Energy and Calorimetry

Calorimetry

The temperature change in the surroundings can be used to calculate the energy change of a chemical process. These changes are described by calorimetry.

Calorimetry is the study of heat exchange between a system and its surroundings. A calorimeter is a device used to measure the heat exchanged between a system and its surroundings. Calorimeters are often used to measure the net heat released or absorbed by a chemical reaction at a constant pressure, called the heat of reaction (ΔHrxn\bold\Delta H_{\rm{rxn}}). This is also called enthalpy of reaction. If the substances absorb heat from the surroundings, the reaction is endothermic and ΔHrxn \Delta H_{\rm {rxn}} is positive. If the substances release heat to their surroundings, the reaction is exothermic and ΔHrxn \Delta H_{\rm {rxn}} is negative.

A bomb calorimeter is a constant-volume calorimeter used to measure the heat of a reaction for combustion reactions. The calorimeter contains a bomb chamber into which the reactants are placed, and it is wired to an electric heater that ignites the reactants and causes the combustion to occur. The rest of the calorimeter is filled with water and pressurized oxygen. A stirrer keeps the water mixed so that the heat is evenly distributed, and a thermometer records the temperature of the water.

Combustion is exothermic, so the heat released by the combustion is absorbed by the water inside the calorimeter and causes the temperature of the water to rise. In calorimetry experiments, the heat capacity of the calorimeter, Ccal—not the specific heat—is used. Each calorimeter has a unique Ccal value that must be determined experimentally. The heat capacity of the calorimeter is the heat required to raise the calorimeter by one degree Celsius.

An important caveat is that the entire bomb calorimeter should not lose any heat to its surroundings. Calorimeters are well insulated against heat transfer. A system that does not exchange heat or matter with the surroundings is an isolated system. Thus, a bomb calorimeter is built as an isolated system to ensure that no heat is released to the environment around the device and all of the energy released by the combustion is used to change the temperature inside the calorimeter. Because a bomb calorimeter is an isolated system the sum of the energies of the reactants and the calorimeter must be zero,
qrxn+qcal=0q_{\rm{rxn}}+q_{\rm{cal}}=0
where qrxn is the energy released or used by the reaction and qcal is the heat absorbed or released by the calorimeter. The value of qcal can be determined by multiplying the heat capacity of the calorimeter by the change in temperature.
qcal=CcalΔTq_{\rm{cal}}=C_{\rm{cal}}\Delta T
Thus, ΔHrxn \Delta H_{\rm {rxn}} can be calculated from the ΔT \mathit\Delta T of the calorimeter. This is because energy is conserved. The energy released by the combustion reaction (qrxn) is equal in magnitude to the energy absorbed by the calorimeter. However, because the calorimeter absorbs heat, its ΔT \mathit\Delta T and energy change are both positive. The energy change for the exothermic combustion is negative.
qcal=qrxn=ΔHrxnq_{\rm{cal}}=-q_{\rm{rxn}}=\Delta H_{\rm{rxn}}
A bomb calorimeter consists of a bomb chamber inside a second chamber of water. The water is stirred and there is a thermometer that measures the water temperature. The test material is placed inside the bomb chamber and ignited by electric leads. The entire system is encased in one or two layers of insulation to ensure that the heat released by the combustion reaction does not escape to the surroundings and is entirely absorbed by the calorimeter.
A bomb calorimeter can be used to find the energy available in food. A calorie (cal) is the amount of energy required to change the temperature of one gram of water by 1°C. It is equal to 4.184 J. The chemical energy of food is usually reported in calories. In the United States, calories in food are described by the unit kilocalorie (kcal), but the kilo part is often not used. Some brands report calories in food as "Cal," with a capital c to distinguish it from the scientific definition of calories.
Step-By-Step Example
Bomb Calorimeter Calculation
Given a calorimeter with a heat capacity of 4.2 kJ/°C, calculate the chemical energy in kcal stored in 1.0 g of sucrose (C12H22O11) if the combustion causes the temperature of the calorimeter to rise from 25.0°C to 28.9°C.
Step 1
The first step is to use the specific heat and ΔT\Delta T of the calorimeter to calculate the heat absorbed by the calorimeter.
qcal=CcalΔT=(4.2kJ°C)(28.9°C25.0°C)=16.4kJq_{\rm{cal}}=C_{\rm{cal}}\Delta T=\left(4.2\,\frac{\rm{kJ}}{{}\degree\rm C}\right)(28.9\degree\rm C-25.0\degree\rm C)=16.4\;\rm{kJ}
Step 2
The change in energy of the reaction has the same magnitude but opposite sign as the heat absorbed by the calorimeter.
ΔHrxn=qcal=16.4kJ\Delta H_{\rm{rxn}}=-q_{\rm{cal}}=-16.4\;\rm{kJ}
Solution
Thus, the heat released by the combustion of 1.0 g of sucrose—and therefore the heat stored as chemical energy in 1.0 g of sucrose—is 16.4 kJ. To convert this to kcal, use the conversion factor 1cal=4.184J1\rm\;{cal}=4.184\rm\;{J}.
ΔHrxn=(16.4kJ)(1,000J1kJ)(1cal4.184J)(1kcal1,000cal)=3.9kcal\begin{aligned}\Delta H_{\rm{rxn}}&=(16.4\rm\;{kJ})\!\left(\frac{1\rm{,}000\;{J}}{1\rm\;{kJ}}\right)\!\left(\frac{1\rm\;{cal}}{4.184\rm\;J}\right)\!\left(\frac{1\rm\;{kcal}}{1\rm{,}000\;{cal}}\right)\\&=3.9\rm\;{kcal}\end{aligned}
Therefore 1.0 g of sucrose contains 3.9 kcal of energy.

Bomb calorimeters are useful in industrial laboratories, but the more common device in a general chemistry lab is the simple calorimeter, sometimes called a "coffee cup calorimeter," which is a constant-pressure calorimeter. A simple calorimeter works on the same principle as a bomb calorimeter but is usually made of two nested foam cups. A simple calorimeter is typically used to measure the heats of reaction for other types of processes besides combustion reactions. To measure ΔHrxn \Delta H_{\rm {rxn}} , the reactants are placed inside the inner cup, usually in an aqueous solution, and a stirrer and thermometer are inserted through a cover. The stirrer ensures the heat is distributed throughout the reaction system, and the thermometer measures the temperature of the solution.

Because foam is a good insulator, there is essentially no heat exchange with the surroundings of the calorimeter. Thus, any temperature change of the solution is a result of energy either being absorbed or released by the chemical reaction.

In a coffee cup calorimeter,
qrxn+qsoln=0q_{\rm{rxn}}+q_{\rm{soln}}=0
where qrxn is the energy released or used by the reaction (the system), and qsoln is the energy transferred to or from the solution (the surroundings). Therefore,
qrxn=qsolnq_{\rm{rxn}}=-q_{\rm{soln}}
The coffee cup calorimeter measures changes in the solution (the surroundings), not the reaction itself. Because ΔHrxn\Delta H_{\rm{rxn}} is equal to –qsoln, a negative qsoln means ΔHrxn\Delta H_{\rm{rxn}} is positive. A positive ΔHrxn\Delta H_{\rm{rxn}} means the reaction has absorbed energy from the solution, so the reaction is endothermic.
ΔHrxn=qsoln\Delta H_{\rm{rxn}}=-q_{\rm{soln}}
For an exothermic reaction, ΔHrxn\Delta H_{\rm{rxn}} is negative, so qsoln will be positive and the temperature of the solution will rise.
A simple calorimeter is made of two foam cups, one nested inside the other. The test solution is place in the inner cup, a stirrer keeps the solution mixed, and a thermometer measures the temperature of the solution. Any heat transfer that results from a chemical reaction is entirely absorbed by or released by the solution because the cups act as insulation through which heat cannot pass, so measuring the temperature change of the solution allows for the calculation of the amount of energy that has been transferred.
Simple calorimeters can also measure the heat change of processes other than chemical reactions. For example, in order for salt to dissolve, the lattice structure forming the salt must break down. This dissociation has an energy change associated with it, called the heat of solution, or ΔHsoln \Delta H_{\rm{soln}} . For example, the ΔHsoln \Delta H_{\rm{soln}} of NaCl is +3.88 kJ/mol.
Step-By-Step Example
Heat Change for a Physical Change
How much NaCl, in grams, is needed to change the temperature of 125 mL of water in a simple calorimeter by 0.50°C? Will the temperature of the water increase or decrease?
Step 1
First, calculate the heat energy required to change the temperature of 125.0 mL of water by 0.5°C. For this, use the specific heat of water of 4.184 J/(g°C) and assume the density of water is 1.0 g/mL.
mwater=(125.0mL)(1g1mL)=125.0gq=mcΔT=(125.0g)(4.184J(g°C))(0.50°C)=261.5J=0.2615kJ\begin{aligned}m_{\rm{water}}&=(125.0\rm\;{mL})\!\left(\frac{1\rm\;{g}}{1\rm\;{mL}}\right)\\&=125.0 \rm\;{g}\\\\q&=mc\Delta T\\&=(125.0\rm\;{g})\!\left(4.184\frac {\rm{J}}{{\rm{(g}}\,\degree\rm{C)}}\right)\!\left(0.50\degree\rm C\right)\\&=261.5\rm\;{J}\\&=0.2615\rm\;{kJ}\end{aligned}
Step 2
Next, calculate the mass of NaCl required for a q=2.615kJq=2.615\;\rm{kJ}. Note that ΔHsoln\Delta H_{\rm{soln}} is reported in kJ/mol, so the mass required to achieve a given q is as follows:
mNaCl=qΔHsoln=0.2615kJ3.88kJmol=0.0674molm_{\rm{NaCl}}=\frac q{\Delta H_{\rm{soln}}}=\frac{0.2615\;\rm{kJ}}{3.88\;{\displaystyle\frac{\rm{kJ}}{\rm{mol}}}}=0.0674\;\rm{mol}
Solution
Use the molecular weight of NaCl, 58.44 g/mol, to convert moles to mass.
mNaCl=(0.0674mol)(58.44g/mol)=3.9g\begin{aligned}m_{\rm{NaCl}}&=(0.0674\;\rm{mol})(58.44\rm\;{g/mol})\\&=3.9\;\rm{g}\end{aligned}
Thus, if 3.9 g of NaCl is dissolved in 125 mL of water, the temperature of the water will change by 0.50°C. This calculation has taken into account the effect the NaCl has on the specific heat of the water, but that effect is negligible because the mass of the water is much greater than the mass of the NaCl. The second question is whether the water temperature will go up or down by 0.50°C. Because the ΔHsoln>0 \Delta H_{\rm{soln}}\;>\;0 for NaCl, the dissolution process absorbs heat and is endothermic; qNaCl>0q_{\rm{NaCl}}\gt{0}. This means the NaCl uses heat energy from the water to break its ionic bonds, and qwater<0q_{\rm{water}}\lt{0}. In fact, the heat released by the water has the same magnitude but opposite sign as the heat gained by the NaCl.
qwater=qNaClq_{\rm{water}}=-q_{\rm{NaCl}}
Heat energy leaves the water and the temperature of the water decreases by 0.50°C.