Calorimetry

Calorimetry lets chemists measure heat with a thermometer. It avoids using a bomb calorimeter's namesake explosion. Whenever a reaction runs inside an insulated container, this happens. The heat it releases or absorbs shows up as a temperature change. This change occurs in the surrounding water or metal. That temperature change is easy to measure precisely. Because of this, calorimetry turns an abstract quantity into a measurable number. That quantity is enthalpy change. You can read it off a thermometer. This article builds the two central tools of experimental thermochemistry. These are the coffee-cup calorimeter for reactions at constant pressure. The other is the bomb calorimeter for reactions at constant volume. Both rest on one simple accounting idea. Heat lost by one part of the system equals heat gained by another. This idea is dressed up with the heat capacity equation $q = mc\Delta T$ .

Key takeaways

Heat capacity $C$ and specific heat $c$ quantify how much heat changes a substance's temperature; $q = mc\Delta T$ connects mass, specific heat, and temperature change to heat flow.
In an isolated calorimeter, heat lost by the hot object equals heat gained by the cold surroundings: $q_{released} = -q_{absorbed}$ .
Coffee-cup calorimeters run at constant pressure and measure $q_p = \Delta H$ directly — ideal for solution reactions.
Bomb calorimeters run at constant volume and measure $q_v = \Delta E$ , using a calibrated heat capacity $C_{cal}$ for the whole device.
Careful calorimetry underlies tables of enthalpies of reaction and feeds directly into Hess's Law and Enthalpies of Formation.
Key idea: temperature change is a messenger — measure $\Delta T$ precisely, and heat capacity translates it into $q$ , and ultimately into $\Delta H$ .

The core insight

Calorimetry never measures heat directly. It measures a temperature change, then converts that change into heat using a known heat capacity, exploiting the fact that within an isolated system, energy is conserved.

The physical picture

Imagine dropping a hot piece of metal into cool water inside a well-insulated cup. The metal cools, the water warms, and the two settle at a common final temperature. No energy has left the cup — it has only moved from the metal to the water. That single sentence is the entire logic of calorimetry:

q_{metal} = -q_{water}

The negative sign says the two heat flows are equal in magnitude but opposite in direction: whatever energy the metal gives up is exactly the energy the water takes in. This works because the cup is (approximately) an isolated system — no heat escapes to the lab, and none enters from outside. The same logic scales up to chemical reactions. Instead of a hot metal block, picture a neutralization reaction happening inside a cup of water. The reaction mixture is both reactant and surroundings — as bonds break and form, the released or absorbed heat is captured by the solution itself, and the solution's temperature change reveals how much heat was involved. To connect temperature change to heat, every substance has a heat capacity $C$ — the heat required to raise its temperature by one degree Celsius (equivalently, one kelvin, since the two scales share step size). Heat capacity is an extensive property: it scales with the amount of substance. Dividing by mass gives the specific heat $c$ , an intensive property characteristic of the substance itself, with units $\text{J}\,\text{g}^{-1}\,^\circ\text{C}^{-1}$ . Water's specific heat, $4.184\ \text{J g}^{-1\,\circ}\text{C}^{-1}$ , is unusually large — one reason large bodies of water moderate coastal climates, and why water is the workhorse fluid of calorimetry.

Figure

Heat flows from the warmer reaction mixture into the surrounding water bath until both reach a common final temperature

The core relations

q = mc\Delta T

where:

$q$ = heat absorbed (positive) or released (negative), in joules
$m$ = mass of the substance, in grams
$c$ = specific heat of the substance, in $\text{J g}^{-1\,\circ}\text{C}^{-1}$
$\Delta T = T_{final} - T_{initial}$ , in $^\circ$ C or K

q = C\Delta T

where:

$C = mc$ is the total heat capacity of the object or calorimeter (J/°C), used when a device's heat capacity is calibrated as a whole rather than per gram

q_{system} = -q_{surroundings}

where:

this is the conservation-of-energy statement for an isolated calorimeter; heat is neither created nor destroyed, only transferred

q_p = \Delta H \qquad q_v = \Delta E

where:

$q_p$ = heat measured at constant pressure (coffee-cup calorimeter), equal to the enthalpy change of the reaction
$q_v$ = heat measured at constant volume (bomb calorimeter), equal to the change in internal energy $\Delta E$

Derivation AFrom measured ΔT to reaction ΔH in a coffee-cup calorimeter

Goal. Show how a measured temperature change in a dilute aqueous solution converts into the enthalpy of reaction per mole of reactant.

Step 1. Treat the solution (reactants plus water) as the "surroundings" that absorb whatever heat the reaction releases. Because the cup is open to the atmosphere but insulated from heat loss, pressure is constant, so the heat measured is directly the enthalpy change of the reaction: $q_p = \Delta H_{rxn}$ in magnitude, with sign convention to be fixed in Step 4.

Step 2. Approximate the dilute solution's mass and specific heat with those of pure water — a standard assumption for dilute aqueous reactions, since the solute's contribution to total heat capacity is small.

q_{soln} = m_{soln}\, c_{water}\, \Delta T

Step 3. Measure $\Delta T = T_{final} - T_{initial}$ for the solution. If the reaction is exothermic, the solution warms up, so $\Delta T > 0$ and $q_{soln} > 0$ — heat flowed into the solution from the reaction.

Step 4. Apply energy conservation between the reaction and the solution it sits in:

q_{rxn} = -q_{soln}

An exothermic reaction ( $q_{rxn} < 0$ ) heats the solution ( $q_{soln} > 0$ ), consistent with the minus sign.

Step 5. Convert to a per-mole quantity by dividing by moles of limiting reactant $n$ , since $\Delta H$ is usually reported as an intensive molar quantity:

\Delta H_{rxn} = \frac{q_{rxn}}{n}

The final expression, combining all steps, is

\Delta H_{rxn} = -\frac{m_{soln}\, c_{water}\, \Delta T}{n}

This single line is the working equation for every coffee-cup calorimetry problem: measure mass, specific heat, temperature change, and moles, and the molar enthalpy of reaction falls out.

Go deeper · Derivation B · 1 minCalibrating and using a bomb calorimeter

Derivation BCalibrating and using a bomb calorimeter

Goal. Find the internal energy change of a combustion reaction using a bomb calorimeter's total heat capacity, which must first be calibrated.

Step 1. A bomb calorimeter is a sealed, rigid steel vessel submerged in a known mass of water, itself inside an insulated jacket. Because the vessel's volume cannot change, no pressure–volume work is done: $w = -P\Delta V = 0$ . The first law then gives $\Delta E = q_v + w = q_v$ , so the heat measured at constant volume equals the internal energy change directly.

Step 2. The calorimeter itself (steel bomb, stirrer, thermometer, surrounding water) has a combined heat capacity $C_{cal}$ , which is not simply "mass of water times $c_{water}$ " because the metal parts also absorb heat. $C_{cal}$ must be measured once, experimentally, using a reaction of precisely known heat output — commonly the combustion of benzoic acid, for which $\Delta E_{comb}$ is tabulated.

Step 3. Burn a known mass of benzoic acid, record $\Delta T_{cal}$ , and back out the calorimeter constant:

C_{cal} = \frac{q_{known}}{\Delta T_{cal}} = \frac{-n_{acid}\,\Delta E_{comb,\,acid}}{\Delta T_{cal}}

Step 4. With $C_{cal}$ now known, burn the unknown sample in the same device and record its temperature rise $\Delta T_{sample}$ . The heat absorbed by the calorimeter is

q_{cal} = C_{cal}\,\Delta T_{sample}

Step 5. Apply the same conservation logic as before — heat released by combustion is absorbed by the calorimeter:

q_{rxn} = -q_{cal} = -C_{cal}\,\Delta T_{sample}

Step 6. Since this heat was measured at constant volume, it equals $\Delta E$ for the combustion, not $\Delta H$ . Dividing by moles of sample burned gives the molar internal energy of combustion:

\Delta E_{comb} = \frac{q_{rxn}}{n_{sample}} = \frac{-C_{cal}\,\Delta T_{sample}}{n_{sample}}

Converting this $\Delta E$ to $\Delta H$ (if needed) requires the relation $\Delta H = \Delta E + \Delta n_{gas}RT$ , which belongs to a broader treatment of gas-phase thermochemistry — but the calorimetry itself stops at $\Delta E$ .

Worked examples

Example 1 — Coffee-cup calorimeter, neutralization reaction. A student mixes $50.0\ \text{mL}$ of $1.20\ M$ HCl with $50.0\ \text{mL}$ of $1.20\ M$ NaOH in a coffee-cup calorimeter. The temperature rises from $22.3\,^\circ\text{C}$ to $30.7\,^\circ\text{C}$ . Assume the combined solution has density $1.02\ \text{g/mL}$ and specific heat $4.02\ \text{J g}^{-1\,\circ}\text{C}^{-1}$ (typical adjustments for a dilute salt solution rather than pure water). Find $\Delta H$ per mole of water formed. Mass of solution:

m_{soln} = (50.0 + 50.0)\ \text{mL} \times 1.02\ \text{g/mL} = 102\ \text{g}

Temperature change:

\Delta T = 30.7 - 22.3 = 8.4\,^\circ\text{C}

Heat absorbed by the solution:

q_{soln} = (102\ \text{g})(4.02\ \text{J g}^{-1\,\circ}\text{C}^{-1})(8.4\,^\circ\text{C}) = 3.44 \times 10^3\ \text{J}

Heat released by the reaction:

q_{rxn} = -q_{soln} = -3.44 \times 10^3\ \text{J} = -3.44\ \text{kJ}

Moles of limiting reactant (HCl and NaOH are stoichiometrically equal here, both $0.0500\ \text{L} \times 1.20\ M = 0.0600\ \text{mol}$ ):

n = 0.0600\ \text{mol}

Molar enthalpy of neutralization:

\Delta H = \frac{-3.44\ \text{kJ}}{0.0600\ \text{mol}} = -57.3\ \text{kJ/mol}

This lands close to the accepted value for strong acid–strong base neutralization, around $-57.1$ kJ/mol, confirming the method. As a limiting-case check: if the reaction released no heat at all ( $\Delta T \to 0$ ), the formula correctly gives $\Delta H \to 0$ — no reaction, no temperature change, no enthalpy signal, exactly as expected.

Example 2 — Bomb calorimeter, calibration then combustion. A calorimeter is first calibrated by burning $0.7500\ \text{g}$ of benzoic acid ( $\Delta E_{comb} = -26.38\ \text{kJ/g}$ , a per-gram value often quoted directly for this standard). The temperature rises by $2.98\,^\circ\text{C}$ . Heat released by the known combustion:

q_{known} = (0.7500\ \text{g})(-26.38\ \text{kJ/g}) = -19.79\ \text{kJ}

Calorimeter constant:

C_{cal} = \frac{-q_{known}}{\Delta T_{cal}} = \frac{19.79\ \text{kJ}}{2.98\,^\circ\text{C}} = 6.640\ \text{kJ/}^\circ\text{C}

Now a $1.100\ \text{g}$ sample of glucose ( $M = 180.16\ \text{g/mol}$ ) is burned in the same calorimeter, producing a temperature rise of $3.65\,^\circ\text{C}$ . Heat absorbed by the calorimeter:

q_{cal} = (6.640\ \text{kJ/}^\circ\text{C})(3.65\,^\circ\text{C}) = 24.24\ \text{kJ}

Heat released by combustion:

q_{rxn} = -24.24\ \text{kJ}

Moles of glucose burned:

n = \frac{1.100\ \text{g}}{180.16\ \text{g/mol}} = 6.105 \times 10^{-3}\ \text{mol}

Molar internal energy of combustion:

\Delta E_{comb} = \frac{-24.24\ \text{kJ}}{6.105 \times 10^{-3}\ \text{mol}} = -3971\ \text{kJ/mol}

This is close to the literature value near $-2803$ kJ/mol for $\Delta H_{comb}$ of glucose — the two numbers are not expected to match exactly, since one is $\Delta E$ at constant volume and the other is the tabulated $\Delta H$ at constant pressure; the gap illustrates precisely why the derivation stopped short of converting between them. As a check on the calibration step alone: if the known reaction had released exactly zero heat, $C_{cal}$ would be undefined (division by $\Delta T_{cal} = 0$ ), correctly signaling that calibration requires a genuine, measurable temperature change.

Limiting cases & common traps

Situation	What actually happens	Common mistake
$\Delta T \to 0$	$q \to 0$ regardless of mass or specific heat	Forgetting sign conventions and reporting $\Delta H$ with the wrong sign for exothermic reactions
Very dilute aqueous solution	$m$ and $c$ of pure water are good approximations	Using the solute's mass alone instead of total solution mass
Bomb calorimeter data used for $\Delta H$	Only $\Delta E$ is measured directly; converting to $\Delta H$ needs $\Delta n_{gas}RT$	Reporting bomb-calorimeter $q_v$ as if it were $\Delta H$ without conversion
Coffee-cup calorimeter with a metal or plastic cup	The cup itself absorbs some heat, causing small systematic error	Assuming the cup has zero heat capacity when high precision is needed
Reaction run open to air, heat escapes	Measured $\Delta T$ underestimates true heat released	Not correcting for heat loss to surroundings beyond the calorimeter

The chemist behind it

Calorimetry as a rigorous quantitative technique traces to the late eighteenth century, when the concept of heat itself was still contested — was heat a substance ("caloric") or a form of motion? Antoine Lavoisier and Pierre-Simon Laplace built an early ice calorimeter in 1782–1783, measuring heat by how much ice a process could melt, and used it to compare the heat released by respiration in a guinea pig against that from burning carbon, an experiment that helped establish respiration as a form of slow combustion. Their device measured heat quantitatively without settling the caloric debate, but it established the core principle still used today: surround a process with a measurable, calibrated medium, and let its response reveal the heat involved. Later work, particularly by Pierre Berthelot in the 1880s with his development of the bomb calorimeter, refined the constant-volume technique used for combustion reactions into the precise tool described here.

Origins · 1 minAntoine Lavoisier and how this was discovered

Antoine Lavoisier

1743–1794 · French chemist working at the close of the Enlightenment, active as both scientist and tax administrator; guillotined during the French Revolution. With Laplace, built an ice calorimeter to quantify heat released by combustion and respiration, helping to found quantitative thermochemistry and, more broadly, modern chemistry's insistence on mass and energy accounting.

Ask the tutor: If a reaction in a coffee-cup calorimeter releases enough heat to raise the solution's temperature above the cup's insulation limit, how would heat loss to the surroundings bias the calculated ΔH — would the magnitude come out too large or too small?

Why is water used as the medium in most calorimeters?

Water has an unusually high specific heat, $4.184\ \text{J g}^{-1\,\circ}\text{C}^{-1}$ , meaning it absorbs a large amount of heat per degree of temperature rise. This makes temperature changes in a water bath small and precisely measurable relative to the heat involved, and it makes water an excellent, cheap, and chemically inert medium for capturing heat from a reaction without itself reacting.

What's the practical difference between a coffee-cup calorimeter and a bomb calorimeter?

A coffee-cup calorimeter is open to the atmosphere, so it measures reactions at constant pressure — appropriate for solution-phase reactions like acid–base neutralizations or dissolution. A bomb calorimeter is a sealed, rigid vessel used for combustion reactions, especially ones involving gases, and measures heat at constant volume. The coffee-cup gives $\Delta H$ directly; the bomb gives $\Delta E$ directly.

Why must a bomb calorimeter be calibrated with a known reaction before use?

The calorimeter's total heat capacity $C_{cal}$ includes contributions from the steel bomb, the stirrer, the thermometer, and the surrounding water — quantities that are hard to calculate separately from first principles. Burning a substance with a precisely known $\Delta E_{comb}$ , most commonly benzoic acid, lets the whole device be treated as one calibrated instrument with a single known heat capacity.

Does calorimetry ever measure heat capacity itself as the primary unknown?

Yes — the calibration step in bomb calorimetry is exactly this: a known amount of heat is released by a standard reaction, and the resulting $\Delta T$ is used to solve for the calorimeter's unknown heat capacity $C_{cal}$ , which is then treated as fixed for later experiments using the same apparatus.

How does calorimetry connect to Hess's Law and enthalpies of formation?

Calorimetry supplies the experimentally measured $\Delta H$ values that fill thermochemical tables. Once enough reaction enthalpies are known from calorimetry, Hess's Law allows combining them algebraically to find enthalpies of reactions that are difficult or dangerous to run directly, and standard Enthalpies of Formation are themselves built up from a network of calorimetric measurements.