CHEMISTRY THE CENTRAL SCIENCE

5 THERMOCHEMISTRY

5.5 CALORIMETRY

The value of ΔH can be determined experimentally by measuring the heat flow accompanying a reaction at constant pressure. Typically, we can determine the magnitude of the heat flow by measuring the magnitude of the temperature change the heat flow produces. The measurement of heat flow is calorimetry; a device used to measure heat flow is a calorimeter.

Heat Capacity and Specific Heat

The more heat an object gains, the hotter it gets. All substances change temperature when they are heated, but the magnitude of the temperature change produced by a given quantity of heat varies from substance to substance. The temperature change experienced by an object when it absorbs a certain amount of heat is determined by its heat capacity, denoted C. The heat capacity of an object is the amount of heat required to raise its temperature by 1 K (or 1 °C). The greater the heat capacity, the greater the heat required to produce a given increase in temperature.

For pure substances the heat capacity is usually given for a specified amount of the substance. The heat capacity of one mole of a substance is called its molar heat capacityCm. The heat capacity of one gram of a substance is called its specific heat capacity, or merely its specific heat. The specific heat, Cs, of a substance can be determined experimentally by measuring the temperature change, Δ T, that a known mass m of the substance undergoes when it gains or loses a specific quantity of heat q:

For example, 209 J is required to increase the temperature of 50.0 g of water by 1.00 K. Thus, the specific heat of water is

A temperature change in kelvins is equal in magnitude to the temperature change in degrees Celsius: ΔT in K = Δ T in °C.  (Section 1.4) Therefore, this specific heat for water can also be reported as 4.18 J/g-°C.

Because the specific heat values for a given substance can vary slightly with temperature, the temperature is often precisely specified. The 4.18 J/g-K value we use here for water, for instance, is for water initially at 14.5 °C (FIGURE 5.17). Water's specific heat at this temperature is used to define the calorie at the value given in Section 5.1: 1 cal = 4.184 J exactly.

GO FIGURE

Is the process shown in the figure endothermic or exothermic?

FIGURE 5.17 Specific heat of water

When a sample absorbs heat (positive q), its temperature increases (positive ΔT). Rearranging Equation 5.21, we get

Thus, we can calculate the quantity of heat a substance gains or loses by using its specific heat together with its measured mass and temperature change.

TABLE 5.2 lists the specific heats of several substances. Notice that the specific heat of liquid water is higher than those of the other substances listed. The high specific heat of water affects Earths climate because it makes the temperatures of the oceans relatively resistant to change.

GIVE IT SOME THOUGHT

Which substance in Table 5.2 undergoes the greatest temperature change when the same mass of each substance absorbs the same quantity of heat?

TABLE 5.2 • Specific Heats of Some Substances at 298 K

SAMPLE EXERCISE 5.5 Relating Heat, Temperature Change, and Heat Capacity

(a) How much heat is needed to warm 250 g of water (about 1 cup) from 22 °C (about room temperature) to 98 °C (near its boiling point)? (b) What is the molar heat capacity of water?

SOLUTION

Analyze In part (a) we must find the quantity of heat (q) needed to warm the water, given the mass of water (m), its temperature change (Δ T), and its specific heat (Cs). In part (b) we must calculate the molar heat capacity (heat capacity per mole, Cm) of water from its specific heat (heat capacity per gram).

Plan (a) Given Cs, m, and Δ T, we can calculate the quantity of heat, q, using Equation 5.22. (b) We can use the molar mass of water and dimensional analysis to convert from heat capacity per gram to heat capacity per mole.

Solve

(a) The water undergoes a temperature change of

Using Equation 5.22, we have

(b) The molar heat capacity is the heat capacity of one mole of substance. Using the atomic weights of hydrogen and oxygen, we have

From the specific heat given in part (a), we have

PRACTICE EXERCISE

(a) Large beds of rocks are used in some solar-heated homes to store heat. Assume that the specific heat of the rocks is 0.82 J>g-K. Calculate the quantity of heat absorbed by 50.0 kg of rocks if their temperature increases by 12.0 °C. (b) What temperature change would these rocks undergo if they emitted 450 kJ of heat?

Answers: (a) 4.9 × 105 J, (b) 11 K decrease = 11 °C decrease

Constant-Pressure Calorimetry

The techniques and equipment employed in calorimetry depend on the nature of the process being studied. For many reactions, such as those occurring in solution, it is easy to control pressure so that ΔH is measured directly. Although the calorimeters used for highly accurate work are precision instruments, a simple “coffee-cup” calorimeter (FIGURE 5.18) is often used in general chemistry laboratories to illustrate the principles of calorimetry. Because the calorimeter is not sealed, the reaction occurs under the essentially constant pressure of the atmosphere.

GO FIGURE

Propose a reason for why two Styrofoam® cups are often used instead of just one.

FIGURE 5.18 Coffee-cup calorimeter. This simple apparatus is used to measure temperature changes of reactions at constant pressure.

Imagine adding two aqueous solutions, each containing a reactant, to a coffee-cup calorimeter. Once mixed, the reactants can react to form products. In this case there is no physical boundary between the system and the surroundings. The reactants and products of the reaction are the system, and the water in which they are dissolved is part of the surroundings. (The calorimeter apparatus is also part of the surroundings.) If we assume that the calorimeter is perfectly insulated, then any heat released or absorbed by the reaction will raise or lower the temperature of the water in the solution. Thus, we measure the temperature change of the solution and assume that any changes are due to heat transferred from the reaction to the water (for an exothermic process) or transferred from the water to the reaction (endothermic). In other words, by monitoring the temperature of the solution, we are seeing the flow of heat between the system (the reactants and products in the solution) and the surroundings (the water that forms the bulk of the solution).

For an exothermic reaction, heat is “lost” by the reaction and “gained” by the water in the solution, so the temperature of the solution rises. The opposite occurs for an endothermic reaction: Heat is gained by the reaction and lost by the water in the solution, and the temperature of the solution decreases. The heat gained or lost by the solution, qsoln, is therefore equal in magnitude but opposite in sign to the heat absorbed or released by the reaction, qrxnqsoln = –qrxn. The value of qsoln is readily calculated from the mass of the solution, its specific heat, and the temperature change:

For dilute aqueous solutions we usually assume that the specific heat of the solution is the same as that of water, 4.18 J/g-K.

Equation 5.23 makes it possible to calculate qrxn from the temperature change of the solution in which the reaction occurs. A temperature increase (ΔT> 0) means the reaction is exothermic (qrxn < 0).

SAMPLE EXERCISE 5.6 Measuring ΔH Using a Coffee-Cup Calorimeter

When a student mixes 50 mL of 1.0 M HCl and 50 mL of 1.0 M NaOH in a coffee-cup calorimeter, the temperature of the resultant solution increases from 21.0 °C to 27.5 °C. Calculate the enthalpy change for the reaction in kJ/mol HCl, assuming that the calorimeter loses only a negligible quantity of heat, that the total volume of the solution is 100 mL, that its density is 1.0 g/mL, and that its specific heat is 4.18 J/g-K.

SOLUTION

Analyze Mixing solutions of HCl and NaOH results in an acid-base reaction:

We need to calculate the heat produced per mole of HCl, given the temperature increase of the solution, the number of moles of HCl and NaOH involved, and the density and specific heat of the solution.

Plan The total heat produced can be calculated using Equation 5.23. The number of moles of HCl consumed in the reaction must be calculated from the volume and molarity of this substance, and this amount is then used to determine the heat produced per mol HCl.

Solve

Because the total volume of the solution is 100 mL, its mass is

The temperature change is

Using Equation 5.23, we have

Because the process occurs at constant pressure,

To express the enthalpy change on a molar basis, we use the fact that the number of moles of HCl is given by the product of the volume (50 mL = 0.050 L) and concentration (1.0 M = 1.0 mol/L) of the HCl solution:

Thus, the enthalpy change per mole of HCl is

Check ΔH is negative (exothermic), which is expected for the reaction of an acid with a base and evidenced by the fact that the reaction causes the temperature of the solution to increase. The magnitude of the molar enthalpy change seems reasonable.

PRACTICE EXERCISE

When 50.0 mL of 0.100 M AgNO3 and 50.0 mL of 0.100 M HCl are mixed in a constant-pressure calorimeter, the temperature of the mixture increases from 22.30 °C to 23.11 °C. The temperature increase is caused by the following reaction:

Calculate ΔH for this reaction in kJ/mol AgNO3, assuming that the combined solution has a mass of 100.0 g and a specific heat of 4.18 J/g °C.

Answer: –68,000 J/mol = –68 kJ/mol

Bomb Calorimetry (Constant-Volume Calorimetry)

An important type of reaction studied using calorimetry is combustion, in which a compound reacts completely with excess oxygen. (Section 3.2) Combustion reactions are most accurately studied using a bomb calorimeter (FIGURE 5.19). The substance to be studied is placed in a small cup within an insulated sealed vessel called a bomb. The bomb, which is designed to withstand high pressures, has an inlet valve for adding oxygen and electrical leads. After the sample has been placed in the bomb, the bomb is sealed and pressurized with oxygen. It is then placed in the calorimeter and covered with an accurately measured quantity of water. The combustion reaction is initiated by passing an electrical current through a fine wire in contact with the sample. When the wire becomes sufficiently hot, the sample ignites.

The heat released when combustion occurs is absorbed by the water and the various components of the calorimeter (which all together make up the surroundings), causing the water temperature to rise. The change in water temperature caused by the reaction is measured very precisely.

To calculate the heat of combustion from the measured temperature increase, we must know the total heat capacity of the calorimeter, C cal. This quantity is determined by combusting a sample that releases a known quantity of heat and measuring the temperature change. For example, combustion of exactly 1 g of benzoic acid, C6H5COOH, in a bomb calorimeter produces 26.38 kJ of heat. Suppose 1.000 g of benzoic acid is combusted in a calorimeter, leading to a temeperature increase of 4.857 °C. The heat capacity of the calorimeter is then Ccal = 26.38 kJ/4.857 °C = 5.431 kJ/°C. Once we know C cal, we can measure temperature changes produced by other reactions, and from these we can calculate the heat evolved in the reaction, qrxn:

Measurements made with a bomb calorimeter are generally more precise than those made with a coffee-cup calorimeter.

GO FIGURE

Why is a stirrer used in calorimeters?

FIGURE 5.19 Bomb calorimeter.

SAMPLE EXERCISE 5.7 Measuring qrxn Using a Bomb Calorimeter

The combustion of methylhydrazine (CH6N2), a liquid rocket fuel, produces N2, CO2 (g), and H2O(l):

When 4.00 g of methylhydrazine is combusted in a bomb calorimeter, the temperature of the calorimeter increases from 25.00 °C to 39.50 °C. In a separate experiment the heat capacity of the calorimeter is measured to be 7.794 kJ/°C. Calculate the heat of reaction for the combustion of a mole of CH6N2.

SOLUTION

Analyze We are given a temperature change and the total heat capacity of the calorimeter. We are also given the amount of reactant combusted. Our goal is to calculate the enthalpy change per mole for combustion of the reactant.

Plan We will first calculate the heat evolved for the combustion of the 4.00-g sample. We will then convert this heat to a molar quantity.

Solve

For combustion of the 4.00-g sample of methylhydrazine, the temperature change of the calorimeter is

We can use ΔT and the value for Ccal to calculate the heat of reaction (Equation 5.24):

We can readily convert this value to the heat of reaction for a mole of CH6N2:

Check The units cancel properly, and the sign of the answer is negative as it should be for an exothermic reaction. The magnitude of the answer seems reasonable.

PRACTICE EXERCISE

A 0.5865-g sample of lactic acid (HC3H5O3) is burned in a calorimeter whose heat capacity is 4.812 kJ/°C. The temperature increases from 23.10 °C to 24.95 °C. Calculate the heat of combustion of lactic acid (a) per gram and (b) per mole.

Answers: (a) –15.2 kJ/g, (b) –1370 kJ/mol

Because reactions in a bomb calorimeter are carried out at constant volume, the heat transferred corresponds to the change in internal energy, Δ E, rather than the change in enthalpy, ΔH (Equation 5.14). For most reactions, however, the difference between Δ E and ΔH is very small. For the reaction discussed in Sample Exercise 5.7, for example, the difference between Δ E and ΔH is about 1 kJ/mol—a difference of less than 0.1%. It is possible to correct the measured heat changes to obtain ΔH values, and these form the basis of the tables of enthalpy used in the following sections. We need not concern ourselves with how these small corrections are made.

CHEMISTRY AND LIFE
THE REGULATION OF BODY TEMPERATURE

For most of us, being asked the question “Are you running a fever?” was one of our first introductions to medical diagnosis. Indeed, a deviation in body temperature of only a few degrees indicates something amiss. In the laboratory you may have observed how difficult it is to maintain a solution at a constant temperature. Yet our bodies maintain a near-constant temperature in spite of widely varying weather, levels of physical activity, and periods of high metabolic activity (such as after a meal).

Maintaining a near-constant temperature is one of the primary physiological functions of the human body. Normal body temperature generally ranges from 35.8 °C to 37.2 °C (96.5 °F to 99 °F). This very narrow range is essential to proper muscle function and to control of the rates of the biochemical reactions in the body. You will learn more about the effects of temperature on reaction rates in Chapter 14.

The portion of the human brain stem called the hypothalamus regulates body temperature—in essence, the hypothalamus acts as a thermostat for the body. When body temperature rises above the normal range, the hypothalamus triggers mechanisms to lower the temperature. It likewise triggers mechanisms to increase the temperature if body temperature drops too low.

To understand how the body's heating and cooling mechanisms operate, we can view the body as a thermodynamic system. The body increases its internal energy content by ingesting foods from the surroundings. The foods, such as glucose (C6H12O6), are metabolized—a process that is essentially controlled oxidation to CO2 and H2O:

Roughly 40% of the energy produced is ultimately used to do work in the form of muscle contractions and nerve cell activities. The remainder is released as heat, part of which is used to maintain body temperature. When the body produces too much heat, as in times of heavy physical exertion, it dissipates the excess to the surroundings.

Heat is transferred from the body to its surroundings primarily by radiation, convection, and evaporation. Radiation is the direct loss of heat from the body to cooler surroundings, much as a hot stovetop radiates heat to its surroundings. Convection is heat loss by virtue of heating air that is in contact with the body. The heated air rises and is replaced with cooler air, and the process continues. Warm clothing, which usually consists of insulating layers of material with “dead air” in between, decreases convective heat loss in cold weather. Evaporative cooling occurs when perspiration is generated at the skin surface by the sweat glands (FIGURE 5.20). Heat is removed from the body as the perspiration evaporates into the surroundings. Perspiration is predominantly water, so the process is the endothermic conversion of liquid water into water vapor:

FIGURE 5.20 Perspiration!

The speed with which evaporative cooling occurs decreases as the atmospheric humidity increases, which is why we feel more sweaty and uncomfortable on hot, humid days.

When the hypothalamus senses that body temperature has risen too high, it increases heat loss from the body in two principal ways. First, it increases blood flow near the skin surface, which allows for increased radiational and convective cooling. The reddish, “flushed” appearance of a hot individual is the result of this increased subsurface blood flow. Second, the hypothalamus stimulates secretion of perspiration from the sweat glands, which increases evaporative cooling. During extreme activity, the amount of liquid secreted as perspiration can be as high as 2 to 4 liters per hour. As a result, the body's water supply must be replenished during these periods. If the body loses too much liquid through perspiration, it will no longer be able to cool itself and blood volume decreases, which can lead to either heat exhaustion or the more serious and potentially fatalheat stroke, during which the body temperature can rise to as high as 41 °C to 45 °C (106 °F to 113 °F). However, replenishing water without replenishing the electrolytes lost during perspiration can also lead to serious problems, as pointed out in the “Chemistry and Life” box in Section 4.5.

When body temperature drops too low, the hypothalamus decreases blood flow to the skin surface, thereby decreasing heat loss. It also triggers small involuntary contractions of the muscles; the biochemical reactions that generate the energy to do this work also generate heat for the body. When these contractions get large enough—as when the body feels a chill—a shiver results. If the body is unable to maintain a temperature above 35 °C (95 °F), the very dangerous condition called hypothermia can result.

The ability of the human body to maintain its temperature by “tuning” the amount of heat it transfers to and from its surroundings is truly remarkable. If you take courses in human anatomy and physiology, you will see many other applications of thermochemistry and thermodynamics to the ways in which the human body works.