6.2: Calorimetry

Learning Objectives

• Explain the technique of calorimetry
• Calculate and interpret heat and related properties using typical calorimetry data

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 change in temperature of the measuring part of the calorimeter is converted into the amount of heat (since the previous calibration was used to establish its heat capacity). 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 (Figure \(\PageIndex{1}\)). 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.

Figure \(\PageIndex{1}\): 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.

By convention, q is given a negative (-) sign when the system releases heat to the surroundings (exothermic); q is given a positive (+) sign when the system absorbs heat from the surroundings (endothermic).

Scientists use well-insulated calorimeters that minimize 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 (Figure \(\PageIndex{2}\)). These easy-to-use “coffee cup” calorimeters allow more heat exchange with their surroundings, and therefore produce less accurate energy values.

Figure \(\PageIndex{2}\): 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. 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 as shown in Figure \(\PageIndex{3}\).

Figure \(\PageIndex{3}\): 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 simple example that illustrates the core idea behind calorimetry. Suppose we initially have a high-temperature substance, such as a hot piece of metal and a low-temperature substance, such as cool water. If we place the metal in the water, heat will flow from the metal to the water. The temperature of the metal will decrease, and the temperature of the water will increase, until the two substances have the same temperature—that is, when they reach thermal equilibrium (Figure \(\PageIndex{4}\)). 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:

\[q_\mathrm{\,metal} + q_\mathrm{\,water}=0 \label{5.3.1}\]

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

\[q_\mathrm{\,metal}=-q_\mathrm{\,water} \label{5.3.2}\]

The magnitude of the heat (change) is therefore the same for both substances, and the negative sign merely shows that \(q_{metal}\) and \(q_{water}\) 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, \(q_{metal}\) is a negative value and \(q_{water}\) is positive, since heat is transferred from the metal to the water.

Figure \(\PageIndex{4}\): 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 \(\PageIndex{1}\): Heat Transfer between Substances at Different Temperatures

A hot 360-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 is 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 (Table T4), 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 surroundings, then heat given off by rebar = − heat taken in by water, or:

\[q_\ce{rebar}=−q_\ce{water} \nonumber \]

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

\[(c×m×ΔT)_\ce{rebar}=−(c×m×ΔT)_\ce{water} \nonumber \]

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

\[ c_\ce{rebar}×m_\ce{rebar}×(T_\mathrm{f,rebar}−T_\mathrm{i,rebar})=−c_\ce{water}×m_\ce{water}×(T_\mathrm{f,water}−T_\mathrm{i,water}) \nonumber\]

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:

\[ \mathrm{(0.449\:J/g\: °C)(360g)(42.7°C−\mathit T_\mathrm{i,rebar})=-(4.184\:J/g\: °C)(425\:g)(42.7°C−24.0°C)} \nonumber\]

\[\mathrm{\mathit T_{i,rebar}=\dfrac{(4.184\:J/g\: °C)(425\:g)(42.7°C−24.0°C)}{(0.449\:J/g\: °C)(360\:g)}+42.7°C} \nonumber\]

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

Exercise \(\PageIndex{1A}\)

A 248-g piece of copper is dropped into 390 mL of water at 22.6 °C. The final temperature of the water was measured as 39.9 °C. Calculate the initial temperature of the piece of copper. Assume that all heat transfer occurs between the copper and the water.

Answer

The initial temperature of the copper was 335.6 °C.

Exercise \(\PageIndex{1B}\)

A 248-g piece of copper initially at 314 °C is dropped into 390 mL of water initially at 22.6 °C. Assuming that all heat transfer occurs between the copper and the water, calculate the final temperature.

Answer

The final temperature (reached by both copper and water) is 38.7 °C.

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

When we use calorimetry to determine the heat involved in a chemical reaction, the same principles we have been discussing apply. 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”), q_reaction, plus the heat absorbed or lost by the solution (the “surroundings”), \(q_{solution}\), must add up to zero:

\[q_\ce{reaction}+q_\ce{solution}=0\ \label{ 5.3.10}\]

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

\[q_\ce{reaction}=−q_\ce{solution} \label{5.3.11}\]

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

Example \(\PageIndex{2}\): Heat Produced by an Exothermic Reaction

When 50.0 mL of 0.10 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?

\[\ce{HCl}(aq)+\ce{NaOH}(aq)⟶\ce{NaCl}(aq)+\ce{H2O}(l) \nonumber\]

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:

\[q_\ce{reaction}=−q_\ce{solution} \nonumber\]

(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 its surroundings.)

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

\[q_\ce{solution}=(c×m×ΔT)_\ce{solution} \nonumber\]

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 4.18 J/g °C, so we use that for the specific heat of the solution. Substituting these values gives:

\[\mathrm{\mathit q_{solution}=(4.184\:J/g\: °C)(1.0×10^2\:g)(28.9°C−22.0°C)=2.89×10^3\:J}\]

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

\[q_\ce{reaction}=−q_\ce{solution}=−2.89×10^3\:J \nonumber\]

The negative sign indicates that the reaction is exothermic. It produces 2.89 kJ of heat.

Exercise \(\PageIndex{2}\)

When 100 mL of 0.200 M NaCl(aq) and 100 mL of 0.200 M AgNO₃(aq), both at 21.9 °C, are mixed in a coffee cup calorimeter, the temperature increases to 23.5 °C as solid AgCl forms. How much heat is produced by this precipitation reaction? What assumptions did you make to determine your value?

Answer

\(1.34 \times 10^3\; J\); assume no heat is absorbed by the calorimeter, no heat is exchanged between the calorimeter and its surroundings, and that the specific heat and mass of the solution are the same as those for water

Thermochemistry of Hand Warmers

When working or playing outdoors on a cold day, you might use a hand warmer to warm your hands (Figure \(\PageIndex{5}\)). A common reusable hand warmer contains a supersaturated solution of NaC₂H₃O₂ (sodium acetate) and a metal disc. Bending the disk creates nucleation sites around which the metastable NaC₂H₃O₂ quickly crystallizes (a later chapter on solutions will investigate saturation and supersaturation in more detail).

The process \(\ce{NaC2H3O2}(aq)⟶\ce{NaC2H3O2}(s)\) is exothermic, and the heat produced by this process is absorbed by your hands, thereby warming them (at least for a while). If the hand warmer is reheated, the NaC₂H₃O₂ redissolves and can be reused.

Figure \(\PageIndex{5}\): Chemical hand warmers produce heat that warms your hand on a cold day. In this one, you can see the metal disc that initiates the exothermic precipitation reaction. (credit: modification of work by Science Buddies TV/YouTube)

Another common hand warmer produces heat when it is ripped open, exposing iron and water in the hand warmer to oxygen in the air. One simplified version of this exothermic reaction is

\[\ce{2Fe(s) + 3/2 O2(g) ⟶ Fe2O3(s)} \nonumber \]

Salt in the hand warmer catalyzes the reaction, so it produces heat more rapidly; cellulose, vermiculite, and activated carbon help distribute the heat evenly. Other types of hand warmers use lighter fluid (a platinum catalyst helps lighter fluid oxidize exothermically), charcoal (charcoal oxidizes in a special case), or electrical units that produce heat by passing an electrical current from a battery through resistive wires.

Calorimetry Calculations - Video

Link to Video

Conservation of Energy: Coffee Cup Calorimetry: https://youtu.be/FwQcc17PN0k

 

Attribution

• Prof. Steven Farmer (Sonoma State University)

Summary

Calorimetry is used to measure the amount of thermal energy transferred in a chemical or physical process. This requires careful measurement of the temperature change that occurs during the process and the masses of the system and surroundings. These measured quantities are then used to compute the amount of heat produced or consumed in the process using known mathematical relations. Calorimeters are designed to minimize energy exchange between the system being studied and its surroundings. They range from simple coffee cup calorimeters used by introductory chemistry students to sophisticated bomb calorimeters used to determine the energy content of food.

Glossary

calorimeter

device used to measure the amount of heat absorbed or released in a chemical or physical process

calorimetry

process of measuring the amount of heat involved in a chemical or physical process

nutritional calorie (Calorie)

unit used for quantifying energy provided by digestion of foods, defined as 1000 cal or 1 kcal

surroundings

all matter other than the system being studied

system

portion of matter undergoing a chemical or physical change being studied

Contributors and Attributions

• Paul Flowers (University of North Carolina - Pembroke), Klaus Theopold (University of Delaware) and Richard Langley (Stephen F. Austin State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at http://cnx.org/contents/85abf193-2bd...a7ac8df6@9.110).