/*! 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] | 91Ó°ÊÓ

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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

Camels require very little water because they are able to tolerate relatively large changes in their body temperature. While humans keep their body temperatures constant to within one or two Celsius degrees, a dehydrated camel permits its body temperature to drop to 34.0\(^\circ\)C overnight and rise to 40.0\(^\circ\)C during the day. To see how effective this mechanism is for saving water, calculate how many liters of water a 400-kg camel would have to drink if it attempted to keep its body temperature at a constant 34.0\(^\circ\)C by evaporation of sweat during the day (12 hours) instead of letting it rise to 40.0\(^\circ\)C. (Note: The specific heat of a camel or other mammal is about the same as that of a typical human, 3480 J/kg \(\cdot\) K. The heat of vaporization of water at 34\(^\circ\)C is \(2.42 \times 10{^6} J/kg\).)

Two rods, one made of brass and the other made of copper, are joined end to end. The length of the brass section is 0.300 m and the length of the copper section is 0.800 m. Each segment has cross-sectional area 0.00500 m\(^2\). The free end of the brass segment is in boiling water and the free end of the copper segment is in an ice–water mixture, in both cases under normal atmospheric pressure. The sides of the rods are insulated so there is no heat loss to the surroundings. (a) What is the temperature of the point where the brass and copper segments are joined? (b) What mass of ice is melted in 5.00 min by the heat conducted by the composite rod?

An asteroid with a diameter of 10 km and a mass of \(2.60 \times 10{^1}{^5} kg\) impacts the earth at a speed of 32.0 km/s, landing in the Pacific Ocean. If 1.00% of the asteroid's kinetic energy goes to boiling the ocean water (assume an initial water temperature of 10.0\(^\circ\)C), what mass of water will be boiled away by the collision? (For comparison, the mass of water contained in Lake Superior is about \(2 \times 10{^1}{^5} kg\).)

(a) On January 22, 1943, the temperature in Spearfish, South Dakota, rose from -4.0\(^\circ\)F to 45.0\(^\circ\)F in just 2 minutes. What was the temperature change in Celsius degrees? (b) The temperature in Browning, Montana, was 44.0\(^\circ\)F on January 23, 1916. The next day the temperature plummeted to -56\(^\circ\)F. What was the temperature change in Celsius degrees?

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 kg and surface area 1.85 m\(^2\) produces energy at a rate of up to 1300 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\(^\circ\)C instead of the usual 30\(^\circ\)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\(^\circ\)C (104\(^\circ\)F)? (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 \(2.42 \times 10{^6} J/kg\). (e) How many 750-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 kg.

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