Heat Capacity and Calorimetry

AP Chemistry· AP Chemistry CED — Thermodynamics· 14 min read

1. Heat Capacity: Extensive and Intensive Forms★★☆☆☆⏱ 4 min

Heat capacity describes the amount of heat energy required to change the temperature of a given amount of substance by 1°C (or 1 K). Calorimetry is the experimental technique used to measure heat transferred during a chemical or physical change by measuring temperature change of a system with known heat capacity.

📘 Definition

Total Heat Capacity

C

The ratio of heat added to a system to the resulting temperature change: $C = \frac{q}{Δ T}$ . It is an extensive property that scales with the total amount of substance.

Example:

A 1 kg block of iron has a larger total heat capacity than a 1 g block of iron.

Specific heat capacity ( $c$ ): Intensive property normalized by mass: $c = \frac{q}{m Δ T}$ , units $J / (g \cdot^{\circ} C)$ . Core calculation formula: $q = m c Δ T$ .
Molar heat capacity ( $C_{m}$ ): Intensive property normalized by moles: $C_{m} = \frac{q}{n Δ T}$ , units $J / (mol \cdot^{\circ} C)$ .

📐 Worked Example

How much heat is absorbed by 375.0 g of liquid ethanol when its temperature rises from 18.5°C to 62.0°C? The specific heat of ethanol is $2.44 J / (g \cdot^{\circ} C)$ .

1
Identify all known values:
2
$m = 375.0 \text{ g}, c = 2.44 \text{ J}/(\text{g} \cdot ^\circ\text{C}), T_{initial} = 18.5^\circ\text{C}, T_{final} = 62.0^\circ\text{C}$
3
Calculate temperature change $Δ T$ :
4
$\Delta T = T_{final} - T_{initial} = 62.0^\circ\text{C} - 18.5^\circ\text{C} = 43.5^\circ\text{C}$
5
Substitute into the core formula $q = m c Δ T$ :
6
$q = (375.0 \text{ g})(2.44 \text{ J}/(\text{g} \cdot ^\circ\text{C}))(43.5^\circ\text{C})$
7
Calculate the final answer, converting to kilojoules:
8
$q = 39802.5 \text{ J} \approx 39.8 \text{ kJ}$

Exam tip:

Always check that units cancel correctly: if your final answer has leftover mass or mole units, you used the wrong form of heat capacity.

2. Constant-Pressure (Coffee-Cup) Calorimetry★★★☆☆⏱ 4 min

Constant-pressure calorimetry is the most common simple technique for measuring enthalpy change ( $Δ H$ ) for reactions run in open containers at atmospheric pressure, like dissolution, neutralization, or precipitation. By definition, at constant pressure, the heat transferred by the reaction equals the enthalpy change: $Δ H = q_{p}$ .

The core assumption of simple coffee-cup calorimetry is that no heat is exchanged with the environment outside the calorimeter, and the heat capacity of the foam cup itself is negligible. This gives the key relationship:

q_{rxn} = -q_{solution} = -(mc\Delta T)_{solution}

Most AP problems assume dilute aqueous solutions have the same density (1.00 g/mL) and specific heat as pure water, so you can use $c = 4.184 J / (g \cdot^{\circ} C)$ to calculate heat change.

📐 Worked Example

A student dissolves 4.00 g of ammonium nitrate ( $N H_{4} N O_{3}$ , molar mass = 80.04 g/mol) in 125 g of water in a coffee-cup calorimeter. The temperature drops from 24.1°C to 18.7°C. Calculate the molar enthalpy of dissolution of ammonium nitrate.

1
Calculate total mass of the solution (add solute mass to solvent mass):
2
$m_{total} = 4.00 \text{ g} + 125 \text{ g} = 129 \text{ g}$
3
Calculate $Δ T$ :
4
$\Delta T = 18.7^\circ\text{C} - 24.1^\circ\text{C} = -5.4^\circ\text{C}$
5
Calculate $q_{so l u t i o n}$ :
6
$q_{solution} = (129 \text{ g})(4.184 \text{ J}/(\text{g} \cdot ^\circ\text{C}))(-5.4^\circ\text{C}) \approx -2910 \text{ J} = -2.91 \text{ kJ}$
7
Relate $q_{r x n}$ to $q_{so l u t i o n}$ :
8
$q_{rxn} = -q_{solution} = 2.91 \text{ kJ}$
9
Calculate moles of ammonium nitrate:
10
$n = 4.00 g /80.04 g/mol = 0.04998 mol$
11
Calculate molar enthalpy of dissolution:
12
$\Delta H_{diss} = \frac{2.91 \text{ kJ}}{0.04998 \text{ mol}} \approx +58.2 \text{ kJ/mol}$

Exam tip:

If the reaction causes a temperature drop, it is endothermic, so $Δ H$ will be positive — this is a quick sanity check for your final sign.

3. Constant-Volume (Bomb) Calorimetry★★★☆☆⏱ 3 min

Constant-volume (bomb) calorimetry is used for combustion reactions, which require a sealed, high-pressure container. The bomb is filled with oxygen, the sample is ignited, and heat released by combustion raises the temperature of a surrounding water bath. Because volume is constant ( $Δ V = 0$ ), pressure-volume work $w = - P Δ V = 0$ , so by the first law of thermodynamics, $Δ U = q_{v}$ , meaning the heat measured directly equals the change in internal energy.

For bomb calorimetry, we use the pre-calibrated total heat capacity of the entire calorimeter assembly ( $C_{c a l}$ , units $kJ /^{\circ} C$ ), which already accounts for the mass of the bomb, water, and container. The core relationship is:

q_{rxn} = -C_{cal} \Delta T

For most AP problems, you can approximate $Δ H \approx q_{v}$ , so the calculated value is the approximate molar enthalpy of combustion.

📐 Worked Example

Combustion of 0.750 g of caffeine increases the temperature of a bomb calorimeter with $C_{c a l} = 4.22 kJ /^{\circ} C$ by 1.85°C. The molar mass of caffeine is 194.2 g/mol. Calculate the molar enthalpy of combustion of caffeine.

1
Calculate heat gained by the calorimeter:
2
$q_{cal} = C_{cal} \Delta T = (4.22 \text{ kJ}/^\circ\text{C})(1.85^\circ\text{C}) = 7.807 \text{ kJ}$
3
Relate $q_{r x n}$ to $q_{c a l}$ :
4
$q_{rxn} = -q_{cal} = -7.807 \text{ kJ}$
5
Calculate moles of caffeine:
6
$n = 0.750 g /194.2 g/mol = 0.003862 mol$
7
Calculate molar enthalpy of combustion:
8
$\Delta H_{comb} = \frac{-7.807 \text{ kJ}}{0.003862 \text{ mol}} \approx -2020 \text{ kJ/mol}$

Exam tip:

If the problem asks for $Δ U$ instead of $Δ H$ , the answer is just the calculated $q_{v}$ , no further adjustment is needed for AP-level problems.

4. AP Style Concept Check★★★★☆⏱ 3 min

✓ Quick check

Test your understanding with this AP-style multiple choice question:

A student mixes 50 mL of 1.0 M HNO₃(aq) and 50 mL of 1.0 M KOH(aq) in a coffee-cup calorimeter. The initial temperature of both solutions is 22.0°C, and the final temperature is 28.7°C. Assuming the density of the solution is 1.0 g/mL and $c = 4.18 J / (g \cdot^{\circ} C)$ , what is the approximate molar enthalpy of neutralization?
- +57 kJ/mol
- -57 kJ/mol
- +28 kJ/mol
- -28 kJ/mol
Reveal answer
1 —
Correct! Neutralization is exothermic, so $Δ H$ is negative. Calculations give ~-57 kJ/mol, matching this option.

5. Common Pitfalls

Wrong move:

Only using the mass of water to calculate $q_{so l u t i o n}$ in coffee-cup calorimetry, ignoring the mass of the dissolved solute.

Why:

Students memorize the formula as $q = m_{w a t er} c Δ T$ and forget the solute adds to the total mass of the solution that absorbs or releases heat.

Correct move:

Always add the mass of the solute to the mass of the solvent to get the total mass of the solution before calculating $q$ .

Wrong move:

Taking the absolute value of $Δ T$ early, leading to the wrong sign for $Δ H$ .

Why:

Students think 'temperature changed by 8 degrees' so they drop the sign, forgetting the sign encodes direction of heat flow.

Correct move:

Always calculate $Δ T = T_{f ina l} - T_{ini t ia l}$ first, and preserve the sign through all subsequent steps.

Wrong move:

Leaving $Δ H$ in J/mol instead of converting to kJ/mol as requested.

Why:

Specific heat is usually given in J/(g·°C), so $q$ comes out in joules, but AP exam questions almost always request $Δ H$ in kJ/mol.

Correct move:

Convert $q$ from joules to kilojoules immediately after calculation, before finding molar enthalpy.

Wrong move:

Multiplying $C_{c a l}$ by mass in bomb calorimetry problems, leading to an answer multiple orders of magnitude wrong.

Why:

Students confuse total heat capacity of the calorimeter with specific heat, which requires a mass term.

Correct move:

Check the units of the given heat capacity: if units are kJ/°C (no mass term), use $q = C_{c a l} Δ T$ , no mass needed.

Wrong move:

Forgetting to invert the sign between $q_{c a l} / q_{so l u t i o n}$ and $q_{r x n}$ , leading to positive $Δ H$ for exothermic reactions.

Why:

Students mix up which system absorbs vs releases heat: if the reaction releases heat, the calorimeter absorbs it, so $q_{c a l}$ is positive, $q_{r x n}$ must be negative.

Correct move:

Write the relationship 'heat lost by system 1 = heat gained by system 2 → $q_{1} = - q_{2}$ ' and label each system before starting calculations.

6. Quick Reference Cheatsheet

Category	Formula	Notes
Total heat capacity	$C = \frac{q}{Δ T}$	Extensive, for full calorimeter, units J/°C
Specific heat capacity	$c = \frac{q}{m Δ T}$	Intensive, per gram, units J/(g·°C)
Molar heat capacity	$C_{m} = \frac{q}{n Δ T}$	Intensive, per mole, units J/(mol·°C)
Constant-pressure (coffee-cup)	$q_{r x n} = - m c Δ T$	$Δ H = q_{r x n}$ at constant pressure
Constant-volume (bomb)	$q_{r x n} = - C_{c a l} Δ T$	$Δ U = q_{r x n}$ , $Δ H \approx q_{r x n}$ for AP

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · MCQ
Coffee-cup enthalpy calculation
2022 · FRQ
Bomb calorimetry error analysis

What's Next

Heat capacity and calorimetry form the experimental foundation for all enthalpy calculations in AP Chemistry thermodynamics. These skills are frequently combined with other thermodynamics concepts in multi-part FRQ questions, so mastering sign conventions and calculation steps is critical for exam success. Next, you will build on these skills to learn how to combine reaction enthalpies with Hess's law, and calculate reaction enthalpy from standard enthalpy of formation values for more complex processes.