chapter the energy content of food proofs

Food is essential to your enjoyment of life and your very existence. The composition of different foods governs their taste, their appearance, and the amount of nutrients and energy they supply when you eat them. Many people need to monitor their diet carefully, so it is important for this information to be reflected accurately on packaging.

In this chapter, you will compare the energy content of proteins, carbohydrates and fats. You will observe how the relative amounts of these in different foods influence the energy available from foods. You will learn how the energy content of foods is determined experimentally by burning them in a bomb calorimeter and how calorimeters can be calibrated to make the results obtained more accurate. Finally, you will look at how solution and bomb calorimetry can be used to determine the enthalpy change in a range of chemical reactions.

Key knowledge• The comparison of energy values of carbohydrates, proteins and fats and oils

• Glucose as the primary energy source, including a balanced thermochemical equation for cellular respiration

• The principles of calorimetry; solution and bomb calorimetry, including determination of calibration factor and consideration of the effects of heat loss; and analysis of temperature–time graphs obtained from solution calorimetry

VCE Chemistry Study Design extracts © VCAA (2015); reproduced by permission.

The energy content of foodCHAPTER

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AREA OF STUDY 2 | WHAT IS THE CHEMISTRY OF FOOD?2

17.1 Food—an energy sourceFood supplies the energy required for the millions of chemical reactions that occur in your body. Energy is needed for physical activity and functions such as breathing, as well as for the production of materials for growth and repair.

The balance between the energy supplied by the food you eat and the energy you use in everyday activities is a frequent topic of discussion. While thin fashion models are portrayed as having the ‘perfect’ body shape, studies reveal that obesity is on the rise and people are heavier than they used to be. In contrast, in some parts of the world, people can only dream of the opportunity to eat as much as their bodies need every day.

If you take in more energy from food than you use, you will gain weight. Similarly, if you use more energy than is supplied to your body, you will lose weight. It is important to know how much energy is available from particular foods, so processed foods are labelled with their energy content, as well as percentages of carbohydrates, fats and protein.

Table 17.1.1 summarises the nutrient and energy content of some common foods.

TABLE 17.1.1 The nutrient and energy content of some common foods (per 100 g)

Food Carbohydrates (g/100 g)

Fat (g/100 g)

Protein (g/100 g)

Energy (kJ/100 g)

Milk chocolate

52 30 8 2130

Donut

51 23 5 1803

Hot chips

40 19 3 1434

Apples

13.8 0.2 0.3 247

The unit of energy is the joule (J). Larger amounts of energy are commonly expressed in kilojoules (kJ). 1 kJ = 1000 J.

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3CHAPTER 17 | THE ENERGY CONTENT OF FOOD

ENERGY FROM GLUCOSEYou may remember from Chapter 15 that glucose is a monosaccharide, one of the simplest carbohydrates. Fructose, found in fruits and honey, is also a monosaccharide. All monosaccharides are white, crystalline solids with a sweet taste, and all have the formula C6H12O6.

Molecules of the monosaccharides contain a number of polar hydroxyl groups, enabling them to form hydrogen bonds with water. As a result, monosaccharides are highly soluble in water. The structure of glucose is shown in Figure 17.1.1.

CH2OH

HH

HO

H

H

H

O

OHOH

OHFIGURE 17.1.1 Structural formula of glucose. Carbon atoms in the ring have been omitted for clarity.

Remember that the smallest carbohydrates are the monosaccharides. With the molecular formula C6H12O6, glucose, fructose and galactose are isomers. Starch, cellulose and glycogen are polysaccharides—polymers of glucose molecules linked together in different ways by condensation reactions.

Glucose is found in all living things, especially in the juice of fruits, the sap of plants and the blood and tissue of animals. Both glucose and its polymer starch are more rapidly digested than other forms of food. They are the main sources of energy in most diets and the human body uses them for energy in preference to fats and proteins.

The process by which polysaccharides are broken down to form monosaccharides such as glucose is called hydrolysis. This process was described in detail in Chapter 15 on page xx.

Cellular respirationGlucose is the primary energy source for the cells of plants and animals and is used to obtain energy by a process known as cellular respiration. There are two main types of cellular respiration: • Aerobic respiration requires oxygen and is the main source of energy for the

human body.• Anaerobic respiration does not require oxygen and yields less energy.In aerobic respiration, glucose is oxidised to carbon dioxide and water through a sequence of reactions. The overall equation for aerobic respiration is:

C6H12O6(aq) + 6O2(g) → 6CO2(g) + 6H2O(l) ∆H = –2860 kJ mol–1

In humans, an alternative form of respiration, called anaerobic respiration, can occur in muscles during prolonged and vigorous exercise, when the supply of oxygen is limited. It can result in the build-up of lactate ions, CH3CH(OH)COO–, which can cause painful muscular cramping. The overall equation for anaerobic respiration in humans is:

C6H12O6(aq) → 2CH3CH(OH)COO–(aq) + 2H+(aq) ∆H = –120 kJ mol–1

In yeast, anaerobic respiration produces ethanol and carbon dioxide—a process widely used to produce alcoholic beverages. The equation for this process is:

C6H12O6(aq) → 2C2H5OH(aq) + 2CO2(g) ∆H = –69 kJ mol–1

These reactions for anaerobic and aerobic respiration are all exothermic. Although anaerobic respiration yields less energy per mole of glucose, it is often a faster process than aerobic respiration. Short bursts of exercise, such as sprinting, rely on anaerobic processes for energy because the individual steps in the reactions involved in anaerobic respiration occur more rapidly.

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EXTENSION

Exercise and lactate ionsAs athletes turn the final bend in a 400-metre race, they strive to maintain or increase their speed. They gasp for breath, their hearts race, and their legs feel like lead. Then when they finish, the athletes experience severe cramping pain in their muscles.

The experience of these athletes is a consequence of their muscles running out of oxygen to oxidise glucose. Despite the accelerated beating of their hearts and their rapid breathing, the athletes reach a point during the race when they experience ‘oxygen debt’: their blood can no longer provide oxygen at a sufficiently fast rate. From this stage onwards their bodies obtain energy anaerobically.

Unlike aerobic energy production, in which the products of glucose oxidation are water and carbon dioxide, anaerobic energy production involves the breakdown of glucose molecules into the lactate ion (‘lactic acid’), shown in Figure 17.1.2. Lactate production yields much less energy.

HHO CH3

C + 2H+

COO –

2

CH2OH

HH

HO

H

H

H

O

OHOH

OHglucose lactate ion

FIGURE 17.1.2 The anaerobic production of energy involves formation of two lactate ions from each glucose molecule.

The accumulation of lactate ions in the muscles causes much of the pain and fatigue that the athlete experiences (Figure 17.1.3). Once the race is over the discomfort eventually disappears as the lactate ions are oxidised to carbon dioxide and water. The training of elite athletes involves conditioning their bodies to use energy supplies as efficiently as possible.

FIGURE 17.1.3 The fatigue experienced by an athlete is partly due to the formation of lactate ions produced by the anaerobic breakdown of glucose molecules in muscle cells.

ENERGY VALUES OF CARBOHYDRATES, PROTEINS AND FATSA balanced diet is made up of a variety of foods containing carbohydrates, proteins and fats. Each of these three major nutrients provides a different quantity of energy per gram.

The energy content of foods is measured in kJ g–1, kJ/100 g or even kJ mol–1 if the food is a pure substance such as glucose. For most foods, the energy released on combustion is similar to the energy released when the food is oxidised during respiration.

For convenience, each of the major food nutrients—carbohydrates, fats and proteins—are considered to have a particular heat of combustion, although there is a range of values for different members of these food groups. For example, carbohydrates are considered to have a heat of combustion of 17 kJ g–1, whereas the heat of combustion of monosaccharides is 15.7 kJ g–1 and that of polysaccharides is 17.6 kJ g–1.

Table 17.1.2 compares the heats of combustion (energy content) and energy available to the body for each of the nutrients in food. The energy available to the body from a nutrient or food is called its energy value.

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TABLE 17.1.2 Comparison of the heat of combustion and energy available to the human body of the three main nutrients

Nutrient Energy content (heat of combustion)

(kJ g–1)

Energy value (energy available for the body)

(kJ g–1)

Carbohydrates 17 17

Fats and oils 39 37

Proteins 24 17

Note that fats and oils have a significantly higher energy value than carbohydrates and proteins. This is essentially due to the degree to which these molecules can be oxidised.

Carbohydrates tend to contain more oxygen atoms than fats and oils do. At a simple level, the carbon atoms in carbohydrate molecules have a higher ‘degree of oxidation’. Therefore, fats and oils have greater potential for oxidation and release more energy on combustion.

The following equations compare the combustion of a trisaccharide made up of three glucose units, C18H32O16, and the combustion of linoleic acid, C18H32O2, a fatty acid with the same number of carbons and hydrogens. As more oxygen is required to oxidise the fatty acid than the trisaccharide, more energy will be produced. This can be seen in the enthalpy values shown in the equations.

C18H32O16(s) + 18O2(g) → 18CO2(g) + 16H2O(l) ∆H = –6111 kJ mol–1

trisaccharideC18H32O2(s) + 25O2(g) → 18CO2(g) + 16H2O(l) ∆H = –8382 kJ mol–1

linoleic acid

Energy available to the bodyTable 17.1.2 shows that the energy released when food is burned is often greater than the energy that is available for the human body to use after the food has been digested. This can be due to:• incomplete absorption of nutrients by the body after digestion of the food• incomplete oxidation of nutrients, such as proteins and insoluble fibre• heat loss; not all of the energy released by the oxidation of glucose is available

for use in cells as some is lost as waste heat.Even within a particular nutrient class, such as the carbohydrates, the energy

available to the body can vary. Starch is readily digested by humans and is our primary source of glucose, so most of the energy from starch is absorbed during digestion. On the other hand, humans lack the enzyme cellulase, which is required to hydrolyse cellulose. Cellulose in foods such as vegetables is often referred to as dietary fibre and provides little energy.

CHEMFILE

The importance of celluloseEven though humans are unable to digest cellulose, a diet high in cellulose provides ‘bulk’ to aid the passage of food through the digestive system. Such ‘bulk’ helps prevent constipation and reduce the risk of bowel cancer, also known as colorectal cancer. Bowel cancer is the second most common cancer in men and women in Australia and is more common in people over the age of 50. High red meat consumption, especially processed meats, increases the risk of developing bowel cancer. This risk can be reduced by eating a healthy diet with plenty of fresh fruit and vegetables.

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Calculating the energy value of foodsWhen the percentage composition of a food is known, its energy value can be calculated using the percentages and the energy available to the body for each nutrient. The composition and energy value of a range of foods are given in Table 17.1.3.

TABLE 17.1.3 The composition and energy value of a range of foods

Food Carbohydrate (%)

Protein (%)

Fats and oils (%)

Energy value (kJ g–1)

White rice 79 7 Negligible 15.2

Wholemeal bread 39 11 4 9.7

Avocados 6 2 17 7.2

Roasted peanuts 18 26 50 24.3

Pizza (with cheese) 31 11 8 10.0

Apples 53 4 8 11.7

Almonds 19 19 54 25.4

Experimental methods for determining the energy value of foods will be discussed in section 17.2.

Worked example 17.1.1

CALCULATING THE ENERGY VALUE OF FOODS

Labelling on a sample of unsalted cashews indicates they contain 29.0% carbohydrates, 18.0% protein and 46.0% fat. The remaining 7.0% is water, which does not supply energy.

Calculate the energy value of the cashews, in kJ g–1.

Thinking Working

Use Table 17.1.2 to determine the available energy for each nutrient type.

Carbohydrate: 17 kJ g–1

Protein: 17 kJ g–1

Fat: 37 kJ g–1

Assuming that there is 100 g of the sample, multiply each percentage of the nutrient by the available energy per gram for the nutrient type.

Carbohydrate: 29.0 g × 17 kJ g–1 = 493 kJ

Protein: 18.0 g × 17 kJ g–1 = 306 kJ

Fat: 46.0 g × 37 kJ g–1 = 1702 kJ

Find the sum of the energies for the three nutrient types and divide by 100 to find the energy value in kJ g–1.

Energy value = 493 + 306 + 1702100

= 25 kJ g–1 (2 significant figures)

Worked example: Try yourself 17.1.1

CALCULATING THE ENERGY VALUE OF FOODS

Labelling on a sample of white bread indicates it contains 53.0% carbohydrates, 8.0% protein and 4.0% fat. The remaining 35.0% is water, which does not supply energy.

Calculate the energy value of the bread, in kJ g–1.

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• Glucose is a monosaccharide, one of the simplest types of carbohydrates, with the formula C6H12O6.

• Glucose is the primary energy source for humans.

• Cellular respiration is an exothermic process. There are two forms of cellular respiration: aerobic respiration and anaerobic respiration.

• In aerobic respiration, oxygen oxidises glucose according to the equation:

C6H12O6(aq) + 6O2(g) → 6CO2(g) + 6H2O(l) ∆H = –2860 kJ mol–1

• Anaerobic respiration does not require oxygen and yields less energy than aerobic respiration.

• The energy value of foods is measured in kJ g–1.

• The energy value of a food can be calculated using data for the available energy of the various components of the food and the percentages of each component of the food.

17.1 ReviewSUMMARY

KEY QUESTIONS

1 Match each quantity of energy in joules to the equal quantity in kilojoules.

Energy in joules (J) Energy in kilojoules (kJ)

100 1.00

1.0 × 104 1.0 × 10–4

10.0 0.100

1.00 × 103 1.00 × 10–2

0.10 10

2 Which one of the following is not a monosaccharide?A Starch B GlucoseC Fructose D Galactose

3 Use the following terms to complete the sentences about respiration in humans: cellular/intramuscular, endothermic/exothermic, oxidised/reduced, carbon dioxide/oxygen, released/absorbed, aerobic/anaerobic, less/more /the same.In _______________________ respiration, glucose is used by cells to obtain energy. Aerobic respiration is an ________________________ process in which the glucose is ______________________ by ______________________ . A relatively large amount of energy is ______________________ during aerobic respiration and can be used by the cells of the body. Another type of respiration, known as _____________________ respiration, occurs less frequently in human cells and releases ____________________ energy per mole of glucose than aerobic respiration.

4 The heat of combustion of a food is often larger than the energy that is available for the human body to use. Which one of the following reasons does not explain this observation?A The food is not completely absorbed by the body after digestion.B Energy is lost to the surroundings during combustion reactions.C Proteins may not be completely oxidised.D Dietary fibre (cellulose from plant material) is not digested.

5 Labelling on a sample of cheddar cheese indicates it contains 2.0% carbohydrates, 26.7% protein and 33.3% fat. The remaining 38.0% is water. Calculate the energy value of the cheese, in kJ g–1.

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17.2 Introducing calorimetry One of the simplest ways to measure the energy in a sample of food is to burn it. The energy released when the food is burned can be transferred to water and the change in the temperature of the water is measured.

If the food is burned in an enclosed space with an ample supply of oxygen, this energy determination can be very accurate, with little energy lost to the surroundings. For most foods, the energy released during combustion is slightly more than the energy released when the food is oxidised during respiration.

In this section, you will learn the principles of the techniques used to determine the energy content of food.

CALORIMETRYCalorimetry is the experimental method of measuring the heat energy released or absorbed by a chemical reaction or physical process, such as by the combustion of a fuel or a food.

In the simplest form of calorimetry, the heat energy released by a chemical reaction can be transferred to water in a separate container. The quantity of energy transferred can be calculated from the increase in temperature of the water in the calorimeter.

Transferring heat energy to waterWhen an exothermic chemical reaction, such as the combustion of food, is carried out underneath a container of water, such as a test tube, some of the heat released by the combustion reaction is transferred to the water. This can be seen in Figure 17.2.1. (A similar experimental arrangement was described in Chapter 2, on page XXX, for determining the approximate amount of heat energy released by the combustion of fuels.)

The heat energy transferred to the volume of water can be calculated by measuring the:• initial temperature of the water• highest temperature of the water• volume of water.

The relationship used to calculate the energy that has been transferred to the water is:

q = m × C × ∆Twhere q is the energy that is transferred to the water (in J), m is the mass of the water (in g), C is the specific heat capacity of the water (4.18 J °C–1 g–1), and ∆T is the change in temperature of the water (in °C or K):

∆T = ∆Tfinal – ∆Tinitial

Recall that the density of water is 1.0 g mL–1, so 1 mL of water has a mass of 1 g.

Water is distinguished by its very high heat capacity, which is a consequence of the hydrogen bonding between its molecules. See Heinemann Chemistry 1, section 12.2, for more detail on the specific heat capacity of water and calculations using specific heat capacity.

FIGURE 17.2.1 Heat energy from a burning biscuit can be transferred to a measured volume of water in a test tube.Pa

ge Pr

oofs



CALCULATING THE HEAT ENERGY TRANSFERRED TO WATER FROM BURNING FOOD

A sample of burning food is used to heat 100 mL of water. Calculate the heat energy, in kilojoules, that has been transferred if the temperature of the water increases from 18.5°C to 44.0°C.

Thinking Working

Change the volume of water in millilitres to mass of water in grams. Remember that the density of water is 1.00 g mL–1.

1 mL of water has a mass of 1 g, so 100 mL of water has a mass of 100 g.

Calculate the change in temperature, ∆T, by subtracting the initial temperature from the final temperature:



= 44.0 – 18.5

= 25.5°C

Calculate the heat energy transferred to the water, in joules, using the formula:

q = m × C × ∆T

q = m × C × ∆T

= 100 × 4.18 × 25.5

= 10 659 J

Express the quantity of energy, in kJ, to the appropriate number of significant figures. Remember that to convert from joules to kilojoules, you divide by 103.

q = 10 6591000

= 10.7 kJ (3 significant figures)


CALCULATING THE HEAT ENERGY TRANSFERRED TO WATER FROM BURNING FOOD

A sample of burning food is used to heat 200 mL of water. Calculate the heat energy, in kilojoules, that has been transferred if the temperature of the water increases from 15.5°C to 30.0°C.

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Heat lossWhen energy is transferred from a burning piece of food, or another fuel, across an open space, heat will be lost to the surroundings such as the air around the burning food. Similarly if there is no lid on a container of water, heat will be lost from the surface of the water. This heat loss is illustrated in Figure 17.2.3.

thermometer

beaker

burning corn chip

water

heat being lost heat being lost

FIGURE 17.2.3 Heat is lost to the surroundings when burning a piece of food to heat some water.

CHEMFILE

‘Calorie’ or ‘calorie’A unit called the calorie (cal) was used in the old imperial system of measurement to measure the energy present in food. A calorie is the amount of heat required to raise the temperature of 1 g of water by 1°C. 1 cal is equal to 4.18 J.

To make matters more complicated, a unit called the Calorie (with a capital ‘C’) was also used. 1 Cal is equal to 1000 cal.

Although imperial units were withdrawn from general use in Australia between 1970 and 1988 and replaced with metric units, such as the joule, it is still common to find references to calories, or Calories, in dietary advice, as shown in Figure 17.2.2, especially when weight loss is a targeted aim for the use of a food product.

Impo

rtan

t fa

cts

and

figu

res

- Trans

- Polyunsaturated

- Monounsaturated

Energy

Protein

Fat, Total

- Saturated

Carbohydrate

- Sugars

Sodium

Dietary fibre

Quantity per serving

Nutrition informationServings per package: 11 serving size: 11.4 g (2 Biscuits)

181 kJ (43 Cal) 1590 kJ (380 Cal)

11.3 g

1.2 g

0.3 g

0.0 g

0.6 g

0.2 g

77.5 g

3.7 g

4.1 g

849 mg

2.1%

2.6%

0.2%

0.1%

2.8%

0.5%

1.7%

4.2%

1.3 g

0.1 g

0.0 g

0.0 g

8.8 g

0.0 g

0.1 g

0.5 g

97 mg

0.4 g

Quantity per 100 g

* Percentage daily intakes are based on an average adult diet of 8700 kJ. Your daily intakes may be higher or lower depending onyour energy needs. All values considered averages unless otherwise indicated.

% daily intake* (per serving)

FIGURE 17.2.2 Nutrition information from a packet of crispbread biscuits. Notice the energy values in kilojoules are also converted to Calories.

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When some heat energy from the burning food is transferred to the surrounding air, the temperature of the water does not increase as much as it would if all the energy was used to heat the water. A lower change in temperature, ∆T, of the water results in a lower energy value, q.

There are a number of ways to reduce heat loss during the calorimetry experiment shown in Figure 17.2.3, including:• putting a lid on the calorimeter• insulating the beaker of water (with flameproof material)• placing insulation around the burning food, although sufficient oxygen must

reach the food for combustion to be complete.

ESTIMATING THE ENERGY CONTENT OF FOODThe amount of energy transferred to a fixed volume of water when a sample of food is burned can be calculated using the relationship:

q = m × C × ∆TIf the mass of food burned to produce this energy is measured, the energy

content of the food (energy released per gram of food) can be calculated as follows:

Energy content = energy transferred to the water

change in mass of the food during combustion =

q∆m

The mass of food burned is determined by measuring the initial mass of food, and then subtracting the final mass. It is not necessary to burn all the food sample, as long as the mass that is burned is measured.

This calculation of energy content assumes that all the energy is transferred from the burning food to the volume of water. Substantial heat losses will occur when the experiment described above is performed, so the values calculated for the energy content will be less than the actual values. At best, this experiment could only be used to obtain an estimate of a food’s energy content or perhaps to compare the energy released by two or more foods.

As you will see shortly, modifications to the experimental equipment allow much more accurate measurements. However, before looking at these modifications, Worked Example 17.2.2 gives you practice at calculating energy content from simple experimental data.

Worked example 17.2.2ESTIMATING THE ENERGY CONTENT OF A SAMPLE OF FOOD

A 0.850 g plain biscuit was burned under a steel can containing 150 mL of water. After the flame went out, the mass of the biscuit was 0.300 g and the temperature of the water had risen by 22.5°C. Calculate the energy content of the biscuit in kJ g–1.

Thinking Working

Calculate the heat energy absorbed by the water in joules, using the formula:

q = m × C × ∆T

q = m × C × ∆T

= 150 × 4.18 × 22.5

= 14 108 J

Express the quantity of energy in kJ. Remember that to convert from joules to kilojoules, you divide by 103.

q (in kJ) = 14 1081000

= 14.108 kJ

Calculate the mass of the food that was burned by subtracting the final mass from the initial mass:

∆m = ∆minital – ∆mfinal

∆m = ∆minital – ∆mfinal

= 0.850 – 0.300

= 0.550 g

Calculate the energy content of the food by dividing the energy transferred to the water by the change in mass during combustion:

Energy content = q

∆m

Energy content = q

∆m

= 14.1080.550

= 25.7 kJ g–1 (3 significant figures)

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Worked example: Try yourself 17.2.2ESTIMATING THE ENERGY CONTENT OF A SAMPLE OF FOOD

A 2.500 g corn chip was burned under a steel can containing 200 mL of water. After the flame went out, the mass of the corn chip was 1.160 g and the temperature of the water had risen by 35.0°C. Calculate the energy content of the corn chip in kJ g–1.

CALORIMETERSEnergy changes that occur during chemical and physical changes are measured with a device called a calorimeter. Calorimeters are constructed in such a way that the energy losses that occur in the simple experimental apparatus described earlier are minimised, enabling more accurate results. In a calorimeter, almost all of the heat energy released or absorbed is transferred directly to or from a measured volume of water.

Two types of calorimeters are designed for measuring the energy changes in different types of reactions.• In solution calorimeters, the reaction takes place in a solution.• In bomb calorimeters, the reaction takes place in a sealed bomb vessel.

Solution calorimetryEnergy changes that occur in reactions can be measured using a solution calorimeter. A solution calorimeter may be as simple as a polystyrene foam coffee cup with a lid, as shown in Figure 17.2.4.

The insulation provided by the polystyrene foam prevents the transfer of heat to or from the surroundings of the calorimeter. The reaction is carried out in the calorimeter with an accurately known volume of water. The initial and final temperatures are measured and recorded, as are the amounts of reactants used.

Measurement of enthalphy change, ΔH, was explained in detail in Chapter 2, page xx. Reactions that involve the release of energy are exothermic (ΔH is negative); reactions that involve the absorption of energy are endothermic (ΔH is positive).

If the temperature of the water in the calorimeter increases, the reaction occurring in the calorimeter is an exothermic reaction. The reaction has released heat energy, the water in the calorimeter has absorbed that energy and the temperature of the water has increased.

Similarly, if the temperature of the water in the calorimeter decreases, the reaction occurring in the calorimeter has absorbed energy from the water. In this case, the reaction is an endothermic reaction.

A coffee cup calorimeter has some limitations. The polystyrene container absorbs some heat, so the temperature change will be lower than it would otherwise be and the calculated value for the heat released or absorbed by the reaction will be lower than it should be.

A solution calorimeter cannot be used to measure the energy content of food because the reaction in this case is a combustion reaction in which the food burns in oxygen. However, a solution calorimeter can be used to determine the energy change that occurs when a food such as glucose dissolves in water. Solution calorimetry is used in the laboratory to determine the enthalpy changes that occur when acids react with bases and solids dissolve in water.

thermometer

polystyrenecover

polystyrenecups

stirrer

reactants in solution

FIGURE 17.2.4 A simple ‘coffee-cup’ calorimeter.

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The construction of a laboratory solution calorimeter is shown in Figure 17.2.5. The stirrer is used to ensure the temperature of the water is uniform. The use of the electrical heater for calibration of the calorimeter is described in the next section.

electric heater forcalibrating the calorimeterstirrerthermometer

insulatedcontainer

glass bulb containingsecond reactant

solution ofone reactant

FIGURE 17.2.5 A solution calorimeter; breaking the glass bulb starts the reaction.

Bomb calorimetryFigure 17.2.6 shows the components of a bomb calorimeter, which is used for reactions that involve gaseous reactants or products. The reaction vessel in a bomb calorimeter is designed to withstand the high pressures that may build up during reactions.

thermometer

electric heater forcalibrating the

calorimeterelectric heater

to ignite sample stirrer

insulated container

pressurised vessel

oxygen underpressure

water

sample in crucible

FIGURE 17.2.6 A bomb calorimeter used for reactions that involve gases, such as the combustion of foods.

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CHEMISTRY IN ACTION

Food processing and calorimetryA walk down the supermarket aisles reveals the seemingly endless variety of processed foods available to the consumer. Preventing the growth of microorganisms and slowing the oxidation of fats that cause rancidity are two reasons why food requires preserving. Traditional methods of food preservation available to home cooks include drying, heating, freezing, salting, pickling and smoking.

Modern, commercial processing techniques involve heat treatment for preservation of the food product or to develop texture, flavours and colours. Instruments that can simulate the conditions found in food processing, enable food scientists to improve processing techniques and ensure a high-quality product.

Calorimetry can be used to establish a connection between temperature and specific physical properties of substances. Differential scanning calorimetry (DSC) is a thermal analysis technique that measures how physical properties of a sample change, along with temperature, against time. Such measurements can provide both quantitative and qualitative information about physical and chemical changes that involve endothermic and exothermic processes.

A typical differential scanning calorimeter is shown in Figure 17.2.7.

Food processing involves either heating or cooling materials so DSC is particularly suitable for analysis of food systems. The energy information from DSC can be used directly to understand the thermal changes that food may undergo during processing or storage.

DSC can be used to study the effects of temperature changes on proteins, carbohydrates, fats and oils.

• Fats, oils and spreads: oxidation stability, melting and crystallisation behaviour

• Protein: denaturation and thermal stability

• Pastes and gels containing polysaccharides: specific heat capacity, melting and crystallisation behaviour

• Flour and rice starch: changes in the granules of starch due to gelatinisation

Food scientists measure the energy content of food by burning the food in a bomb calorimeter. Sufficient oxygen is required to completely combust the food so that all the available energy is released. Insulation around the calorimeter prevents heat escaping and the change in temperature is measured with a thermometer. The stirrer ensures that the temperature of the water is uniform.

Calorimetry calculations for both bomb and solution calorimeters are explained in more detail in section 17.3.

FIGURE 17.2.7 A differential scanning calorimeter is used to measure the effects of temperature on substances.

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• Calorimetry is the experimental method by which the heat energy released or absorbed in a chemical reaction or physical process is measured.

• During calorimetry, energy is transferred to a fixed volume of water and the temperature change is measured. The quantity of energy that is transferred can be calculated using the equation: q = m × C × ∆T.

• A calorimeter is an instrument that measures energy changes in a reaction. It is made up of an insulated container of water in which the reaction occurs, with a stirrer and thermometer to measure the temperature change during the reaction. A lid is an important part of the insulation.

• If the insulation around a calorimeter is insufficient, or missing, or if there is no lid, heat will be lost from the water to the surroundings and the calculated value measured for ∆T will be lower.

• A solution calorimeter is an insulated container that holds a known volume of water and in which a reaction in solution, such as dissolution of a solid, or a neutralisation reaction, can be carried out.

• A bomb calorimeter is an insulated container in which a sealed, oxygen filled reaction vessel, is surrounded by a known volume of water. Combustion reactions are carried out in the reaction vessel and the heat from the reaction is transferred to the surrounding water.

17.2 ReviewSUMMARY

KEY QUESTIONS

1 Match each definition on the left to a symbol on the right.

Final temperature – initial temperature t

Specific heat capacity q

Initial mass – final mass ∆T

Heat energy C

Time ∆m

2 Calculate the energy required to increase the temperature of:a 100 mL of water by 15.2°b 500 mL of water from 16.0°C to 28.0°Cc 1.50 kg of water from 20.0°C to 30.0°C.

3 Calculate the final temperature of 250 g of water which is initially at 18.0°C if the amount of energy transferred to the water is 4.50 kJ.

4 Calculate the energy content, in kJ g–1, of a wafer biscuit if combustion of 5.00 g of biscuit heats 500 mL of water by 38.3°C.

5 If the energy content of corn chips is 22.0 kJ g–1, what is the mass of corn chips that is burned when 34.7 kJ of energy is transferred to 200 mL of water, assuming all the energy from the burning chips is absorbed by the water?

6 Classify the following investigations according to whether they would be best carried out in a bomb calorimeter or a solution calorimeter.a Determining the energy content of a corn chipb Determining the enthalpy of neutralisation of

hydrochloric acidc Determining the enthalpy of the reaction between

magnesium and hydrochloric acidd Determining the enthalpy of combustion of sucrosee Determining the enthalpy of solution of glucosef Determining the energy content of dried bread.

7 Use the following terms to complete the statements about calorimeters: absorbed, decreases, endothermic, exothermic, increases, insulating, lid, lost, released, required, stays the same, thermometer._______________ a calorimeter improves the accuracy of measurement of the amount of energy _______________ or absorbed by a chemical reaction.Heat energy can be _______________ from a calorimeter, so a(an) _______________ is a useful form of insulation.If the reaction occurring in a calorimeter is exothermic, the temperature of the water _______________. If the reaction occurring is _______________ the temperature of the water in a calorimeter decreases.Pa

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17.3 Calibration of calorimetersA calorimeter is useful for determining how much energy is released or absorbed by chemical reactions. A bomb calorimeter, such as the one shown in Figure 17.3.1, is used to determine the energy released by combustion reactions, including the combustion of foods. Calorimeters can also be used for reactions in solution, including neutralisation reactions and the dissolution of solids in water.

Heat energy is often lost to the surroundings when it is transferred, so it is useful to be able to allow for losses that occur within a calorimeter when measuring the energy change associated with a reaction. A relationship has to be established between the heat energy being transferred and the temperature change that occurs within the calorimeter.

In this section, you will learn how calorimeters can be calibrated, so that the data they supply is more accurate and the effect of heat loss is minimised. You will also learn how a calibrated bomb calorimeter can be used to determine the energy content of food in kJ g–1 or the enthalpy of combustion of a pure food, such as glucose, in kJ mol–1.

FIGURE 17.3.1 Bomb calorimeter temperature changes being observed by a laboratory technician. Bomb calorimeters are used to measure the heat released by combustion reactions. The rise in temperature is used to calculate the energy released.

CHEMFILE

Pierre BerthelotPierre Berthelot (1827–1907) was a French pharmacist and professor of organic chemistry. He was one of the first scientists to use the word ‘synthesis’ to describe the production of organic compounds from their elements. Berthelot is also known for the development of new experimental techniques in the field of thermochemistry. He introduced the terms ‘exothermic’ to describe a reaction that releases heat and ‘endothermic’ for a reaction that absorbs heat.

Berthelot also devised methods for conducting chemical reactions within a closed chamber surrounded by water at a known temperature. The quantity of heat produced by the reaction could then be determined by measuring the rise in temperature of the water at the completion of the reaction.

From these experiments, Berthelot introduced the bomb calorimeter, shown in Figure 17.3.2. For the combustion reactions of gases, the gas under test was mixed with excess oxygen, compressed and then sparked. Using a bomb calorimeter, Berthelot could determine heats of combustion more accurately than previously known. Berthelot, his students and other collaborators compiled reliable data for heats of combustion, solution and neutralisation.

FIGURE 17.3.2 Berthelot’s bomb calorimeter. The reaction vessel was lined with platinum to avoid corrosion from acidic vapours and solutions.

CALIBRATION OF CALORIMETERSWhen a reaction takes place in a calorimeter, the heat change causes a rise or fall in the temperature of the contents of the calorimeter. Although the formula q = m × C × ∆T can be used to determine the relationship between heat energy and temperature change, the heat losses that occur can lead to inaccurate results.

For more accurate measurements of the heat produced in a reaction, you have to determine first how much energy is required to change the temperature of the water by 1°C for the particular calorimeter in use. This is known as the calibration factor (CF) of the calorimeter. For energy measured in joules, the calibration factor has the unit J °C−1. Once the calibration factor of a calorimeter is known, it is said to be calibrated.

Both the solution and bomb calorimeters described in section 17.2 have electric heaters, which can be used for calibration (see Figures 17.2.5 and 17.2.6 on page XX).

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Electrical calibration of calorimetersA calorimeter can be calibrated by using the electric heater to release a known quantity of thermal energy and measuring the resultant rise in temperature of the water in the calorimeter.

The thermal energy released when an electric current passes through the heater can be calculated from the formula:

energy (in joules) = voltage (volts) × current (amps) × time (seconds), orE = V × I × t

If the temperature change, ∆T, caused by the addition of the known amount of energy is measured, the calibration factor for the calorimeter can be calculated from the expression:

CF = E∆T

= VIt∆T

Temperature–time graphs in calorimetryMeasuring the temperature change, ∆T, that occurs during a calorimetry experiment is not always as simple as calculating the difference between the final temperature and the initial temperature of the water in the calorimeter. Because a calorimeter is not perfectly insulated, it slowly loses heat during and after the heater is operating. A more accurate determination of the temperature change (∆T) can be achieved by plotting a graph of temperature against time before, during and after the calibration.

Two temperature–time graphs for the calibration of different calorimeters are shown in Figure 17.3.3. Figure 17.3.3a shows the results for a calorimeter with ‘perfect’ insulation and no heat loss. Figure 17.3.3b is typical of the results obtained using school laboratory calorimeters.

Time (s)

line of best fit

The graph is extrapolated backto when the current is turned on.

instantcurrent

turned on

instantcurrent

turned on

instantcurrent

turned off

ΔT

ΔT

Tem

pera

ture

(°C

)

19.0

20.0

21.0

22.0

23.0

24.0

30 60 90 120 150 180 210 240 270 300 330 360 390 420 450 480 510

ΔT = Tfinal – Tinitial

= 24.4 – 20.0= 4.4°C

ΔT = Tfinal – Tinitial

= 22.0 – 20.0= 2.0°C

Time (s)

Tem

pera

ture

(°C

)

20.0

21.0

22.0

23.0

24.0

25.0

30 60 90 120 150 180 210 240 270 300 330 360 390 420 450 480 510

instantcurrent

turned off

(a)

(b)

FIGURE 17.3.3 Graphs of temperature in a solution calorimeter before, during and after calibration. (a) These results would be obtained when a perfectly insulated calorimeter is calibrated. (b) These results might be obtained when the temperature readings are affected by heat loss and non-instantaneous heat transfer.

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In Figure 17.3.3b, heat loss causes a negative slope to the line of the graph after the heater has been turned off. A more accurate estimate of the value of ∆T can be found by extrapolating the line back to the time when heating commenced. The temperature change between this value and the initial temperature can be measured. A delay in the transfer of heat through the water can be observed by the continuing increase in temperature after the heater has been turned off. This can also be accounted for by the extrapolation method, although it is important to understand that the accuracy of this approach is limited.


CALCULATING THE CALIBRATION FACTOR OF A CALORIMETER BY ELECTRICAL CALIBRATION

A bomb calorimeter was calibrated by passing 1.50 A through the electric heater for 50.5 s at a potential difference of 6.05 V. The temperature of the water in the calorimeter was initially 18.05°C and rose to 19.38°C during the calibration. Determine the calibration factor of the calorimeter.

Thinking Working

Calculate the thermal energy released by the heater in the calorimeter when the electric current was passed through it.

Use the equation:

E = V × I × t

E = V × I × t

= 6.05 × 1.50 × 50.5

= 458 J

Calculate the temperature change during the calibration.

Use the equation:



= 19.38 – 18.05

= 1.33°C

Calculate the calibration factor by dividing the energy by the change in temperature. Use the equation:

CF = E∆T

= VIt∆T

CF = E∆T

= 4581.33

= 344 J °C–1


CALCULATING THE CALIBRATION FACTOR OF A CALORIMETER BY ELECTRICAL CALIBRATION

A bomb calorimeter was calibrated by passing 1.05 A through the electric heater for 120 s at a potential difference of 5.90 V. The temperature of the water in the calorimeter was initially 15.20°C and rose to 17.50°C during the calibration. Determine the calibration factor of the calorimeter.

If you know the calibration factor for a calorimeter, you can use it to determine the energy change that is responsible for a temperature change during a reaction in the calorimeter. These calculations will be discussed later in this section.Page

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Chemical calibration of calorimetersA calorimeter may also be calibrated by performing a chemical reaction in the calorimeter that releases a known quantity of thermal energy and then measuring the resultant rise in temperature.

Solid benzoic acid (C6H5COOH) is commonly used to calibrate bomb calorimeters. The enthalpy of combustion, ∆Hc, of benzoic acid is –3227 kJ mol–1.

Calibration is achieved by burning a known amount (in moles) of benzoic acid in the calorimeter. The change in temperature (∆T) is measured and used to calculate the calibration factor, as shown in Worked Example 17.3.2.

The energy released during calibration is calculated using ∆Hc for the reaction and the amount of reactant (n):

E = n × ∆Hc

The calibration factor is then calculated using:

CF = E∆T


CALCULATING THE CALIBRATION FACTOR OF A CALORIMETER BY CHEMICAL CALIBRATION

A bomb calorimeter was calibrated by the complete combustion of 1.000 g of benzoic acid (M(C6H5COOH) = 122.0 g mol–1; ∆Hc = –3227 kJ mol–1). The temperature of the water in the calorimeter rose from 17.50°C to 21.67°C during the calibration. Determine the calibration factor of the calorimeter.

Thinking Working

Determine the amount, in moles, of benzoic acid that undergoes complete combustion. Use the equation:

n = mM

n = mM

= 1.000122.0

= 0.008 197 mol

Calculate the thermal energy released by the complete combustion of the benzoic acid in the calorimeter.

Use the equation E = n × ∆Hc, noting that the energy will be in kJ.

E = n × ∆Hc

= 0.008 197 × 3227

= 26.35 kJ

Calculate the temperature change during the calibration. Use the equation:



= 21.67 – 17.50

= 4.17°C

Calculate the calibration factor by dividing the energy by the change in temperature. Use the equation:

CF = E∆T

CF = E∆T

= 26.454.17

= 6.34 kJ °C–1


CALCULATING THE CALIBRATION FACTOR OF A CALORIMETER BY CHEMICAL CALIBRATION

A bomb calorimeter was calibrated by the complete combustion of 1.245 g of benzoic acid (M(C6H5COOH) = 122.0 g mol–1; ∆Hc = –3227 kJ mol–1). The temperature of the water in the calorimeter rose from 17.50°C to 24.70°C during the calibration. Determine the calibration factor of the calorimeter.

Enthalpy of combustion was discussed in Chapter 2 on page xx and is defined as the enthalpy change that occurs when one mole of the compound is burned completely in oxygen. It has the symbol ∆Hc and is measured in kJ mol–1.

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USING A CALORIMETER TO DETERMINE THE ENERGY CONTENT OF FOODA bomb calorimeter may be calibrated by either the electrical or the chemical calibration methods described earlier in this section.

When an accurately weighed food sample is burned in a bomb calorimeter as shown in Figure 17.3.4, the heat released by the combustion is transferred into the water surrounding the reaction vessel and the change in temperature is measured. This is a very accurate method of determining the energy content of a food sample.

The calibration factor is used to determine the energy (E) that is responsible for the temperature change that occurs during the combustion reaction:

E = CF × ∆TNote that the value of ∆T is the change in temperature that occurs when the

reaction is carried out in the calorimeter, not when the calorimeter is calibrated.The energy content of the food, in kJ g–1, is calculated by dividing the energy

change, in kJ, by the mass of the food that was combusted.

Energy content of food = Em


CALCULATING THE ENERGY CONTENT OF A SAMPLE OF FOOD USING A BOMB CALORIMETER

A 1.670 g cheese snack was burned in a bomb calorimeter. The calibration factor of the bomb calorimeter was 6.34 kJ °C–1. During the combustion of the snack, the temperature of the water rose by 5.85°C. The mass of the ash left over at the end of the experiment was negligible. Calculate the energy content of the snack in kJ g–1.

Thinking Working

Calculate the heat energy released by the food, in kJ, using the equation:

E = CF × ∆T

E = CF × ∆T

= 6.34 × 5.85

= 37.089 kJ

Calculate the energy content of the food using the equation:

Energy content = Em

Express your answer to the correct number of significant figures.

There was a negligible mass left over after combustion, so the mass reacting was 1.670 g.

Energy content = Em

= 37.0891.670

= 22.2 kJ g–1 (3 significant figures)


CALCULATING THE ENERGY CONTENT OF A SAMPLE OF FOOD USING A BOMB CALORIMETER

7.00 g of pizza biscuits were burned in a bomb calorimeter. The calibration factor of the bomb calorimeter was 8.45 kJ °C–1. During the combustion of the biscuits, the temperature of the water rose by 17.75°C. The mass of the ash left over at the end of the experiment was negligible. Calculate the energy content of the biscuits in kJ g–1.

FIGURE 17.3.4 A food scientist inserts a food sample into a bomb calorimeter.

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CALORIMETRY INVOLVING PURE SUBSTANCESSolution and bomb calorimeters can be used to determine the enthalpy changes, ∆H, for different chemical reactions. Worked Example 17.3.4 shows how the enthalpy of combustion of glucose, ∆Hc in kJ mol–1, can be determined.


CALCULATING THE ENTHALPY OF COMBUSTION OF A PURE COMPOUND USING A BOMB CALORIMETER

9.396 g of glucose (molar mass 180.0 g mol–1) is completely burned in a bomb calorimeter. The calibration factor of the calorimeter was 9.56 kJ °C–1. During the combustion of the glucose, the temperature of the water rose by 15.27°C. Calculate the enthalpy of combustion of glucose in kJ mol–1.

Thinking Working

Calculate the amount of reactant in moles, using the equation:

n = mM

n = mM

= 9.396180.0

= 0.052 20 mol

Calculate the heat energy released, in kJ, using the equation:

E = CF × ∆T

E = CF × ∆T

= 9.56 × 15.27

= 1463 kJ

Calculate the energy released per mole.

Energy released per mole = En

= 14630.052 20

= 2797

= 2.80 × 103 kJ mol–1

State the enthalpy of combustion with the correct sign, remembering that combustion is an exothermic process.

∆Hc(glucose) = –2.80 × 103 kJ mol–1 (3 significant figures)


CALCULATING THE ENTHALPY OF COMBUSTION OF A PURE COMPOUND USING A BOMB CALORIMETER

1.164 g of oleic acid (molar mass 284.0 g mol–1) was completely burned in a bomb calorimeter. The calibration factor of the calorimeter was 10.20 kJ °C–1. During the combustion of the oleic acid, the temperature of the water rose by 4.51°C. Calculate the enthalpy of combustion of oleic acid in kJ mol–1.

A solution calorimeter can be used to determine energy changes that occur for reactions in solution. Worked Example 17.3.5 shows how the heat of solution of glucose can be determined. The heat of solution of glucose is the ∆H when one mole of glucose dissolves in water.Pa

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CALCULATING THE HEAT OF SOLUTION

A solution calorimeter has a calibration factor of 400 J ˚C–1. If the temperature decreases by 1.375˚C when 9.00 g of glucose (M = 180.0 g mol–1) is dissolved in the water of the calorimeter, calculate the enthalpy of solution of glucose.

Thinking Working

Calculate the amount of reactant in moles, using the equation:

n = mM

n = mM

= 9.00180.0

= 0.0500 mol

Calculate the heat energy released or absorbed, in kJ, using the equation:

E = CF × ∆T

E = CF × ∆T

= 400 × 1.375

= 550 J

= 0.550 kJ

Calculate the energy released per mole.

Energy change per mole = En

= 0.5500.0500

= 11.0 kJ mol–1

State the heat of solution with the correct sign, remembering that a temperature increase indicates an exothermic process and a temperature decrease indicates an endothermic process.

∆H(glucose) = +11.0 kJ mol–1 (3 significant figures)


CALCULATING THE HEAT OF SOLUTION

The heat of solution of sucrose, C12H22O11 (M = 342.0 g mol–1) can be determined using solution calorimetry. When 41.587 g of sucrose was dissolved in 200 mL of water in a calorimeter, the temperature decreased by 1.65˚C. The calibration factor of the calorimeter was 396.4 J °C–1. Calculate the heat of solution of sucrose.Page

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• Solution and bomb calorimeters can be calibrated to establish the relationship between the energy transferred to the water and the temperature change in the calorimeter.

• Calibration involves adding a known quantity of heat energy from an electrical source or from a chemical source.

• A calibration factor tells you how much energy is required to change the temperature of the water in a calorimeter by 1°C.

• Electrical energy may be calculated using the equation E = V × I × t, where V is the voltage, in volts, I is the current, in amps, and t is the time, in seconds.

• For electrical calibration of a calorimeter, the calibration factor is calculated using the equation

CF = E∆T

where E = VIt.

• During electrical calibration, the value of the temperature change, ∆T, may be affected by heat

loss from the calorimeter, so a temperature–time graph is used to determine a more accurate value of ∆T.

• For the chemical calibration of a bomb calorimeter using the combustion of benzoic acid, the calibration factor is calculated using the equation

CF = E∆T

, where E = n × ∆Hc. ∆Hc is the heat of

combustion of benzoic acid.

• When a reaction is carried out in a calibrated calorimeter, the energy released by the reaction is determined using the equation E = CF × ∆T.

• The energy content of food, in kJ g–1, is determined

using the equation energy content = Em

, where energy is measured in kJ.

• The enthalpy of combustion, in kJ mol–1, of pure nutrients such as glucose and fatty acids can be determined using a calibrated bomb calorimeter. In this case the enthalpy of combustion is

determined using the equation ∆Hc = En.

17.3 ReviewSUMMARY

KEY QUESTIONS

1 Identify whether the following features of calorimeters apply to bomb calorimeters, solution calorimeters or both.a The calorimeter contains water that is heated up by the reaction.b The reaction occurs in the water.c The initial and final temperatures of water must be known.d Energy changes for gaseous reactions can be measured.e Food can be burned in the calorimeter to find energy content.f A calorimeter factor can be found for the calorimeter.

2 The temperature of a calorimeter increases by 8.5°C when 4.5 A of current is passed through it at a potential of 5.2 V for 4 minutes. Which one of the following is the calibration factor of the calorimeter?A 110 J °C–1 B 661 J °C–1 C 796 J °C–1 D 5616 J °C–1

3 A bomb calorimeter was calibrated by passing 1.25 A through the electric heater for 90.5 s at a potential difference of 4.40 V. The temperature of the water in the calorimeter was initially 16.600°C and rose to 17.500°C during the calibration. Calculate the calibration factor of the calorimeter.

4 A bomb calorimeter containing 100 mL of water is calibrated by passing a 3.00 A current through the instrument for 240 s at a potential difference of 3.50 V. The temperature of the water in the calorimeter rises by 1.80°C.a Calculate the calibration factor for the calorimeter and water.b A piece of fish finger of mass 5.00 g was burned in the calibrated bomb

calorimeter. If the temperature of the water increased by 14.8°C, calculate the energy content of the fish finger in kJ g–1.

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5 A solution calorimeter was calibrated by the electrical method, using a current of 3.80 A for 150 s with a voltage of 6.50 V. The temperature–time graph for the calibration is shown in Figure 17.3.5. The heater was turned on at t = 120 s and was turned off at t = 270 s. Calculate the calibration factor for this calorimeter.

15.5

16

15

16.5

17

17.5

18

18.5

19

19.5

20

20.5

21

21.5

22

120 180Time (s)

Tem

pera

ture

(°C

)

240 30030 60 90 150 210 270 330 360 390 420 450 480 510

FIGURE 17.3.5 Graph of temperature against time during electrical calibration of a solution calorimeter.

6 During the chemical calibration of a bomb calorimeter by the complete combustion of 0.153 g of benzoic acid (M(C6H5COOH) = 122.0 g mol–1; ∆Hc = –3227 kJ mol–1), the temperature of the water in the calorimeter increased by 5.80°C. Calculate the calibration factor for this calorimeter.

7 Which one of the following quantities could be expressed in kJ mol–1?A Heat of combustion of a chicken-flavoured biscuitB Heat of combustion of ascorbic acidC Energy content of an appleD Energy value of a chocolate biscuit

8 1.45 g of citric acid (C6H8O7: molar mass = 192.0 g mol–1) was completely burned in a bomb calorimeter. The calibration factor of the bomb calorimeter was 601 J °C–1. During the combustion of the citric acid, the temperature of the water rose by 24.63°C. Calculate the enthalpy of combustion of citric acid in kJ mol–1.

9 The heat of solution of sucrose, C12H22O11 (M = 342.0 g mol–1) can be determined using solution calorimetry. When 39.94 g of sucrose was dissolved in 200 mL of water in a calibrated calorimeter, the temperature decreased by 1.50°C. The calibration factor of the calorimeter was 420.1 J °C–1. Calculate the heat of solution of sucrose, including the sign to indicate whether it is exothermic or endothermic.

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KEY TERMS

Food—an energy source1 Aerobic cellular respiration involves the oxidation

of glucose by oxygen. Which one of the following alternatives lists the products of aerobic respiration?A Water and glycogenB Water and carbon dioxideC Oxygen and carbon dioxideD Lactic acid and ethanol

2 Which one of the following nutrients will produce the most energy when consumed?A 15 g proteinB 8 g fatC 20 g carbohydrateD 5 g fat and 10 g protein

3 The composition of roasted peanuts and of almonds is given in Table 17.1.3 on page xx. Consider this information as you complete the statements by selecting from the following terms: almonds, roasted peanuts, lower, higher, 1998, 1.998, 1850, 850._________________ have the greater fat content and this results in these nuts having a significantly ________________ energy value. The energy obtained from the fat in almonds would be ________kJ per 100 g of nuts whereas the fat in roasted peanuts would only yield _________ kJ per 100 g of nuts.

4 Labelling for a stuffed pepper indicates it contains 16% carbohydrates, 13% protein and 6.0% fat. The rest of the pepper is water.Calculate the energy value of the stuffed pepper, in kJ g–1.

Introducing calorimetry5 Calculate the amount of energy that has been

transferred to 50.0 mL of water in a test tube when the temperature of the water increases from 18.5°C to 29.8°C.

6 The equation q = m × C × ∆T describes the relationship between heat energy and the temperature change of a substance being heated. It is used to determine how much energy has been absorbed by a material, such as a mass of water, when its temperature increases by a measured amount.

Which one of the following statements about this equation is correct?A The specific heat capacity, C, depends on the

material that is generating the heat energy.B The mass, m, represents the mass of the material

that is being heated.C q is measured in kJ °C–1.D ∆T refers to the change in temperature of the

fuel generating the heat.

7 A 500 mL volume of water in an electric kettle was heated using 136 kJ of electrical energy. If the water was initially at 15.0˚C, calculate its final temperature.

8 A peanut with a mass of 1.200 g was burned under a steel can containing 200 mL of water. After the flame went out the mass of the peanut was 0.750 g and the temperature of the water had risen by 13.2°C. Calculate the energy content of the peanut in kJ g–1, assuming all the energy from the burning peanut is transferred to the water.

9 A biscuit has an energy content of 12.8 kJ g–1. What volume of water, in litres, would be heated from 20.0°C to 60.0°C by the transfer of all of the energy from the combustion of one biscuit with a mass of 17.0 g?

Calibration of calorimeters10 A bomb calorimeter can be used to determine

the energy content of a food such as popcorn.a Rearrange the following steps to describe how a

bomb calorimeter can be electrically calibrated.• Measure and record the highest temperature

reached of the water.• Turn off and disconnect the electrical circuit.• Connect the heater of the bomb calorimeter

into an electirc circuit with a voltmeter and an ammeter in the circuit.

• Add an accurately measured and recorded volume of water to a calorimeter and measure and record the temperature of the water.

• Turn on the power supply and read the current and voltage for a predetermined time, such as 120 s. Record these values.

Chapter review

aerobic respirationanaerobic respirationbomb calorimetercalibratedcalibration factorcaloriecalorimeter

calorimetrycarbohydratecellular respirationdietary fibreendothermic reactionenergy contentenergy value

enthalpy of combustion

exothermic reactionextrapolateheat of combustionheat of solutionhydrolysis

monosaccharidesolution calorimeterspecific heat capacity

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b Rearrange the following steps to describe how a calibrated bomb calorimeter is used to determine the energy content of a food such as popcorn.• Record the highest temperature reached in

the calorimeter.• Add the pre weighed food to be burned into the

calorimeter and connect the oxygen supply and ignition device.

• Monitor the temperature of the water until the temperature no longer increases.

• Add the same volume of fresh water as used for the calibration.

• Record the initial temperature of the water in the calorimeter.

• Ignite the food using the ignition device.

11 A calorimeter is calibrated by the passage of 2.75 A through an electric heater for 150 s at a potential difference of 4.05 V. A temperature increase of 3.40°C occurs during the calibration. Determine the calibration factor.

12 A temperature rise of 1.78°C was observed when 1.00 × 10–3 mol of propane gas was burned in a calorimeter. The calibration factor of the calorimeter was previously determined to be 1250 J °C–1.a Write an equation for the combustion of propane.b Calculate the energy change during the reaction.c Calculate the heat of combustion for propane in

kJ mol–1.d Write a thermochemical equation for the

combustion of propane.

13 When 1.01 g of solid potassium nitrate was dissolved in a calorimeter, the temperature dropped by 0.672°C. When the calorimeter was calibrated, a current of 1.50 A applied at a potential difference of 5.90 V for 60.0 s caused a temperature rise of 0.456°C.a Write an equation for the dissolution of potassium

nitrate.b Determine the calibration factor of the calorimeter.c Calculate the energy change during the reaction.d Write the thermochemical equation for the reaction.

14 A bomb calorimeter was calibrated by the complete combustion of 2.405 g of benzoic acid (M(C6H5COOH) = 122.0 g mol–1; ∆Hc = –3227 kJ mol–1). The temperature of the water in the calorimeter rose from 17.50°C to 26.35°C during the calibration. Determine the calibration factor of the calorimeter.

15 20.0 g of snack food was completely burned in a bomb calorimeter with a calibration factor of 14.32 kJ °C–1. During the combustion of the snack food, the temperature of the water rose by 28.98°C. Calculate the energy content of the snack food in kJ g–1.

16 A bomb calorimeter was calibrated electrically. A potential of 8.20 V was applied for 120 s with a current of 4.25 A and the temperature increased by 1.02°C.4.50 g of glucose (molar mass 180.0 g mol–1) was burned completely in the same calibrated bomb calorimeter. During the combustion of the glucose, the temperature of the water rose by 16.74°C. What is the enthalpy of combustion of glucose in kJ mol–1?

17 A bomb calorimeter has a calibration factor of 15.40 kJ °C–1. Calculate the temperature increase if 25.0 g of caramel popcorn (energy content 18.0 kJ g–1) was completely burned in the calorimeter.

18 A bomb calorimeter was found to have a calibration factor of 7.89 kJ °C–1. It was calibrated by the complete combustion of 1.375 g of benzoic acid (M(C6H5COOH) = 122.0 g mol–1; ∆Hc = –3227 kJ mol–1). The temperature of the water in the calorimeter was initially 16.70°C. Determine the final temperature reached by the calorimeter during the calibration.

19 Palmitic acid (C16H32O2) (M = 256.0 g mol–1; ∆Hc = –10 031 kJ mol–1) was burnt in a bomb calorimeter. Calculate the mass that would cause the temperature of the water in the calorimeter, with calibration factor of 25.3 kJ °C–1, to increase by 10.0°C.

Connecting the main ideas20 Use the following terms to complete the sentences

about calorimeters and their calibration: absorbed, calibrating, combustion, decrease, electrical, energy, heat, increase, neutralisation, released, temperature, water.A bomb calorimeter is suitable for determining the energy released by a __________________ reaction.The heat energy released by the reaction is transferred to a fixed volume of __________________ and the temperature __________________ is measured using a thermometer.__________________ a calorimeter is important because it allows us to relate the change in __________________ of the water in the calorimeter with a specific amount of __________________ that is released into or absorbed from the water.

21 Which one of the following is most likely to be the temperature change in 100.0 mL of water in a beaker with no insulation and no lid, when 2.10 kJ of energy is produced by a burning biscuit under the beaker?A 3.20°CB 5.02°CC 5.50°CD 6.30°C

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22 A bomb calorimeter was calibrated by the passage of 4.50 A for 4.00 minutes at a voltage of 6.15 V. The temperature of the water in the calorimeter increased by 2.00°C during the calibration.This calorimeter was then used to determine the heat content of a sample of potato crisps. 5.00 g of potato crisps was burned in the calorimeter and the temperature of the water in the calorimeter rose by 33.9°C.Calculate the energy content of the potato crisps in kJ g–1.

23 A food van serves cinnamon donuts from the side of the road.a A 70.0 g cinnamon donut was described as

containing 14.0 g fat, 28.0 g carbohydrate and 4.5 g protein.Calculate the theoretical energy value in kJ g–1 of this donut.

b 10.0 g of cinnamon donut was combusted in a calibrated bomb calorimeter. The temperature of the water in the calorimeter increased by 11.2°C. The calibration factor of this calorimeter was 13.2 kJ °C–1. Calculate the energy content in kJ g–1 of this donut using this experimental data.

24 A calorimeter is calibrated by the electrical calibration method. The temperature rise was measured as the difference between the highest temperature reached and the initial temperature.Which one of the following would be the most likely outcome if the insulation surrounding the calorimeter was not adequate?A Heat would escape from the calorimeter, but the

calibration factor would not change.B Heat would escape from the calorimeter and the

calibration factor would be measured as being smaller than it should be.

C Heat would escape from the calorimeter and the calibration factor would be measured as being larger than it should be.

D Heat would transfer into the calorimeter and the calibration factor would be measured as being larger than it should be.

25 A student is having difficulty with the calculations for the calibration of a solution calorimeter and the calculations that follow. The student wants to determine the heat energy per gram required to dissolve honey in water. The student’s data for the calibration are shown in the tables below.

TABLE 1 Calibration data

Volume of water in calorimeter (mL) 100

Voltage (V) 5.40

Current (amps) 2.60

Time (s) 180

Initial temperature (°C) 15.5

Final temperature (°C) 17.8

TABLE 2 Data from experiment in which honey is dissolved in water

Volume of water in calorimeter (mL) 100

Mass of honey (g) 19.80

Initial temperature (°C) 17.8

Final temperature (°C) 16.1

The student’s calculations are as follows:

CF = 5.40 × 2.60 × 180 = 2527.2 JHeat energy per gram to dissolve the honey:E = CF × ∆T = 2527.2 × 1.7 = 4296.24 J

Energy per gram = 4296.24100

= 42.96 J g–1

0.430 kJ of energy is given out when 1 gram of honey dissolves. This is an exothermic reaction.

Describe any errors that the student has made in the calculations, and then complete the calculation correctly.

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26 Figure 17.4.1 shows the temperature–time graph plotted by the data logging equipment used by a student to determine the change in temperature during the calibration of a solution calorimeter. The heater was turned on at t = 120 s and it was turned off at t = 300 s.

18.5

19

19.5

20

20.5

21

21.5

22

18

120 180Time (s)

Tem

pera

ture

(°C

)

240 30030 60 90 150 210 270 330 360 390 420 450 480 510

FIGURE 17.4.1 Temperature–time graph used for calibration of a calorimeter

The student ignored the graph and instead used the formula ∆T = Tfinal – Tinitial to determine ∆T and hence the calibration factor. The student then correctly made the additional measurements needed to determine the enthalpy change of a reaction. Which of the following statements would best describe the student’s results?A The value of ∆T will be too high.B The calibration factor calculated for the calorimeter

will be too high.C The value of the enthalpy change measured by the

calorimeter will be too low.D The effect on the experimental results would

cancel out when the calibration factor was used to calculate energy changes.

27 Several mistakes or errors that could occur when calibrating a calorimeter are listed in the following table. For each mistake or error, indicate the likely effect on the calculated calibration factor by placing a tick in the appropriate column.

Mistake or error Calibration factor is too small

Calibration factor would be correct

Calibration factor is too large

The lid is left off during calibration.

Less water is used in the calorimeter for the calibration than for subsequent measurements.

During electrical calibration, the voltmeter had a systematic error making it read too low.

During chemical calibration, the mass of benzoic acid was recorded as less than its actual value.

28 A student conducted an experiment to determine the energy content of jelly babies. A bomb calorimeter was electrically calibrated by passing a current of 4.05 A at a potential difference of 6.10 V for 240 s. The temperature increased by 13.90°C. Approximately half a jelly baby of mass 5.147 g was placed into the combustion chamber of the calorimeter. The combustion chamber was then filled with oxygen and the jelly baby ignited. The temperature of the water surrounding the combustion chamber increased from 18.50°C to 48.20°C. The remains of the jelly baby weighed 4.207 g.a Determine the calibration factor of the calorimeter.b Calculate the energy content of the jelly baby

in kJ g–1.c State the major sources of error that may have

affected the student’s result.

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chapter the energy content of food proofs

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