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General Chemistry--Unit 4


Thermodynamics 1
Thermodynamics 2


Thermodynamics Part 1: Calorimetry

�Before You Begin:

To be successful at this material you will need to be familiar with the SI units system and be able to perform unit conversions and round the results to the correct number of significant figures. The measurement unit contains a review of unit analysis.

Thermodynamics, Part 1: Calorimetery

Thermodynamics is the study of the energy and entropy transformations that accompany chemical processes. To understand thermodynamics we need to learn some basic terms about the types and components of energy.


Energy, Heat, and Temperature

If we bring two systems together, there will be a transfer of energy and/or matter between the two systems until both systems are at thermodynamic equilibrium, a state of ‘energy equality’ resulting from transfer from the higher energy system to the lower energy system. There are three parts to thermodynamic equilibrium: thermal equilibrium, mechanical equilibrium, and chemical equilibrium. Thermal equilibrium involves the transfer of thermal energy as heat; this is our focus in this part of the unit. Mechanical equilibrium involves work and the transfer of force. Except for a brief discussion of work done by an expanding gas (pressure/volume work), we will leave mechanical equilibrium to the physics professors. Chemical equilibrium involves the transfer of matter and chemical potential. Freshman chemistry courses study this part of thermodynamic equilibrium in gory detail—almost half of the second semester of the year long course. When chemistry students come across the word ‘equilibrium’ naked and alone, it refers to chemical equilibrium.

How does heat flow from one system to another? Within a chemical system, the kinetic energy of the particles is a distribution. At any given instant some particles are moving faster than the average, and others are moving slower than the average, and all are colliding with each other. During collisions, fast particles lose some kinetic energy and slow particles gain some kinetic energy. However, the overall energy distribution remains the same. When two different systems are brought in contact, fast particles from both systems collide with slow particles from both systems and transfer kinetic energy. Unless the thermal energy of both systems happens to be exactly the same to begin with, the net result is that the system with higher kinetic energy will lose energy, and the system with lower kinetic energy will gain energy. This transfer will continue until, eventually, the kinetic energies of both systems match.


Calculating Heat

We observe the flow of thermal energy as a temperature change. The heat capacity, C, of an object is the amount of heat, q, required to raise the temperature of that object by one degree Kelvin. The heat that flows into or out of an object is calculated using the formula:

q = CΔT

where q is the heat in joules, C is the heat capacity of the object in joules/K, and ΔT is the temperature change (ΔT = Tfinal – Tinitial). There are two components to the heat capacity: the size of the object and how well it transfers thermal energy. Some objects, such as metals, transfer heat very effectively. Other objects, such as wood, do not transfer heat very well at all. We can observe this phenomenon in the kitchen. Chefs use wooden spoons when cooking for two reasons: they are less likely to scratch up the sauce pan, and the handle of a wooden spoon doesn’t get hot like the handle of a metal spoon. The specific heat capacity, c, (sometimes called specific heat with a lower case 's' as the variable instead of lower case 'c'), is the amount of heat required to raise the temperature of one gram of a substance by one degree Kelvin. For pure substances or simple mixtures the specific heat capacity (little c) is a constant, but the heat capacity (big C) is unique for every object.


Sign Convention

Before we worry about actual experimental results and sample calculations, we need to digress a moment and think about heat loss and gain. In thermodynamics, we use an algebraic sign to signify direction: energy flow out of the system has a negative algebraic sign--energy flow into the system has a positive algebraic sign.


4Concept Check: the specific heat capacity of water is 4.184 J/g °C. How much heat does it take to raise the temperature of 250 mL of water from room temperature to just below boiling?


Room temperature is about 22 °C, and the normal boiling point of water is 100 °C, so the temperature change is about 78 °C (ΔT = Tfinal – Tinitial = 100 – 22). The density of water is 1.0 g/mL, so 250 mL is 250 g. The heat is equal to the temperature change times the heat capacity which is the specific heat capacity multiplied by the mass:

q = CΔT = cmΔT

= (4.184 J/g °C)(250 g)(78 °C) = 8.2 x 104 J = 82 kJ

4Concept Check: Does the sign make sense?

Answer: This value for heat has a positive algebraic sign in this sample calculation. Positive values for heat imply that thermal energy flowed into the system.  The water got hotter because thermal energy flowed from a Bunsen burner or hotplate into the water. The heat of the water increased, so the sign does make sense.


4Concept Check: Why did we set the calculation so that the temperature was “just below boiling” instead of boiling point or above?

Answer: Recall from the discussion of phase changes in unit 2 that, when a liquid reaches the boiling temperature, any additional input of energy goes to supplying the energy needed overcome intermolecular attractions, to vaporize the liquid, and to supply kinetic energy for the gas. The process of changing phase requires an input of energy over and above what it needed to raise temperature.


Internal Energy, E

Internal energy, E, is the total energy of a system (the matter we are studying). There are two fundamental parts of the internal energy: the kinetic energy and the potential energy. For chemical systems, the kinetic energy is due to the random motion of the particles that make up the system. The potential energy is the energy stored in bonds and in the electronic and nuclear states of the atoms that make up the matter. If we wanted to calculate the internal energy of a beaker of an aqueous solution, we would have to find the kinetic and potential energy of every atom, molecule, and ion in the beaker at a given instant. The kinetic energies would be translational (moving the particles from point to point), vibrational (stretching and shortening the bonds like tiny springs), and rotational (spinning the particles like tiny tops). The potential energy would be all the energies due to all of the bonds and all of the electrons’ quantum mechanical energy states, some in the ground and some in various excited states. The various different components of the internal energy for the various different particles in the system are constantly changing, as particles collide and as electrons change quantum states. Because it is so complicated, the internal energy cannot be calculated, even for the simplest of chemical systems. However, when a chemical change takes place, internal energy is transferred to or from the system. We can observe and measure the internal energy change, ΔE (U is the other symbol for internal energy seen in some chemistry and most physics texts).


Internal Energy Change, ΔE

The internal energy change has two components: thermal energy and mechanical energy. The thermal energy flows into or out of the system as heat, q. The mechanical energy accomplishes work, w. Heat is the flow of thermal energy, and work is the movement of an object over some distance. In chemical systems, work is most often accomplished by an expanding gas (think of rockets and pistons—the mechanical ones not the basketball teams).

ΔE = q + w

The mechanical energy is fairly easy to keep straight: if we see that an object, a cannon ball for example, has moved, we know that mechanical energy was used. We can measure the mass of the cannon ball and the distance to figure out how much work was done. Thermal energy can be more confusing. It is the part of the internal energy of a system due to the random motion of the particles that make up that object. Heat, q, is the flow of thermal energy observed as a temperature change (it is similar to work in that work is the ‘flow’ of mechanical energy). Heat flows between two objects from higher to lower temperature. These three related ideas are often confused, but to keep them straight:

  • temperature is how we observe the change, how we know something happened;
  • heat is the energy flow, what the energy accomplished;
  • and thermal energy is the part of the internal energy change that is not mechanical energy.

State functions are thermodynamic properties that depend only on the initial and final states not on the method by which a change takes place (also called the path of the change). Internal energy is a state function. The energy change for a chemical reaction does not depend on how the reaction takes place; it only depends on the identity and amounts of the reactants and products. There are thermodynamic parameters that do vary depending on how the chemical change occurs: heat, q, and work, w, for example. Think of a chemical combustion reaction that generates a lot of hot gas, like a fire cracker, for example. We can trap the hot gas and use it to blast a bottle-rocket into the air, or we can break open the fire cracker and allow the gas to escape as the gunpowder burns. In both cases, the internal energy change is the same. The difference is that using the reaction to power a bottle rocket produces more work and less heat.


Variable Convention

A brief digression: By convention, we use lower case letters, like q and w, to represent properties that are not state functions, and we use upper case letters, like E or U, to represent properties that are state functions.


System and Surroundings

Thermodynamics is concerned with energy changes and transfer. To keep track of transfer of energy we use a concept called the system. A system is the part of the universe that we are observing. The surroundings are the rest of the universe. A typical system in freshman chemistry lab consists of chemicals that we combine and observe, an acid and base, for example. The surroundings for this type of system would be everything else, but for practicality’s sake we just observe the immediate surroundings: the beaker, the air around the beaker, a stirring rod, a thermometer, and the water in which the acid and base are dissolved (though not the water that is a product of the reaction—it is part of the system!).



Calorimetry is the study of the heat flow that accompanies physical and/or chemical changes. The apparatus used to measure heat is a calorimeter.  The fundamental parts of a calorimeter are: a thermometer to measure temperature change, a medium to transfer the heat into (or out of), and a way of insulating the system and its immediate surroundings from the rest of the universe so that we can slow the thermal equilibrium process down enough to measure temperature changes. The point to measuring heat is to find the internal energy changes (and other thermodynamic properties) for physical and chemical processes. These properties, called thermodynamic state functions, can be related to each other and to the heat as long as we keep either the volume or the pressure of the system constant. There are two basic types of calorimeters: constant volume calorimeters and constant pressure calorimeters.


Constant Pressure Calorimeter

The typical freshman chemistry lab experiments in calorimetry use a constant pressure calorimeter made from Styrofoam© cups. The system is open to the air, and the pressure is kept constant at atmospheric pressure. A thermometer is used to measure the temperature change. Water in the coffee cups (or an aqueous solution) is the medium into (or out of) which the heat is transferred. The water and the system are insulated from the rest of the room by air trapped in the Styrofoam© and an air space between cups. Some sort of lid keeps the heat from leaking too quickly out of the top of the calorimeter. The lid never insulated too well, though, because it has holes to accommodate the thermometer and a stirrer to mix the water as it heats up (or cools down).


This is a diagram of a constant pressure calorimeter as used in the introductory chemistry laboratory. It consists of styrofoam cups and a lid with a thermometer and stirrer. Water is added to the cups so that the thermometer will be able to record temperature changes. If a system that is not a thermal equilibrium with the water is added to the calorimeter, heat will be transferred to or from the water causing a termperature change.



Heat Capacity of a Constant Pressure Calorimeter

In a typical experiment to find the heat capacity of a calorimeter, a hot object with known thermal properties is added to cold water inside a calorimeter. The hot object acts as the system. The cold water and calorimeter are the surroundings. The heat lost by the system equals the heat gained by the surroundings, so we can calculate the heat capacity of the calorimeter. Often, water is used as the system because it is safe and easy to handle and it has a known specific heat capacity, 4.184 J/g °C. This can be confusing to students, though, because water is also used inside the “coffee cup” calorimeter as the thermal medium.

Suppose we add roughly 75 mL of hot water (80 °C) to roughly 75 mL of cold water (20 °C) in our calorimeter. The hot water (which is our system) would transfer thermal energy to the cold water (the surroundings) as they reach thermal equilibrium. The heat lost by the hot water should equal the heat gained by the cold water. Because the specific heat capacity of hot water is the same as the specific heat capacity of cold water and because we used equal masses, the temperature changes for the hot and cold water should be equal. If the calorimeter was a perfect insulator, we would end up with 150 mL of warm water (50 °C). Well, nothing is a perfect insulator, and freshman chemistry lab equipment is less perfect than most. A typical experiment like this will result in 150 mL of tepid water (40 °C). Where did the extra heat go? Some of it leaked out into the lab, but the missing heat also warmed up the thermometer, the stirrer, the air inside the calorimeter, and the Styrofoam© of which the calorimeter is made. As long as we don’t make any drastic mistakes (like poking a hole in the cup or losing the lid) the thermal properties of the calorimeter should be fairly constant–at least for one lab period. In order to use the calorimeter to measure heat for unknown systems, we must first perform a series of measurements to find the heat capacity of the calorimeter. From that point on, the heat lost by the system will equal (except for the sign) the heat gained by the calorimeter and the water inside.

If the system loses thermal energy, the surroundings gain thermal energy. The thermometer is always in the surroundings. If the temperature of the thermometer goes up, the system lost heat, and the sign of q for the system is negative.


4Concept Check: A measured amount (72.0 grams) of hot water is added to a measured amount (78.1 grams) of cold water in a "coffee cup" calorimeter. The following data was recorded:

Experimental Data Table for Heat Capacity of Calorimeter






Mass = 72.0 g


Mass = 78.1 g


c = 4.184 J/g °C


 c = 4.184 J/g °C


Tinitial = 82.9 °C


Tinitial = 22.0 °C


Tfinal = 44.1 °C


Tfinal = 44.1 °C




C = ?




Tinitial = 22.0 °C




Tfinal = 44.1 °C

What is the heat capacity of the calorimeter?


qsystem = -qsurroundings = -(qwater + qcalorimeter)

Find the heat lost for the system:

qsystem = mcΔT = (72.0 g)(4.184J/g °C)(44.1 – 82.9°C) = -11700 J


Find the heat gained for the water in the calorimeter:

qwater = 78.1 g(4.184J/g °C)(44.1 – 22.0 °C) = 7220 J


Solve for the heat capacity of the calorimeter:

qsystem = -(qwater + qcalorimeter)

-11700 J = -(7220J + CΔT)

C = (11700J – 7220J) ÷ (44.1 – 22.0) = 2.0 x 102 J/°C (two significant figures by the subtraction rule on the numerator)

Save this Value. We will need to know the heat capacity of the calorimeter in the next couple of calculations.


Specific Heat Capacity of an Unknown Solid

If we use this calorimeter (or another with a known heat capacity) we can find the specific heat capacity of a solid. The experimental procedure would be similar to the previous example. The calorimeter and water are the surroundings. We heat a solid to a known temperature (using a boiling water bath, for example). Then we add the hot solid to the calorimeter.


4Concept Check: A weighed amount of metal is heated in a water bath, quickly dried, then added to a "coffee cup" calorimeter (the same one used in the previous experiment). The lid was replaced and the water stirred. The highest temperature was noted. The following data was recorded:

Experimental Data Table for Heat Capacity of a Solid






Mass = 135.7 g


Mass = 145.9 g


c = ? J/g °C


 c = 4.184 J/g °C


Tinitial = 99.5 °C


Tinitial = 22.2 °C


Tfinal = 32.4 °C


Tfinal = 60.7 °C




C = 2.0 x 102J/°C




Tinitial = 22.2 °C




Tfinal = 32.4 °C

What is the specific heat capacity of the metal?


qsystem = -qsurroundings = -(qwater + qcalorimeter) where the system is the unknown solid and the surroundings are the water and calorimeter. Substituting in the formulas for the heats in terms of the specific heat capacities of the pure substances (water and unknown) and the heat capacity of the calorimeter gives the following equation:

mcΔT = -(mcΔT + CΔT)

We know all of these variables except the specific heat capacity for the solid. Substituting values gives:

(135.7 g)(c)(32.4 – 99.5 °C) = - [(145.9 g)(4.184 J/g °C)(32.4 – 22.2°C) + (2.0 x102 J/g°C)(32.4 – 22.2 °C)] 

(135.7 g)(c)(32.4 – 99.5 °C) = - 8270 J

c = -8270J/(135.7 g)(-39.1 °C) =  0.91 J/g °C


4Concept Check: In this type of experiment, the unknown solid is usually heated in a water bath. What would be the result if the students didn’t dry off the solid before adding it to the calorimeter?

Answer: Water has a specific heat capacity of 4.184 J/mol °C, more than four times higher than this metal. If the metal was not dried off very well, hot water would be added to the calorimeter along with the metal. The hot water would transfer heat better than the metal, so the specific heat capacity would be higher than the true value. This is an example of a systematic error.


Heat of Solution

Crystalline solid can give off or absorb heat when it dissolves. An ionic solid requires energy to disrupt the ionic lattice. Partial opposite charges on water molecules stabilize the oppositely charged ions. The thermal energy of the dissolution process depends on which of these situations is more stable (and lower in energy). If the ionic crystal structure makes the compound more stable, the dissolution process will require energy from the surroundings and the temperature on the thermometer will drop (note that the thermometer is in the surroundings). If the aqueous solution has more stability, the solid will give excess energy to the surroundings and the temperature on the thermometer will rise.

As before, the heat given off or absorbed by the system is equal to the sum of the heats absorbed or given off by the water and the calorimeter. In these calculations, though, we are trying to find the heat rather than heat capacity or specific heat capacity.


4Concept Check: A sample of an ionic compound was added to a measured amount of water in a "coffee cup" calorimeter (the same one used in the previous experiments) and stirred until the temperature reached a new, stable reading. The following data was collected:

Experimental Data Table for Heat of Solution





Unknown  ionic solid

Mass =   5.170 g


Mass =75.162 g




 c = 4.184 J/g °C




Tinitial = 22.5 °C




Tfinal = 33.2 °C




C = 2.0 x 102J/°C




Tinitial = 22.5 °C




Tfinal = 33.2 °C

What is the heat of solution for the compound?


qsystem = -qsurroundings

= -[(75.162 g)(4.184 J/g°C)(10.7 °C) + 2.0 x 102 J/°C(10.7 °C)]

= -5500 J

4Concept Check: If the unknown ionic solid is sodium hydroxide, what is the heat evolved per mole?

Answer: The molar mass of sodium hydroxide is 40.00 g/mol, so 5.170 g is 0.1293 mole. The amount of heat given off is about 43 kJ/mol.


Heat of Reactions

Aqueous reactions can give off or absorb heat which can be measured in a simple calorimeter like the one above. A typical calorimetry experiment is to use a “coffee cup” calorimeter to determine the heat of an acid/base neutralization reaction. The calorimeter is set up as before except that it is filled with a measured amount of an aqueous base. A measured amount of aqueous acid is added and the maximum temperature is recorded. The calculations are similar to those of the heat of solution experiment. This time, however, the acid and base solutes are the system, and the water in the surroundings is the solvent in the acid and the base solutions. We assume that the specific heat capacity of the water in the solution is roughly the same as the specific heat capacity of pure water. We also assume that the density of the solution is roughly the same as the density of the water.


4Concept Check:

If we add 55.5 mL of 6.0 M hydrochloric acid to 47.4 mL of 6.0 M sodium hydroxide in the same calorimeter as before we get the following data:

Experimental Data Table for Heat of Reaction





Acid solute

55.5 mL of 6.0 M = 0.33 mole

Water (solvent and surroundings)

Mass = 103 g

Base solute

47.4 mL of 6.0 M = 0.28 mole (limiting)


 c = 4.184 J/g °C




Tinitial = 22.1 °C




Tfinal = 47.5 °C




C = 2.0 x 102J/°C




Tinitial = 22.1 °C




Tfinal = 47.5 °C

How much heat is given off or absorbed by this reaction?


qsystem = -qsurroundings

= -[(103 g)(4.184 J/g°C)(25.5 °C) + 2.0 x 102 J/°C(25.5 °C)]

= -16000 J

The sodium hydroxide is the limiting reactant, so there are 0.28 moles in the system. The heat evolved in the reaction is -57 kJ/mol.



Constant Volume Calorimetry

Recall that the internal energy change is equal to the heat absorbed or given off added to the work done on or by the system. Work in chemical systems is accomplished by the expansion or compression of a gas. At constant volume, work is zero, because the gas can’t expand (or be compressed) without changing its volume. If a calorimetry experiment is performed so as to keep the volume constant (and work zero), the internal energy change is equal to the heat ΔE = qv (the subscript refers to constant volume--this relation is only true at constant volume).

Note that the “coffee cup” calorimeter is a constant pressure apparatus because the calorimeter is open to the air. The “coffee cup” calorimeter is not a constant volume apparatus because, if one of the products of the reaction is a gas, the volume of the gas changes during the experiment (from zero to some volume based on the stoichiometry of the reaction). If a gas is produced, the system has to do work on the surroundings to push against the air as the gas expands into the room.


The ‘Bomb’ Calorimeter

In order to keep the volume constant, even if gases are formed during the reaction, the part of the calorimeter that contains the reaction has to be rigid. A “bomb” calorimeter is an apparatus designed to measure the heat given off during of combustion reactions at constant volume. The reaction takes place in a heavy-duty stainless steel cylinder called a “bomb.” The “bomb” has a threaded lid so that it can be sealed (like screwing two pieces of metal pipe together). The reactants are placed in a pan that is wired to a high voltage power supply and sealed inside the “bomb.” A high pressure of oxygen is added to the “bomb” through a gas inlet valve in the lid. Because the calorimeter is designed to measure the heats of combustion reactions, the high pressure of oxygen gas insures that there will be enough oxygen for the reaction to go to completion. An added bonus is that the pressure inside the “bomb” is nearly constant even though gases are formed (a small number added to a very large number is approximately equal to the very large number).

The “bomb” is lowered into an insulated water tank that has a thermometer and a stirrer. A high voltage electrical current is sent through the chemical pan causing the reactant to combust. The heat given off by the system is transferred to the stainless steel and the water in the calorimeter. We measure the temperature change of the water in the calorimeter, so the heat capacity of that particular calorimeter has to be determined before we can use it to find an unknown heat of combustion.

This is a diagram of a constant volume calorimeter. The reaction takes place in a stainless steel "bomb" so that the volume cannot change. The bomb is inside an insulated water tank with a thermometer and a mechanical stirrer. An electrical current is used to ignite a compound. Heat is given off during combustion. The heat is transferred to the water and the temperature change recorded.


Heat Capacity of a Constant Volume Calorimeter

To find the heat capacity of a calorimeter, perform an experiment to measure the heat given off by combusting a known amount of a standard. Benzoic acid, HC7H5O2, is often used to find the heat capacity of a “bomb” calorimeter. It is a stable solid that is easy to handle. Unlike the “coffee cup” calorimeter calculations, we tend to keep the water contribution as part of the overall heat capacity of the calorimeter. For this to work, we need to make sure that we use the same amount of water in the insulated water bath during the next part of the experiment.


4Concept Check: 1.7735 grams of benzoic acid are reacted in a bomb calorimeter. The temperature of the water bath changes from 22.11 °C to 25.08 °C. What is the heat capacity of the calorimeter? The NIST data base states that the heat of combustion of benzoic acid is -3228 kJ/mol.


qsystem = -qsurroundings = -CΔT

[(1.7735 g)/(122.12 g/mol)](-3228 kJ/mol) = -C(2.97 °C)

C = 15.8 kJ/°C


Heat of Combustion

After you determine the heat capacity of a “bomb” calorimeter, you can use it to determine the heat of combustion of any other compound.


4Concept Check: Sucrose has a molar mass of 342.3 g/mol. 1.028 grams of sucrose are burned in the “bomb” calorimeter above. The temperature rises from 22.21 to 23.29 °C. What is the molar heat of combustion of sucrose?


qsystem = -qsurroundings = -CΔT = -15.8kJ/°C(1.07 °C) = -16.9 kJ

1.028 grams is 0.003003 mole of sucrose, so the heat of combustion is -5630 kJ/mol.

So, why do we care about the heat of combustion of sucrose? The combustion of sucrose is a chemical reaction with this reaction equation: C12H22O11(s) + 12O2(g) 12CO2(g) + 11H2O(g)

We can use heat measurements to find thermodynamic state function information. Then we can use the state function properties to find out more about the energy changes for all sucrose reactions.


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This site was last updated 07/05/05