to heat. The jogger radiates heat but actually absorbs more from the hot air than he radiates away. At such high levels of activity, the skin's temperature can be elevated to around \(33^{\circ} \mathrm{C}\) instead of the usual \(30^{\circ} \mathrm{C}\). (Ignore conduction, which would bring even more heat into his body.) The only way for the body to get rid of this extra heat is by evaporating water (sweating). (a) How much heat per second is produced just by the act of jogging? (b) How much net heat per second does the runner gain just from radiation if the air temperature is \(40.0^{\circ} \mathrm{C}\left(104^{\circ} \mathrm{F}\right) ?\) (Remember: He radiates out, but the environment radiates back in.) (c) What is the total amount of excess heat this runner's body must get rid of per second? (d) How much water must his body evaporate every minute due to his activity? The heat of vaporization of water at body temperature is \(2.42 \times 10^{6} \mathrm{~J} / \mathrm{kg} .\) (e) How many \(750 \mathrm{~mL}\) bottles of water must he drink after (or preferably before!) jogging for a half hour? Recall that a liter of water has a mass of \(1.0 \mathrm{~kg}\).

A Styrofoam bucket of negligible mass contains \(1.75 \mathrm{~kg}\) of water and \(0.450 \mathrm{~kg}\) of ice. More ice, from a refrigerator at \(-15.0^{\circ} \mathrm{C},\) is added to the mixture in the bucket, and when thermal equilibrium has been reached, the total mass of ice in the bucket is \(0.884 \mathrm{~kg} .\) Assuming no heat exchange with the surroundings, what mass of ice was added?

There is \(0.050 \mathrm{~kg}\) of an unknown liquid in a plastic container of negligible mass. The liquid has a temperature of \(90.0^{\circ} \mathrm{C}\). To measure the specific heat capacity of the unknown liquid, you add a mass \(m_{w}\) of water that has a temperature of \(0.0^{\circ} \mathrm{C}\) to the liquid and measure the final temperature \(T\) after the system has reached thermal equilibrium. You repeat this measurement for several values of \(m_{\mathrm{w}},\) with the initial temperature of the unknown liquid always equal to \(90.0^{\circ} \mathrm{C}\). The plastic container is insulated, so no heat is exchanged with the surroundings. You plot your data as \(m_{\mathrm{w}}\) versus \(T^{-1}\), the inverse of the final temperature \(T\). Your data points lie close to a straight line that has slope \(2.15 \mathrm{~kg} \cdot \mathrm{C}^{\circ} .\) What does this result give for the value of the specific heat capacity of the unknown liquid?

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