The Calories printed on a box of crackers were measured by burning a small dried sample in a sealed steel container under water and reading how much the water warmed. The same idea, catching the heat of a process in a measured amount of water, works for reactions in a foam cup on a lab bench.

This lesson shows how to turn a temperature change in a calorimeter into the heat of a reaction. You will find the heat capacity of a calorimeter by calibrating it, then use it with a coffee-cup calorimeter at constant pressure and a bomb calorimeter at constant volume (Figure 1).

1. ΔT of the calorimeterfinal − initial, in °C
2. $\times$mc for the solution, C_cal for the cup
3. Heat gained$q_\text{soln} + q_\text{cal}$by the solution and the calorimeter
4. $\times (-1)$equal and opposite
5. q for the reactionnegative if the calorimeter warms

Calibrating? $C_\text{cal} = \dfrac{q_\text{cal}}{ΔT}$.Bomb calorimeter? $q_\text{rxn} = -C_\text{cal}\,ΔT$.

Learning Objectives

After this lesson, you should be able to:

1. Analyse a coffee-cup (constant-pressure) calorimetry experiment, including the calorimeter heat capacity.
2. Analyse a bomb (constant-volume) calorimetry experiment.

Key Ideas

1. What a Calorimeter Measures

Calorimetry is the measurement of heat. A calorimeter is an insulated container in which a reaction or a physical process takes place; the insulation keeps heat from entering or leaving, so the heat of the process stays inside. The process is the system; the solution or water around it and the container itself are the surroundings that we measure.

A simple coffee-cup calorimeter is two nested foam cups with a lid, a thermometer and a stirrer (Figure 2). The thermometer reads to 0.01 °C, because the temperature changes are often only a few degrees.

Because no heat escapes, the heat released or absorbed by the reaction is exactly balanced by the heat gained or lost by everything else in the calorimeter:

Key idea: a calorimeter never measures the heat of the reaction directly. It measures a temperature change of the surroundings, and the heat of the reaction is equal in size and opposite in sign.

2. Coffee-Cup Calorimetry at Constant Pressure

A coffee-cup calorimeter is open to the air through its loose lid, so the reaction takes place at constant pressure. The heat measured this way is written $q_\text{p}$; a later lesson connects it to the enthalpy change of the reaction.

The solution gains heat $q_\text{soln} = mcΔT$, where m is the mass of the whole solution. For dilute aqueous solutions we use the specific heat of water, 4.184 J/(g·°C), and find the mass from the total volume and the density. The cup itself also warms; its share is $q_\text{cal} = C_\text{cal}ΔT$, where $C_\text{cal}$ is the heat capacity of the calorimeter. Rearranging the balance gives the heat of the reaction:

In Worked Example 2 the cup takes 5.4% of the heat: small, but large enough to change the third significant figure (Figure 3).

The sign of q for the reaction follows from the temperature. If the solution warms, the reaction released heat: it is exothermic and $q_\text{rxn}$ is negative. If the solution cools, the process absorbed heat: it is endothermic and $q_\text{rxn}$ is positive (Figure 4: Panel (a) warms, Panel (b) cools).

Everyday note: an instant cold pack is an endothermic dissolving process: ammonium nitrate takes heat from the water around it, so the pack feels cold.

Exam tip: use the mass of the whole solution, both volumes or the solid plus the water, not only one of the parts.

3. The Heat Capacity of the Calorimeter

The heat capacity $C_\text{cal}$ of a calorimeter is the heat it needs to warm by 1 °C, in J/°C or kJ/°C. It is found by calibration: put a known amount of heat into the calorimeter and measure the temperature rise. An electric heater is a convenient source because the energy it delivers is known accurately.

The heater’s heat warms both the water and the calorimeter. Subtract the water’s share, then divide what is left by the temperature change:

In Worked Example 1 the calorimeter takes 7.9% of the heater’s heat (Figure 5). Once $C_\text{cal}$ is known, the same calorimeter can be used for any reaction.

Key idea: calibrate first, measure second: every calorimeter result needs the heat capacity of that particular calorimeter.

4. Bomb Calorimetry at Constant Volume

Combustion reactions are measured in a bomb calorimeter: a sealed steel container (the bomb) holds the sample and oxygen under pressure and sits in a measured amount of water inside an insulated jacket. An electric spark ignites the sample, and the heat released warms the bomb and the water (Figure 6). The bomb cannot expand, so the reaction takes place at constant volume; the heat measured is written $q_\text{v}$, and a later lesson connects it to the internal energy change.

The bomb, the water and the jacket are treated together as one calorimeter with a single heat capacity. It is calibrated by burning a known mass of a standard, usually benzoic acid, which releases 26.46 kJ/g of heat. Then, for any other sample burned in the same calorimeter:

Dividing by the mass of the sample gives the heat released in kJ/g, the way fuel and food energies are reported.

Key idea: in a bomb calorimeter all the heat goes into one calorimeter of known heat capacity, so q for the reaction is simply −$C_\text{cal}ΔT$.

Worked Example 1: Calibrating a Coffee-Cup Calorimeter

Problem: To calibrate a coffee-cup calorimeter, a student adds 75.0 g of water at 21.45 °C and runs an electric heater in it. The heater supplies 3.977 kJ of heat, and the temperature rises to 33.12 °C. The specific heat of water is 4.184 J/(g·°C). What is the heat capacity of the calorimeter, in J/°C?

Step 1. Find the temperature change.

Step 2. Find the heat gained by the water.

Step 3. The rest of the heater’s heat went into the calorimeter.

Step 4. Divide by the temperature change; write q in J.

Final Answer: C_cal = 27.0 J/°C

The subtraction in Step 3 keeps three decimal places, so the heat capacity has three significant figures. The calorimeter takes only a small part of the heat, but leaving it out of later measurements would make every result too small.

1. Heater$3.977\,\text{kJ}$
2. $-\, 3.662\,\text{kJ}$Steps 2 and 3: subtract the water’s share
3. Calorimeter’s share$q_\text{cal} = 0.315\,\text{kJ}$
4. $\times \dfrac{1}{11.67\,\text{°C}}$Step 4: divide by ΔT, kJ to J
5. Heat capacity$C_\text{cal} = 26.99\,\text{J/°C} \approx 27.0\,\text{J/°C}$3 significant figures

Your turn: calibrate a calorimeter

Same four steps as Worked Example 1: ΔT, the heat gained by the water, the calorimeter’s share by subtraction, then divide by ΔT and convert to J/°C.

Worked Example 2: Heat of a Neutralization Reaction

Problem: In a coffee-cup calorimeter, 60.0 mL of hydrochloric acid and 40.0 mL of sodium hydroxide solution, both at 22.40 °C, are mixed. The neutralization raises the temperature to 28.85 °C. The final solution has a density of 1.02 g/mL and the specific heat of water, 4.184 J/(g·°C); the heat capacity of the calorimeter is 24.5 J/°C. What is q for the reaction, in kJ?

Step 1. Find the mass of the solution from its total volume and density. Both solutions end up in the cup.

Step 2. Find the temperature change. The temperature rises, so ΔT is positive.

Step 3. Find the heat gained by the solution and by the calorimeter.

Step 4. The reaction’s heat is equal and opposite.

Final Answer: q = −2.91 kJ

The solution warmed, so the reaction released heat: q is negative and the neutralization is exothermic. The heat is for the amounts mixed here; a later lesson turns it into a heat in kJ/mol of reaction.

1. Mass of the solution$m = 102.0\,\text{g}$both volumes
2. $\times\, c\,ΔT$Step 3: and C_calΔT for the cup
3. Heat gained$q_\text{soln} = 2.753\,\text{kJ},\ q_\text{cal} = 0.1580\,\text{kJ}$
4. $\times (-1)$Step 4: add, then change the sign
5. q for the reaction$q_\text{rxn} = -2.911\,\text{kJ} \approx -2.91\,\text{kJ}$Negative: exothermic

Your turn: coffee-cup calorimetry

Same four steps as Worked Example 2: the mass of the whole solution, ΔT with its sign, the heat gained by the solution and by the calorimeter, then change the sign of their sum. Some problems dissolve a solid in water; add the mass of the solid to the mass of the water. Type a plus or minus sign with your answer.

Worked Example 3: Calibrating a Bomb Calorimeter

Problem: A bomb calorimeter is calibrated by burning 0.875 g of benzoic acid, which releases 26.46 kJ/g of heat. The temperature of the calorimeter rises from 23.18 °C to 26.04 °C. What is the heat capacity of the calorimeter, in kJ/°C?

Step 1. Find the heat released by the benzoic acid. The calorimeter absorbs all of it.

Step 2. Find the temperature change.

Step 3. Divide the heat by the temperature change.

Final Answer: C_cal = 8.09 kJ/°C

A bomb calorimeter holds much more water and steel than a foam cup, so its heat capacity is in kJ/°C rather than J/°C.

1. Benzoic acid burned$0.875\,\text{g}$
2. $\times\, 26.46\,\text{kJ/g}$Step 1: known heat of the standard
3. Heat released$q = 23.15\,\text{kJ}$
4. $\times \dfrac{1}{2.86\,\text{°C}}$Step 3: divide by ΔT
5. Heat capacity$C_\text{cal} = 8.094\,\text{kJ/°C} \approx 8.09\,\text{kJ/°C}$kJ/°C for a bomb

Your turn: calibrate a bomb calorimeter

Same three steps as Worked Example 3: the heat from the mass of benzoic acid, ΔT, then divide.

Worked Example 4: Energy Content of a Fuel

Problem: A 0.650 g sample of coal is burned in the bomb calorimeter of Worked Example 3, whose heat capacity is 8.09 kJ/°C. The temperature rises from 22.65 °C to 25.36 °C. What is q for the combustion, in kJ/g of coal?

Step 1. Find the temperature change.

Step 2. Find the heat absorbed by the calorimeter; the reaction released the same amount. The calorimeter warms, so the combustion is exothermic and q is negative.

Step 3. Divide by the mass of the sample.

Final Answer: q = −33.7 kJ/g

Every gram of this coal releases about 33.7 kJ when it burns. Reporting the heat in kJ/g lets you compare samples of different sizes.

1. Temperature change$ΔT = 2.71\,\text{°C}$
2. $\times\, (-8.09\,\text{kJ/°C})$Step 2: q for the reaction is −C_calΔT
3. Heat of combustion$q_\text{rxn} = -21.92\,\text{kJ}$
4. $\times \dfrac{1}{0.650\,\text{g}}$Step 3: divide by the mass of the sample
5. q in kJ/g$-33.72\,\text{kJ/g} \approx -33.7\,\text{kJ/g}$Negative: heat released

Your turn: heat released in kJ/g

Same three steps as Worked Example 4: ΔT, then q for the reaction as −C_calΔT, then divide by the mass of the sample. Type the minus sign.

Common Mistakes

Throughout, keep 4 significant figures in intermediate values and round only the final answer to 3.

Common mistake: Leaving out the calorimeter. The cup warms too. In Worked Example 2, using only the solution gives −2.75 kJ instead of −2.91 kJ. Add $q_\text{cal} = C_\text{cal}ΔT$ before you change the sign.

Common mistake: Giving q for the reaction the sign of the temperature change. The thermometer measures the surroundings. When the solution warms, the reaction released heat, so q for the reaction is negative; when it cools, q is positive.

Common mistake: Using only part of the solution’s mass. Both solutions are in the cup. In Worked Example 2, using only the 60.0 mL of acid (61.2 g) gives −1.81 kJ. When a solid dissolves, add its mass to the mass of the water.

Common mistake: Using a temperature instead of the temperature change. Every heat in calorimetry uses ΔT = final − initial, never one of the temperatures.

Common mistake: Stopping before dividing by the mass. In Worked Example 4, −21.9 kJ is the heat for the whole 0.650 g sample. The heat in kJ/g is −33.7 kJ/g.

Common mistake: Mixing J and kJ. A coffee cup’s heat capacity is usually given in J/°C and a bomb’s in kJ/°C. In Worked Example 1, dividing 0.315 kJ by 11.67 °C gives 0.0270 kJ/°C; reported as J/°C it would be 1000 times too small. Convert to J first to get 27.0 J/°C.

Key Takeaway

A calorimeter is insulated, so the heat of a process shows up as a temperature change of the solution and the calorimeter: $q_\text{rxn} = -(q_\text{soln} + q_\text{cal})$. The heat capacity of a calorimeter is found by calibration, with a heater or with benzoic acid (26.46 kJ/g). A coffee-cup calorimeter works at constant pressure; a bomb calorimeter works at constant volume, and there $q_\text{rxn} = -C_\text{cal}ΔT$. A rise in temperature means an exothermic process with a negative q.

Practice: randomized problems

Each problem below uses fresh numbers every time and mixes every kind of problem in this lesson: work it out, check your answer, then open the worked solution to compare your steps.

Your results

This summary collects every practice problem you try on this page and suggests what to review next.

Figures: original diagrams by Dr. Ray’s Learning Center, licensed CC BY 4.0. Data: Whitten, Davis, Peck and Stanley, Chemistry, 9th ed. (Brooks/Cole, Cengage Learning, 2010), Section 15-4 (liquid water) and Section 15-11 (benzoic acid, p. 572). The practice problems are generated and checked by an audited engine; see the lesson audit log.