# Heat Capacity and Calorimetry

> AP Chemistry · Unit 6: Thermodynamics
> Source: https://www.owlsprep.com/study/ap-chemistry-u6-heat-capacity-and-calorimetry/

This module covers core concepts of heat capacity and calorimetry for AP Chemistry Unit 6, including different forms of heat capacity, constant-pressure and constant-volume calorimetry, and calculation of reaction enthalpy from experimental temperature data.

**Prerequisites:** [First law of thermodynamics](https://www.owlsprep.com/study/ap-chemistry-u6-first-law-of-thermodynamics/); [Enthalpy as a state function](https://www.owlsprep.com/study/ap-chemistry-u6-enthalpy-definition/)

## Learning objectives

- Distinguish between extensive and intensive forms of heat capacity
- Calculate heat transfer and reaction enthalpy from calorimetry experimental data
- Apply correct sign conventions for heat flow in calorimetry problems
- Solve problems for both constant-pressure and constant-volume calorimetry

## Heat Capacity: Extensive and Intensive Forms

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.

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

*Notation:* C

*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: $c = \frac{q}{m\Delta T}$, units $\text{J}/(\text{g} \cdot ^\circ\text{C})$. Core calculation formula: $q = mc\Delta T$.
- **Molar heat capacity ($C_m$)**: Intensive property normalized by moles: $C_m = \frac{q}{n\Delta T}$, units $\text{J}/(\text{mol} \cdot ^\circ\text{C})$.

> **info**
>
> A high specific heat means a substance resists temperature change when heat is added or removed. Liquid water has an unusually high specific heat of $4.184 \text{ J}/(\text{g} \cdot ^\circ\text{C})$, which is why large bodies of water moderate regional climate.

**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 $2.44 \text{ J}/(\text{g} \cdot ^\circ\text{C})$.

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 $\Delta T$:
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 $q = mc\Delta T$:
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.

## Constant-Pressure (Coffee-Cup) Calorimetry

Constant-pressure calorimetry is the most common simple technique for measuring enthalpy change ($\Delta H$) 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: $\Delta H = q_p$.

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 $c = 4.184 \text{ J}/(\text{g} \cdot ^\circ\text{C})$ to calculate heat change.

**Worked example:** A student dissolves 4.00 g of ammonium nitrate ($NH_4NO_3$, 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 $\Delta T$:
4. $$\Delta T = 18.7^\circ\text{C} - 24.1^\circ\text{C} = -5.4^\circ\text{C}$$
5. Calculate $q_{solution}$:
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 $q_{rxn}$ to $q_{solution}$:
8. $$q_{rxn} = -q_{solution} = 2.91 \text{ kJ}$$
9. Calculate moles of ammonium nitrate:
10. $n = 4.00 \text{ g} / 80.04 \text{ g/mol} = 0.04998 \text{ mol}$
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 $\Delta H$ will be positive — this is a quick sanity check for your final sign.

## Constant-Volume (Bomb) Calorimetry

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 ($\Delta V = 0$), pressure-volume work $w = -P\Delta V = 0$, so by the first law of thermodynamics, $\Delta U = q_v$, 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 ($C_{cal}$, units $\text{kJ}/^\circ\text{C}$), 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 $\Delta H \approx q_v$, 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 $C_{cal} = 4.22 \text{ kJ}/^\circ\text{C}$ 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 $q_{rxn}$ to $q_{cal}$:
4. $$q_{rxn} = -q_{cal} = -7.807 \text{ kJ}$$
5. Calculate moles of caffeine:
6. $n = 0.750 \text{ g} / 194.2 \text{ g/mol} = 0.003862 \text{ mol}$
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 $\Delta U$ instead of $\Delta H$, the answer is just the calculated $q_v$, no further adjustment is needed for AP-level problems.

## AP Style Concept Check

**Check your understanding**

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

1. 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 $c = 4.18 \text{ J}/(\text{g} \cdot ^\circ\text{C})$, what is the approximate molar enthalpy of neutralization?

   - +57 kJ/mol
   - -57 kJ/mol
   - +28 kJ/mol
   - -28 kJ/mol

   *Answer:* -57 kJ/mol

   *Why:* Correct! Neutralization is exothermic, so $\Delta H$ is negative. Calculations give ~-57 kJ/mol, matching this option.

## Common pitfalls

- **Wrong:** Only using the mass of water to calculate $q_{solution}$ in coffee-cup calorimetry, ignoring the mass of the dissolved solute.
  - Why it fails: Students memorize the formula as $q = m_{water}c\Delta T$ and forget the solute adds to the total mass of the solution that absorbs or releases heat.
  - Correct: 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:** Taking the absolute value of $\Delta T$ early, leading to the wrong sign for $\Delta H$.
  - Why it fails: Students think 'temperature changed by 8 degrees' so they drop the sign, forgetting the sign encodes direction of heat flow.
  - Correct: Always calculate $\Delta T = T_{final} - T_{initial}$ first, and preserve the sign through all subsequent steps.
- **Wrong:** Leaving $\Delta H$ in J/mol instead of converting to kJ/mol as requested.
  - Why it fails: Specific heat is usually given in J/(g·°C), so $q$ comes out in joules, but AP exam questions almost always request $\Delta H$ in kJ/mol.
  - Correct: Convert $q$ from joules to kilojoules immediately after calculation, before finding molar enthalpy.
- **Wrong:** Multiplying $C_{cal}$ by mass in bomb calorimetry problems, leading to an answer multiple orders of magnitude wrong.
  - Why it fails: Students confuse total heat capacity of the calorimeter with specific heat, which requires a mass term.
  - Correct: Check the units of the given heat capacity: if units are kJ/°C (no mass term), use $q = C_{cal}\Delta T$, no mass needed.
- **Wrong:** Forgetting to invert the sign between $q_{cal}/q_{solution}$ and $q_{rxn}$, leading to positive $\Delta H$ for exothermic reactions.
  - Why it fails: Students mix up which system absorbs vs releases heat: if the reaction releases heat, the calorimeter absorbs it, so $q_{cal}$ is positive, $q_{rxn}$ must be negative.
  - Correct: Write the relationship 'heat lost by system 1 = heat gained by system 2 → $q_1 = -q_2$' and label each system before starting calculations.

## Cheatsheet

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

## 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.

- [Standard Enthalpy of Formation](https://www.owlsprep.com/study/ap-chemistry-u6-enthalpy-of-formation/)
- [Unit 6 Thermodynamics Overview](https://www.owlsprep.com/study/ap-chemistry-u6-overview/)
- [Energy of Phase Changes](https://www.owlsprep.com/study/ap-chemistry-u6-energy-of-phase-changes/)

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