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Calorimetry

Calorimetry

One technique we can use to measure the amount of heat involved in a chemical or physical process is known as calorimetry. Calorimetry is used to measure amounts of heat transferred to or from a substance. To do so, the heat is exchanged with a calibrated object (calorimeter). The temperature change measured by the calorimeter is used to derive the amount of heat transferred by the process under study.

The measurement of heat transfer using this approach requires the definition of a system (the substance or substances undergoing the chemical or physical change) and its surroundings (the other components of the measurement apparatus that serve to either provide heat to the system or absorb heat from the system).

Knowledge of the heat capacity of the surroundings, and careful measurements of the masses of the system and surroundings and their temperatures before and after the process allows one to calculate the heat transferred as described in this section.

A calorimeter is a device used to measure the amount of heat involved in a chemical or physical process. For example, when an exothermic reaction occurs in solution in a calorimeter, the heat produced by the reaction is absorbed by the solution, which increases its temperature.

When an endothermic reaction occurs, the heat required is absorbed from the thermal energy of the solution, which decreases its temperature (see the figure below). The temperature change, along with the specific heat and mass of the solution, can then be used to calculate the amount of heat involved in either case.

In a calorimetric determination, either (a) an exothermic process occurs and heat, q, is negative, indicating that thermal energy is transferred from the system to its surroundings, or (b) an endothermic process occurs and heat, q, is positive, indicating that thermal energy is transferred from the surroundings to the system.

Scientists use well-insulated calorimeters that all but prevent the transfer of heat between the calorimeter and its environment. This enables the accurate determination of the heat involved in chemical processes, the energy content of foods, and so on. General chemistry students often use simple calorimeters constructed from polystyrene cups (see the figure below). These easy-to-use “coffee cup” calorimeters allow more heat exchange with the outside environment, and therefore produce less accurate energy values.

A simple calorimeter can be constructed from two polystyrene cups. A thermometer and stirrer extend through the cover into the reaction mixture.

Commercial solution calorimeters are also available. Relatively inexpensive calorimeters often consist of two thin-walled cups that are nested in a way that minimizes thermal contact during use, along with an insulated cover, handheld stirrer, and simple thermometer. More expensive calorimeters used for industry and research typically have a well-insulated, fully enclosed reaction vessel, motorized stirring mechanism, and a more accurate temperature sensor (see the figure below).

Commercial solution calorimeters range from (a) simple, inexpensive models for student use to (b) expensive, more accurate models for industry and research.

Before we practice calorimetry problems involving chemical reactions, consider a simpler example that illustrates the core idea behind calorimetry. Suppose we initially have a high-temperature substance, such as a hot piece of metal (M), and a low-temperature substance, such as cool water (W).

If we place the metal in the water, heat will flow from M to W. The temperature of M will decrease, and the temperature of W will increase, until the two substances have the same temperature—that is, when they reach thermal equilibrium (see the figure below). If this occurs in a calorimeter, ideally all of this heat transfer occurs between the two substances, with no heat gained or lost by either the calorimeter or the calorimeter’s surroundings. Under these ideal circumstances, the net heat change is zero:

$${q}_{\text{substance M}}+{q}_{\text{substance W}}=0$$

This relationship can be rearranged to show that the heat gained by substance M is equal to the heat lost by substance W:

$${q}_{\text{substance M}}=\text{−}{q}_{\text{substance W}}$$

The magnitude of the heat (change) is therefore the same for both substances, and the negative sign merely shows that qsubstance M and qsubstance W are opposite in direction of heat flow (gain or loss) but does not indicate the arithmetic sign of either q value (that is determined by whether the matter in question gains or loses heat, per definition). In the specific situation described, qsubstance M is a negative value and qsubstance W is positive, since heat is transferred from M to W.

In a simple calorimetry process, (a) heat, q, is transferred from the hot metal, M, to the cool water, W, until (b) both are at the same temperature.

Example: Heat Transfer between Substances at Different Temperatures

A 360.0-g piece of rebar (a steel rod used for reinforcing concrete) is dropped into 425 mL of water at 24.0 °C. The final temperature of the water was measured as 42.7 °C. Calculate the initial temperature of the piece of rebar. Assume the specific heat of steel is approximately the same as that for iron (see previous lesson), and that all heat transfer occurs between the rebar and the water (there is no heat exchange with the surroundings).

Solution

The temperature of the water increases from 24.0 °C to 42.7 °C, so the water absorbs heat. That heat came from the piece of rebar, which initially was at a higher temperature. Assuming that all heat transfer was between the rebar and the water, with no heat “lost” to the outside environment, then heat given off by rebar = −heat taken in by water, or:

$${q}_{\text{rebar}}=\text{−}{q}_{\text{water}}$$

Since we know how heat is related to other measurable quantities, we have:

$${\left(c\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}m\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{Δ}\text{T}\right)}_{\text{rebar}}={-\left(c\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}m\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{Δ}\text{T}\right)}_{\text{water}}$$

Letting f = final and i = initial, in expanded form, this becomes:

$${c}_{\text{rebar}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{m}_{\text{rebar}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\left({T}_{\text{f,rebar}}-{T}_{\text{i,rebar}}\right)=\text{−}{c}_{\text{water}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{m}_{\text{water}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\left({T}_{\text{f,water}}-{T}_{\text{i,water}}\right)$$

The density of water is 1.0 g/mL, so 425 mL of water = 425 g. Noting that the final temperature of both the rebar and water is 42.7 °C, substituting known values yields:

$$\left(0.449\phantom{\rule{0.2em}{0ex}}\text{J/g °C}\right)\left(360.0\phantom{\rule{0.2em}{0ex}}\text{g}\right)\left(42.7\phantom{\rule{0.2em}{0ex}}\text{°C}-{T}_{\text{i,rebar}}\right)=-\left(4.184\phantom{\rule{0.2em}{0ex}}\text{J/g °C}\right)\left(425\text{g}\right)\left(42.7\phantom{\rule{0.2em}{0ex}}\text{°C}-24.0\phantom{\rule{0.2em}{0ex}}\text{°C}\right)$$

$${T}_{\text{i,rebar}}=\phantom{\rule{0.2em}{0ex}}\cfrac{\left(4.184\phantom{\rule{0.2em}{0ex}}\text{J/g °C}\right)\left(425\phantom{\rule{0.2em}{0ex}}\text{g}\right)\left(42.7\phantom{\rule{0.2em}{0ex}}\text{°C}-24.0\phantom{\rule{0.2em}{0ex}}\text{°C}\right)}{\left(0.449\phantom{\rule{0.2em}{0ex}}\text{J/g °C}\right)\left(360.0\phantom{\rule{0.2em}{0ex}}\text{g}\right)}\phantom{\rule{0.2em}{0ex}}+42.7\phantom{\rule{0.2em}{0ex}}\text{°C}$$

Solving this gives Ti,rebar= 248 °C, so the initial temperature of the rebar was 248 °C.

This method can also be used to determine other quantities, such as the specific heat of an unknown metal.

Example: Identifying a Metal by Measuring Specific Heat

A 59.7 g piece of metal that had been submerged in boiling water was quickly transferred into 60.0 mL of water initially at 22.0 °C. The final temperature is 28.5 °C. Use these data to determine the specific heat of the metal. Use this result to identify the metal.

Solution

Assuming perfect heat transfer, heat given off by metal = −heat taken in by water, or:

$${q}_{\text{metal}}=\text{−}{q}_{\text{water}}$$

In expanded form, this is:

$${c}_{\text{metal}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{m}_{\text{metal}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\left({T}_{\text{f,metal}}-{T}_{\text{i, metal}}\right)=\text{−}{c}_{\text{water}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{m}_{\text{water}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\left({T}_{\text{f,water}}-{T}_{\text{i,water}}\right)$$

Noting that since the metal was submerged in boiling water, its initial temperature was 100.0 °C; and that for water, 60.0 mL = 60.0 g; we have:

$$\left({c}_{\text{metal}}\right)\left(59.7\phantom{\rule{0.2em}{0ex}}\text{g}\right)\left(28.5\phantom{\rule{0.2em}{0ex}}\text{°C}-100.0\phantom{\rule{0.2em}{0ex}}\text{°C}\right)=-\left(4.18\phantom{\rule{0.2em}{0ex}}\text{J/g °C}\right)\left(60.0\phantom{\rule{0.2em}{0ex}}\text{g}\right)\left(28.5\phantom{\rule{0.2em}{0ex}}\text{°C}-22.0\phantom{\rule{0.2em}{0ex}}\text{°C}\right)$$

Solving this:

$${c}_{\text{metal}}=\phantom{\rule{0.2em}{0ex}}\cfrac{-\left(4.184\phantom{\rule{0.2em}{0ex}}\text{J/g °C}\right)\left(60.0\phantom{\rule{0.2em}{0ex}}\text{g}\right)\left(6.5\phantom{\rule{0.2em}{0ex}}\text{°C}\right)}{\left(59.7\phantom{\rule{0.2em}{0ex}}\text{g}\right)\left(-71.5\phantom{\rule{0.2em}{0ex}}\text{°C}\right)}\phantom{\rule{0.2em}{0ex}}=0.38\phantom{\rule{0.2em}{0ex}}\text{J/g °C}$$

Comparing this with values in table of specific heats from the previous lesson, our experimental specific heat is closest to the value for copper (0.39 J/g °C), so we identify the metal as copper.

When we use calorimetry to determine the heat involved in a chemical reaction, the same principles we have been discussing apply. The amount of heat absorbed by the calorimeter is often small enough that we can neglect it (though not for highly accurate measurements, as discussed later), and the calorimeter minimizes energy exchange with the outside environment. Because energy is neither created nor destroyed during a chemical reaction, there is no overall energy change during the reaction. The heat produced or consumed in the reaction (the “system”), qreaction, plus the heat absorbed or lost by the solution (the “surroundings”), qsolution, must add up to zero:

$${q}_{\text{reaction}}+{q}_{\text{solution}}=0$$

This means that the amount of heat produced or consumed in the reaction equals the amount of heat absorbed or lost by the solution:

$${q}_{\text{reaction}}=\text{−}{q}_{\text{solution}}$$

This concept lies at the heart of all calorimetry problems and calculations.

Example: Heat Produced by an Exothermic Reaction

When 50.0 mL of 1.00 M HCl(aq) and 50.0 mL of 1.00 M NaOH(aq), both at 22.0 °C, are added to a coffee cup calorimeter, the temperature of the mixture reaches a maximum of 28.9 °C. What is the approximate amount of heat produced by this reaction?

$$\text{HCl}\left(aq\right)+\text{NaOH}\left(aq\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{NaCl}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)$$

Solution

To visualize what is going on, imagine that you could combine the two solutions so quickly that no reaction took place while they mixed; then after mixing, the reaction took place. At the instant of mixing, you have 100.0 mL of a mixture of HCl and NaOH at 22.0 °C. The HCl and NaOH then react until the solution temperature reaches 28.9 °C.

The heat given off by the reaction is equal to that taken in by the solution. Therefore:

$${q}_{\text{reaction}}=\text{−}{q}_{\text{solution}}$$

(It is important to remember that this relationship only holds if the calorimeter does not absorb any heat from the reaction, and there is no heat exchange between the calorimeter and the outside environment.)

Next, we know that the heat absorbed by the solution depends on its specific heat, mass, and temperature change:

$${q}_{\text{solution}}={\left(c\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}m\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{Δ}T\right)}_{\text{solution}}$$

To proceed with this calculation, we need to make a few more reasonable assumptions or approximations. Since the solution is aqueous, we can proceed as if it were water in terms of its specific heat and mass values. The density of water is approximately 1.0 g/mL, so 100.0 mL has a mass of about 1.0 $$×$$ 102 g (two significant figures). The specific heat of water is approximately 4.184 J/g °C, so we use that for the specific heat of the solution. Substituting these values gives:

$${q}_{\text{solution}}=\left(4.184\phantom{\rule{0.2em}{0ex}}\text{J/g °C}\right)\left(1.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{2}\phantom{\rule{0.2em}{0ex}}\text{g}\right)\left(28.9\phantom{\rule{0.2em}{0ex}}\text{°C}-22.0\phantom{\rule{0.2em}{0ex}}\text{°C}\right)=2.9\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}\phantom{\rule{0.2em}{0ex}}\text{J}$$

Finally, since we are trying to find the heat of the reaction, we have:

$${q}_{\text{reaction}}=\text{−}{q}_{\text{solution}}=-2.9\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}\phantom{\rule{0.2em}{0ex}}\text{J}$$

The negative sign indicates that the reaction is exothermic. It produces 2.9 kJ of heat.

Resource:

This demonstration shows the precipitation reaction that occurs when the disk in a chemical hand warmer is flexed.

Example: Heat Flow in an Instant Ice Pack

When solid ammonium nitrate dissolves in water, the solution becomes cold. This is the basis for an “instant ice pack” (see the figure below). When 3.21 g of solid NH4NO3 dissolves in 50.0 g of water at 24.9 °C in a calorimeter, the temperature decreases to 20.3 °C.

Calculate the value of q for this reaction and explain the meaning of its arithmetic sign. State any assumptions that you made.

An instant cold pack consists of a bag containing solid ammonium nitrate and a second bag of water. When the bag of water is broken, the pack becomes cold because the dissolution of ammonium nitrate is an endothermic process that removes thermal energy from the water. The cold pack then removes thermal energy from your body.

Solution

We assume that the calorimeter prevents heat transfer between the solution and its external environment (including the calorimeter itself), in which case:

$${q}_{\text{rxn}}=\text{−}{q}_{\text{soln}}$$

with “rxn” and “soln” used as shorthand for “reaction” and “solution,” respectively.

Assuming also that the specific heat of the solution is the same as that for water, we have:

$$\begin{array}{}{q}_{\text{rxn}}=\text{−}{q}_{\text{soln}}={-\left(c×m×\Delta T\right)}_{\text{soln}}\\ = −\left[\left(4.184\text{J/g °C}\right)×\left(53.2\text{g}\right)×\left(20.3\text{°C}-24.9\text{°C}\right)\right]\\ =−\left[\left(4.184\text{J/g °C}\right)×\left(53.2\text{g}\right)×\left(-4.6\text{°C}\right)\right]\\ \text{+}1.0×{10}^{3}\text{J}=+1.0\text{kJ}\end{array}$$

The positive sign for q indicates that the dissolution is an endothermic process.

If the amount of heat absorbed by a calorimeter is too large to neglect or if we require more accurate results, then we must take into account the heat absorbed both by the solution and by the calorimeter.

The calorimeters described are designed to operate at constant (atmospheric) pressure and are convenient to measure heat flow accompanying processes that occur in solution.

A different type of calorimeter that operates at constant volume, colloquially known as a bomb calorimeter, is used to measure the energy produced by reactions that yield large amounts of heat and gaseous products, such as combustion reactions. (The term “bomb” comes from the observation that these reactions can be vigorous enough to resemble explosions that would damage other calorimeters.)

This type of calorimeter consists of a robust steel container (the “bomb”) that contains the reactants and is itself submerged in water (see the figure below). The sample is placed in the bomb, which is then filled with oxygen at high pressure. A small electrical spark is used to ignite the sample. The energy produced by the reaction is absorbed by the steel bomb and the surrounding water. The temperature increase is measured and, along with the known heat capacity of the calorimeter, is used to calculate the energy produced by the reaction.

Bomb calorimeters require calibration to determine the heat capacity of the calorimeter and ensure accurate results. The calibration is accomplished using a reaction with a known q, such as a measured quantity of benzoic acid ignited by a spark from a nickel fuse wire that is weighed before and after the reaction. The temperature change produced by the known reaction is used to determine the heat capacity of the calorimeter. The calibration is generally performed each time before the calorimeter is used to gather research data.

(a) A bomb calorimeter is used to measure heat produced by reactions involving gaseous reactants or products, such as combustion. (b) The reactants are contained in the gas-tight “bomb,” which is submerged in water and surrounded by insulating materials. (credit a: modification of work by “Harbor1”/Wikimedia commons)

You might watch the YouTube video below to view how a bomb calorimeter is prepared for action.

Resource:

This website shows calorimetric calculations using sample data.

Example: Bomb Calorimetry

When 3.12 g of glucose, C6H12O6, is burned in a bomb calorimeter, the temperature of the calorimeter increases from 23.8 °C to 35.6 °C. The calorimeter contains 775 g of water, and the bomb itself has a heat capacity of 893 J/°C. How much heat was produced by the combustion of the glucose sample?

Solution

The combustion produces heat that is primarily absorbed by the water and the bomb. (The amounts of heat absorbed by the reaction products and the unreacted excess oxygen are relatively small and dealing with them is beyond the scope of this text. We will neglect them in our calculations.)

The heat produced by the reaction is absorbed by the water and the bomb:

$$\begin{array}{l}{q}_{\text{rxn}}=-\left({q}_{\text{water}}+{q}_{\text{bomb}}\right)\\ =-\left[\left(4.184\phantom{\rule{0.2em}{0ex}}\text{J/g °C}\right)\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\left(775\phantom{\rule{0.2em}{0ex}}\text{g}\right)\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\left(35.6\phantom{\rule{0.2em}{0ex}}\text{°C}-23.8\phantom{\rule{0.2em}{0ex}}\text{°C}\right)+893\phantom{\rule{0.2em}{0ex}}\text{J/°C}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\left(35.6\phantom{\rule{0.2em}{0ex}}\text{°C}-23.8\phantom{\rule{0.2em}{0ex}}\text{°C}\right)\right]\\ =-\left(38,300\phantom{\rule{0.2em}{0ex}}\text{J}+10,500\phantom{\rule{0.2em}{0ex}}\text{J}\right)\\ =\text{−48,800 J}=\text{−48.8 kJ}\end{array}$$

This reaction released 48.7 kJ of heat when 3.12 g of glucose was burned.

Since the first one was constructed in 1899, 35 calorimeters have been built to measure the heat produced by a living person. These whole-body calorimeters of various designs are large enough to hold an individual human being. More recently, whole-room calorimeters allow for relatively normal activities to be performed, and these calorimeters generate data that more closely reflect the real world.

These calorimeters are used to measure the metabolism of individuals under different environmental conditions, different dietary regimes, and with different health conditions, such as diabetes. In humans, metabolism is typically measured in Calories per day. A nutritional calorie (Calorie) is the energy unit used to quantify the amount of energy derived from the metabolism of foods; one Calorie is equal to 1000 calories (1 kcal), the amount of energy needed to heat 1 kg of water by 1 °C.