Calculating Heat Energy Released Revision Notes for VCE SSCE Chemistry

Calculating Heat Energy Released

Understanding the energy released by different fuels helps us compare their suitability for specific purposes. For example, aircraft fuel differs from car or bus fuel because they have different energy requirements and operating conditions.

When fuels undergo combustion, they release heat energy. This energy release is described in thermochemical equations using the enthalpy change ( $\Delta H$ ) value, which is based on the stoichiometric ratios in the equation. By measuring how much the temperature of water increases when heated by a burning fuel, we can experimentally determine the heat energy released.

Specific heat capacity of water

What is specific heat capacity?

Specific heat capacity tells us how much energy is required to increase the temperature of a substance. It measures the amount of energy (in joules) needed to raise the temperature of a specific quantity of that substance (usually $1$ gram) by $1°\text{C}$ .

Symbol: $c$

Units: joules per gram per degree Celsius ( $\text{J g}^{-1} \text{ °C}^{-1}$ ) or joules per gram per kelvin ( $\text{J g}^{-1} \text{ K}^{-1}$ )

infoNote

An increase of $1°\text{C}$ is the same as an increase of $1 \text{ K}$ , so both units are interchangeable for temperature changes.

Specific heat capacities of common liquids

Different substances have different specific heat capacities. Here are values for some common liquids:

Substance	Specific heat capacity ( $\text{J g}^{-1} \text{ °C}^{-1}$ )
Water	$4.18$
Glycerine	$2.43$
Ethanol	$2.46$
Hexane	$2.26$
Olive oil	$1.97$
Paraffin oil	$2.13$

Water has a relatively high specific heat capacity compared to other common substances. This means water requires more energy to increase its temperature by $1°\text{C}$ compared to most other liquids.

Why does water have a high specific heat capacity?

The specific heat capacity of a substance depends on the types of bonds holding the molecules, ions, or atoms together. Water has a specific heat capacity of 4.18 J g⁻¹ °C⁻¹, which means $4.18$ joules of heat energy are needed to increase the temperature of $1$ gram of water by $1°\text{C}$ .

chatImportant

This relatively high value is due to the hydrogen bonds between water molecules. These bonds must absorb significant energy before the temperature increases. The higher the specific heat capacity, the more effectively a material can store heat energy.

Comparing water and glycerine

When substances are heated, their temperature rises at different rates depending on their specific heat capacity. The graph below shows how water and glycerine respond differently when the same amount of heat energy is supplied:

The graph demonstrates that:

$1$ gram of water increases by $1°\text{C}$ when supplied with $4.18 \text{ J}$ of heat energy
$1$ gram of glycerine increases by $1°\text{C}$ when supplied with $2.43 \text{ J}$ of heat energy
When equal amounts of heat energy are supplied, glycerine's temperature rises more rapidly than water's temperature

Real-world application: Hot beach, cool bay

The different heat capacities of water and sand are evident at the beach on a hot summer's day. Sand that is not in contact with water can heat very quickly to the point where it can burn your feet. In contrast, the water, even in rock pools, remains cool and refreshing.

infoNote

The specific heat capacity of water is $4.18 \text{ J g}^{-1} \text{ °C}^{-1}$ , while that of sand is $0.48 \text{ J g}^{-1} \text{ °C}^{-1}$ . This means water can absorb nearly 10 times as much energy as sand for the same temperature increase. Each gram of water absorbs $4.18 \text{ J}$ before it increases by $1°\text{C}$ , while each gram of sand absorbs only $0.48 \text{ J}$ before it increases by $1°\text{C}$ .

Calculating heat energy transferred to water

The specific heat capacity of water allows us to calculate the heat energy (in joules) needed to increase the temperature of a given mass of water by a specific amount.

The heat energy formula

The heat energy transferred to water can be calculated using measurements of:

Initial temperature of the water
Highest (final) temperature of the water
Mass of water

The relationship used to calculate the energy transferred is:

$\text{Heat energy} = \text{mass of water} \times \text{specific heat capacity} \times \text{temperature change}$

Using symbols, the equation is written as:

$q = m \times c \times \Delta T$

or simply:

chatImportant

$q = mc\Delta T$

where:

$q$ is the amount of heat energy (in $\text{J}$ )
$m$ is the mass (in $\text{g}$ )
$c$ is the specific heat capacity of water ( $4.18 \text{ J g}^{-1} \text{ °C}^{-1}$ )
$\Delta T$ is the temperature change (in $°\text{C}$ or $\text{K}$ )

The temperature change is calculated as:

$\Delta T = T_{\text{final}} - T_{\text{initial}}$

Converting volume to mass for water

Remember that the density of water is 1.0 g mL⁻¹, so $1 \text{ mL}$ of water has a mass of $1 \text{ g}$ .

infoNote

Sometimes the density of water is quoted as $0.997 \text{ g mL}^{-1}$ ; however, this value (to $3$ significant figures) is only accurate at $25°\text{C}$ . For most calculations, using $1.0 \text{ g mL}^{-1}$ is appropriate.

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Worked Example: Calculating energy to heat water

Question: Calculate the heat energy, in $\text{kJ}$ , needed to increase the temperature of $500 \text{ mL}$ of water by $15.0°\text{C}$ .

Solution:

Step 1: Convert volume to mass

$500 \text{ mL}$ of water has a mass of $500 \text{ g}$ (using density $= 1.0 \text{ g mL}^{-1}$ )

Step 2: Find the specific heat capacity

The specific heat capacity of water is $4.18 \text{ J g}^{-1} \text{ °C}^{-1}$

Step 3: Calculate heat energy using $q = mc\Delta T$

$q = 500 \times 4.18 \times 15.0$

$q = 3.14 \times 10^4 \text{ J}$

Step 4: Convert to kilojoules

To convert from $\text{J}$ to $\text{kJ}$ , multiply by $10^{-3}$

$q = 3.14 \times 10^4 \times 10^{-3}$

$q = 31.4 \text{ kJ}$

Experimental determination of heat of combustion

When an exothermic chemical reaction, such as the combustion of a pure substance, is carried out underneath a container of water, some of the heat released by the combustion reaction is transferred to the water. By measuring the temperature change of the water, we can determine the approximate amount of energy released by the substance.

Apparatus setup

The diagram below shows an experimental arrangement for estimating the heat of combustion of an organic liquid, such as ethanol:

This experimental method is suitable for many organic liquids, including:

Alcohols (e.g., methanol, ethanol)
Some alkanes
Some alkenes

Because these liquids release significant energy when burnt, some of them can be used as fuels.

Experimental procedure

The flowchart below summarises the steps followed in this experiment:

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The procedure involves:

Measuring a volume of water (e.g., $100 \text{ mL}$ ) and placing it in the metal can
Measuring the initial temperature of the water
Measuring the mass of the spirit burner and fuel
Lighting the burner and heating the water while stirring continuously
After some time, extinguishing the burner and recording the highest temperature reached by the water
Measuring the mass of the burner and remaining fuel
Calculating the mass of fuel consumed

Key measurements

Three key pieces of information collected from this procedure are:

Mass of water
Change in temperature of the water ( $\Delta T$ )
Mass of organic liquid fuel consumed (from which the amount in moles can be calculated)

Calculating heat of combustion

The heat of combustion of a fuel is the heat energy released when a specified amount of the substance burns completely. It is reported as a positive value. In contrast, the enthalpy of combustion ( $\Delta H_c$ ) reflects the exothermic nature with a negative sign and appears in thermochemical equations.

The heat of combustion can be calculated using:

$\text{heat of combustion} = \frac{q}{n}$

where:

$q$ is the energy absorbed by the water (calculated using $q = mc\Delta T$ )
$n$ is the amount of fuel (in moles) that has been burnt

To find the amount in moles:

$n = \frac{m}{M}$

where $m$ is the mass of fuel burnt and $M$ is the molar mass.

Important limitation

chatImportant

This calculation assumes that all the energy is transferred from the burning fuel to the water. In reality, substantial heat losses occur, so the calculated values for heat of combustion will be less than the actual values. This experiment can only provide an estimate of the heat of combustion or can be used to compare the energy released by different substances.

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Worked Example: Heat of combustion of methanol

Question: $0.355 \text{ g}$ of methanol ( $\text{CH}_3\text{OH}$ ) undergoes complete combustion in a spirit burner. The heat energy released is used to heat $100 \text{ g}$ of water. The temperature of the water rose from $20.24°\text{C}$ to $37.65°\text{C}$ . Calculate the heat of combustion of methanol in $\text{kJ mol}^{-1}$ and write the thermochemical equation for the reaction.

Solution:

Step 1: Calculate the temperature change

$\Delta T = 37.65 - 20.24 = 17.41°\text{C}$

Step 2: Calculate energy absorbed by water using $q = mc\Delta T$

$q = 100 \times 4.18 \times 17.41$

$q = 7277 \text{ J}$

Step 3: Convert to kilojoules

$q = 7277 \times 10^{-3} = 7.277 \text{ kJ}$

Step 4: Calculate amount of methanol in moles

Molar mass of $\text{CH}_3\text{OH} = 32.0 \text{ g mol}^{-1}$

$n = \frac{0.355}{32.0} = 0.0111 \text{ mol}$

Step 5: Calculate heat of combustion

$\text{Heat of combustion} = \frac{q}{n} = \frac{7.277}{0.0111} = 656 \text{ kJ mol}^{-1}$

Step 6: Write thermochemical equation

The thermochemical equation includes a negative sign for $\Delta H$ because combustion is exothermic:

$\text{CH}_3\text{OH(l)} + \frac{3}{2}\text{O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)} + 2\text{H}_2\text{O}$ ; $\Delta H = -656 \text{ kJ}$

or (multiplied by 2 to avoid fractions):

$2\text{CH}_3\text{OH(l)} + 3\text{O}_2\text{(g)} \rightarrow 2\text{CO}_2\text{(g)} + 4\text{H}_2\text{O}$ ; $\Delta H = -1312 \text{ kJ}$

Estimating energy content of fuels and foods

Energy content vs heat of combustion

Foods and some fuels (like diesel) are mixtures rather than pure substances. For these materials, the heat energy released during combustion of a known mass is best described as energy content, measured in $\text{kJ g}^{-1}$ (kilojoules per gram).

infoNote

This differs from heat of combustion (measured in $\text{kJ mol}^{-1}$ ), which applies to pure substances where we can calculate the number of moles.

Calculating energy content

If we measure the mass of fuel or food burned to produce energy, the energy content can be calculated as:

$\text{energy content} = \frac{\text{energy transferred to the water}}{\text{change in mass of the fuel/food during combustion}} = \frac{q}{\Delta m}$

The mass of fuel or food burned is determined by:

$\Delta m = m_{\text{initial}} - m_{\text{final}}$

It is not necessary to burn all the fuel or food sample, as long as the mass that is burned is determined.

Limitation of this method

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Heat loss is a consistent problem in this type of experiment, so the calculated values for energy content are estimates only and will be less than the actual values.

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Worked Example: Energy content of food

Question: A $1.670 \text{ g}$ sample of cheese biscuit was burned under a steel can containing $200 \text{ g}$ of water. After the flame went out, the mass of the cheese biscuit was $0.300 \text{ g}$ and the temperature of the water had risen by $29.8°\text{C}$ . Calculate the energy content of the biscuit in $\text{kJ g}^{-1}$ .

Solution:

Step 1: Calculate heat energy absorbed by water

$q = mc\Delta T$

$q = 200 \times 4.18 \times 29.8$

$q = 24913 \text{ J}$

Step 2: Convert to kilojoules

$q = \frac{24913}{1000} = 24.913 \text{ kJ}$

Step 3: Calculate mass of food burned

$\Delta m = m_{\text{initial}} - m_{\text{final}}$

$\Delta m = 1.670 - 0.300 = 1.370 \text{ g}$

Step 4: Calculate energy content

$\text{Energy content} = \frac{q}{\Delta m}$

$\text{Energy content} = \frac{24.913}{1.370}$

$\text{Energy content} = 18.2 \text{ kJ g}^{-1}$ (to $3$ significant figures)

Heat loss in calorimetry experiments

Sources of heat loss

When energy is transferred from burning a sample of fuel or food across an open space, heat is lost to the surroundings (such as the air around the burning fuel). Similarly, if there is no lid on a container of water, heat will be lost from the surface of the water.

When some heat energy from the burning fuel or 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 temperature change ( $\Delta T$ ) of the water results in a lower calculated energy value ( $q$ ).

Methods to reduce heat loss

There are several ways to reduce heat loss during calorimetry experiments:

Put a lid on the container holding the water
Insulate the beaker of water (with flameproof material)
Place insulation around the burning fuel (although sufficient oxygen must reach the fuel for combustion to be complete)

Impact on experimental results

chatImportant

This loss of heat energy represents a significant systematic error in the calculation of heat of combustion by simple experimental methods. The calculated values will always be lower than the actual values because not all the heat energy reaches the water.

More accurate determinations of energy content require more sophisticated equipment, such as a bomb calorimeter.

Bomb calorimetry

What is a bomb calorimeter?

A bomb calorimeter is specialized equipment used for measuring the energy released by combustion reactions that involve gaseous reactants or products. It provides much more accurate measurements than simple calorimetry methods.

Components and operation

The diagram below shows the main components of a bomb calorimeter:

Key features include:

Pressurised vessel: The reaction vessel is designed to withstand the high pressures that may build up during reactions. This is why it's called a "bomb" calorimeter.

Oxygen supply: Sufficient oxygen under pressure is provided to completely combust the fuel, ensuring all available energy is released.

Insulated container: Insulation around the calorimeter prevents heat escaping to the surroundings, making measurements more accurate.

Thermometer: Measures the change in temperature of the water surrounding the reaction vessel.

Stirrer: Ensures the temperature of the water is uniform throughout.

Electric heater: Used both to ignite the sample and to calibrate the calorimeter.

Sample in crucible: The fuel or food sample is placed in a small crucible at the bottom of the pressurised vessel.

Advantages over simple calorimetry

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The bomb calorimeter provides more accurate results because:

Heat loss is minimized through insulation
Complete combustion is ensured through excess oxygen
Pressure containment allows for reactions involving gases
Temperature measurements are more precise

Case study: The energy of candlelight

Candles have been used for over $5000$ years of human civilization. Candle wax is made of large molecules composed of carbon and hydrogen atoms, such as eicosane ( $\text{C}_{20}\text{H}_{42}$ ), and has an approximate energy content of $47.1 \text{ kJ g}^{-1}$ .

Candles were used by the ancient Romans as early as $500$ BCE. However, Romans primarily used oil lamps burning olive oil as their main light source. The burning of whale oil was common practice in lamps from the 16th to the 19th centuries. Whale oil is a mixture containing large quantities of oleic acid, which has an energy content of $39 \text{ kJ g}^{-1}$ .

One such lamp was developed in $1780$ by François-Pierre-Amédée (also known as Ami) Argand, a student of Antoine Lavoisier. These lamps were considered to have about the same light output as $6$ to $8$ candles.

infoNote

In fact, a "candela" or candle power is a measure of the intensity of light, now known as lumens. A modern $10$ Watt LED globe has an output of over $300$ lumens. Argand further developed his lamp into a burner that was used until the development of the Bunsen burner.

Cheaper alternatives to whale oil were available at that time, but they did not burn as cleanly, producing large quantities of carbon soot. Demand for whale oil vastly increased in the 17th century, and whale hunting reached its peak in the 1820s. After that time, there was a rapid decline in use of whale oil as cheaper and cleaner-smelling combustible materials became more common.

Camphine (also known as "oil of turpentine"), a mixture of turpentine and alcohol, was used until kerosene rapidly dominated the market from the 1860s, making the use of camphine insignificant by 1866.

Now, in the 21st century, whale hunting is banned across most of the globe, and we use electricity to power our lights. Just as humans have moved away from the use of whale oil for lighting, we have the potential to explore more renewable fuel sources for our energy needs, so that we can preserve our environment.

bookmarkSummary

Key Points to Remember:

Specific heat capacity ( $c$ ) measures the energy needed to raise the temperature of $1 \text{ g}$ of a substance by $1°\text{C}$ . Water has a high value ( $4.18 \text{ J g}^{-1} \text{ °C}^{-1}$ ) due to hydrogen bonding.
Heat energy transferred to water is calculated using $q = mc\Delta T$ , where $q$ is heat energy (J), $m$ is mass (g), $c$ is specific heat capacity, and $\Delta T$ is temperature change (°C).
Heat of combustion (in $\text{kJ mol}^{-1}$ ) is calculated by dividing the energy released by the number of moles of fuel burned: $\text{heat of combustion} = \frac{q}{n}$
Energy content (in $\text{kJ g}^{-1}$ ) is used for mixtures like foods and some fuels, calculated as: $\text{energy content} = \frac{q}{\Delta m}$
Heat loss is a major source of error in simple calorimetry experiments, causing calculated values to be lower than actual values. This can be reduced by using lids, insulation, and more sophisticated equipment like bomb calorimeters.

Calculating Heat Energy Released (VCE SSCE Chemistry): Revision Notes