5.2 Calorimetry

Calorimetry Crash Course
Calorimeters, q = mCΔT, and Heating Curves
General Chemistry 5.2
Lesson Overview
Calorimetry is the measurement of heat produced or consumed by chemical reactions, carried out using insulated devices called calorimeters and governed by the equation q = CΔT. The specific heat equation q = mCΔT extends this concept to any substance of known mass, relating heat transfer to mass, specific heat capacity, and temperature change. Heating curves illustrate how temperature changes with continuous heat addition, including the constant-temperature plateaus that occur during phase changes such as melting and boiling.
What Is Calorimetry?
Calorimetry is the study of how much heat is produced or consumed by chemical reactions. Notice the word's resemblance to "calorie" — both relate to measuring heat energy. In this general chemistry lesson, we explore 2 primary approaches to measuring heat: using a calorimeter (a specialized insulated device) and applying the specific heat capacity equation, q = mCΔT.
At its core, calorimetry answers one question: how much heat does a given reaction release or absorb? This information is essential for everything from calculating nutritional calories in food to understanding industrial chemical processes. Mastering calorimetry calculations is a key skill in any general chemistry course.
How Calorimeters Work
A calorimeter is a heavily insulated container — often with water in the walls — designed to minimize heat exchange with the surroundings. The principle is straightforward: place a sample inside, trigger a chemical reaction, and measure how the calorimeter's temperature changes. From that temperature change, you can calculate exactly how much heat the reaction produced or consumed using the heat capacity equation.
How Scientists Measure Calories in Food
This is precisely how the caloric value printed on nutritional labels is determined. To measure the calories in a Snickers bar, scientists place the bar inside a calorimeter, ignite it, let it burn completely, and then record how much the calorimeter's temperature increased. Using the calorimeter's known heat capacity, they calculate the total heat released by the combustion reaction.
Why Burning Equals Eating (Thermodynamically)
Enthalpy change (ΔH) is a state function — it depends only on the initial and final states, not the path taken. When you eat a Snickers bar, your body combines its hydrocarbons with oxygen (O2) and ultimately produces CO2 and H2O through many enzymatically catalyzed steps. Combustion in a calorimeter reaches the same products in a single step. Because the initial state and final state are identical, the ΔH is the same regardless of the path.
The Heat Capacity Equation for Calorimeters
For a calorimeter, the governing equation is:
where q is the heat absorbed or released, C is the calorimeter's heat capacity (how much heat it takes to change the calorimeter's temperature by one degree), and ΔT is the temperature change. This equation has 3 variables — you are given 2 and solve for the third.
How Heat Capacity Is Determined
The heat capacity of a calorimeter is established through a controlled calibration experiment. Once calibrated, that heat capacity value can be used for many subsequent experiments. Periodic recalibration ensures the value remains accurate over time.
Calorimeter Calculations
Example: Solving for Heat Capacity
Problem
If 3,000 joules of heat causes a temperature increase of 3.0°C for a calorimeter, what is its heat capacity?
Intuitively, if 3,000 J raises the temperature by 3 degrees, it takes 1,000 J to raise it by one degree. Let's confirm with the heat capacity equation:
C = q ÷ ΔT = 3,000 J ÷ 3.0°C = 1,000 J/°C
With this heat capacity established, any future reaction in this calorimeter becomes easy to analyze. If a combustion reaction later raises the temperature by 5°C, the heat released is simply 5,000 J.
Example: Enthalpy of Combustion of Methane
Problem
If 8.0 g of methane (CH4) is combusted in a calorimeter with a heat capacity of 29.7 kJ/°C, and its temperature rises from 25°C to 40°C, what is the enthalpy of combustion of methane in kJ/mol?
Step 1 — Calculate q from the calorimeter data:
ΔT = 40°C − 25°C = 15°C
q = C × ΔT = 29.7 kJ/°C × 15°C = 445.5 kJ
Step 2 — Convert to kJ/mol:
The molar mass of CH4 is 16 g/mol (carbon = 12, four hydrogens = 4). Converting grams to moles:
8.0 g ÷ 16 g/mol = 0.50 mol CH4
445.5 kJ ÷ 0.50 mol = 891 kJ/mol
Sign Convention: qreaction vs. qcalorimeter
There is one crucial detail remaining. The 445.5 kJ we calculated is the heat absorbed by the calorimeter, but where did the calorimeter get that heat? The reaction supplied it. Because the system (reaction) and surroundings (calorimeter) must sum to zero in any calorimetry calculation:
Don't Forget the Sign
The calorimeter gained +445.5 kJ, so the reaction released −445.5 kJ. This gives ΔH = −891 kJ/mol for methane combustion — an exothermic reaction, as the negative sign confirms.
Quick Sign Check for Calorimetry Problems
Calorimeter temperature goes up? The reaction was exothermic (ΔH is negative).
Calorimeter temperature goes down? The reaction was endothermic (ΔH is positive).
Specific Heat Capacity and q = mCΔT
What Is Specific Heat Capacity?
Specific heat capacity is the amount of heat required to raise the temperature of one gram of a substance by one degree Celsius. In fact, the calorie was originally defined this way: one calorie is the heat needed to raise one gram of liquid water by one degree Celsius.
This gives us the specific heat equation:
where q is heat, m is mass, c is the specific heat capacity of the substance, and ΔT is the temperature change. Think of specific heat capacity as heat capacity per gram — you then multiply by mass to get the total heat capacity for your particular sample.
2 Approaches to Measuring Heat Transfer: Calorimeter vs. Specific Heat
| Feature | Calorimeter (q = CΔT) | Specific Heat Capacity (q = mcΔT) |
|---|---|---|
| What you know | Overall heat capacity of the device | Identity and mass of a specific substance |
| Variables | 3 (q, C, ΔT) | 4 (q, m, c, ΔT) |
| When to use | You have a calorimeter with a known heat capacity | You have a known substance with a known mass |
Memory Trick for q = mcΔT
Many students notice that "mcΔT" looks like "MCAT." If you are pre-health, you now have one more reason to remember this specific heat capacity equation.
What Specific Heat Capacity Values Tell You
Specific heat capacity values differ by substance and by phase. Liquid water has a specific heat capacity of 4.18 J/(g·°C), meaning one gram of water requires 4.18 J to increase its temperature by one degree. Want to raise two grams by one degree? That takes twice as much heat — hence the multiplication by mass in q = mcΔT. Want to raise one gram by five degrees? That takes five times as much heat — hence the multiplication by ΔT.
Conceptual Example: Water vs. Aluminum Specific Heat Capacity
Which Gets Hotter?
Imagine a kiddie pool of water and an aluminum park bench sitting side by side in the Arizona sun, both receiving the same amount of heat. Liquid water has a specific heat capacity of 4.18 J/(g·°C), while solid aluminum has a specific heat capacity of only 0.90 J/(g·°C).
Because aluminum requires far less heat per gram per degree, the same amount of solar energy raises its temperature much more dramatically. That is why the park bench will burn you while the pool water feels pleasant.
Interpreting Specific Heat Capacity
A lower specific heat capacity means the substance heats up (and cools down) more quickly. A higher specific heat capacity means the substance resists temperature change. This is why water is so effective as a coolant and why coastal climates are more temperate than inland ones.
Heating Curves and Phase Changes

| THERMODYNAMIC QUANTITIES FOR WATER | |
|---|---|
| Cs,ice | 2.09 J/g•°C |
| Cs,liquid water | 4.18 J/g•°C |
| Cs,steam | 2.03 J/g•°C |
| ΔH°fus | 6.01 kJ/mol |
| ΔH°vap | 40.65 kJ/mol |
What Is a Heating Curve?
A heating curve is a graph showing how a substance's temperature changes as heat is continuously added. In general, adding heat raises the temperature — but there is an important exception: during a phase change, temperature remains constant even as heat continues to flow in.
Consider the heating curve for water, starting from ice below 0°C. There are 5 distinct segments to understand:
- Solid phase: Temperature rises as heat is added. Apply q = mcΔT using the specific heat capacity of ice.
- Melting (0°C): Temperature holds steady while ice converts to liquid water. Extra heat is required just to break intermolecular forces — this is the ΔH of fusion.
- Liquid phase: Temperature rises again. Apply q = mcΔT with the specific heat capacity of liquid water.
- Boiling (100°C): Temperature holds steady while liquid converts to gas. This requires the ΔH of vaporization.
- Gas phase: Temperature rises once more. Apply q = mcΔT with the specific heat capacity of steam.
When q = mcΔT Is Not Enough for Phase Changes
The specific heat capacity equation handles temperature changes within a single phase, but it cannot handle phase changes because ΔT = 0 during those transitions. For phase changes, use:
where n is the number of moles and ΔH is the enthalpy of the phase change (fusion or vaporization) in kJ/mol. If the ΔH value is given in kJ/g instead, multiply by mass rather than moles.
Check Your Units
The most common form for ΔH of fusion and ΔH of vaporization is kJ/mol, but some problems give kJ/g. Always check the units before setting up your calorimetry calculation — use moles with kJ/mol and grams with kJ/g.
Counting the Steps on a Heating Curve
Every corner on a heating curve represents a new step in your calculation. Moving from solid to gas passes through 2 phase changes, meaning the total heat is the sum of up to 5 separate calculations.
3-Step Heating Curve Problem: Ice to Liquid Water
Problem
How much heat is required to raise 90 g of ice at −20°C to liquid water at 50°C?
Given: csolid = 2.09 J/(g·°C), cliquid = 4.18 J/(g·°C), ΔHfusion = 6.01 kJ/mol
On the heating curve, the path from −20°C (solid) to 50°C (liquid) crosses the melting point at 0°C. This creates 3 distinct steps that each require a separate calorimetry calculation:
Step 1 — Heat the Ice from −20°C to 0°C
q1 = mcΔT = 90 g × 2.09 J/(g·°C) × 20°C = 3,762 J
The ΔT of 20°C is positive (0°C minus −20°C), confirming the temperature is increasing.
Step 2 — Melt the Ice at 0°C (Enthalpy of Fusion)
First, convert grams to moles. The molar mass of water (H2O) is 18 g/mol:
n = 90 g ÷ 18 g/mol = 5.0 mol
q2 = n × ΔHfusion = 5.0 mol × 6.01 kJ/mol = 30.05 kJ
Step 3 — Heat the Liquid from 0°C to 50°C
q3 = mcΔT = 90 g × 4.18 J/(g·°C) × 50°C = 18,810 J
Combining the Steps
Steps 1 and 3 yield joules; step 2 yields kilojoules. Convert everything to kilojoules before adding:
q1 = 3,762 J = 3.762 kJ
q2 = 30.05 kJ
q3 = 18,810 J = 18.81 kJ
qtotal = 3.762 + 30.05 + 18.81 = 52.62 kJ
Exam Tip: Watch the Units in Multi-Step Calorimetry
The mcΔT equation typically produces joules, while ΔH values are usually in kilojoules. Before adding heating curve steps together, convert everything to the same unit. On multiple-choice exams, check whether the answer choices are in joules or kilojoules and convert accordingly.
Don't Be Fooled by Extra Data
Professors often provide more thermodynamic data than you need. In this problem, the ΔH of vaporization and the specific heat capacity of steam might also be given — but since we never reach the boiling point, those values are irrelevant. Identify which steps your path actually crosses on the heating curve and use only the data required for those steps.
Reading a Heating Curve: Key Observations
ΔH of Fusion vs. ΔH of Vaporization
On a heating curve, the plateau for boiling is noticeably longer than the plateau for melting. This reflects the fact that vaporization requires significantly more energy than fusion. For water, the 2 phase change enthalpies are:
| Phase Change | ΔH Value | Plateau Length |
|---|---|---|
| Melting (fusion) | 6.01 kJ/mol | Shorter |
| Boiling (vaporization) | 40.65 kJ/mol | Much longer |
It takes nearly seven times more energy to boil water than to melt ice, which is why the vaporization plateau dominates the heating curve.
Slope and Specific Heat Capacity on a Heating Curve
The slope of the temperature-rising segments reveals each phase's specific heat capacity. A shallower slope indicates a higher specific heat capacity — meaning it takes more heat to raise the temperature by each degree.
Water's Heating Curve: 3 Slopes Compared
For water, the liquid phase has the shallowest slope because liquid water's specific heat capacity (~4.18 J/(g·°C)) is roughly double that of ice or steam (~2 J/(g·°C)). The steeper slopes for the solid and gas phases reflect their lower specific heat capacities — less heat is needed per degree of temperature increase.
8 Key Takeaways: Calorimetry and q = mcΔT
- Calorimetry is the measurement of heat produced or consumed by chemical reactions, using the equation q = CΔT for calorimeters.
- Enthalpy change is a state function — it depends only on initial and final states, which is why burning food in a calorimeter gives the same caloric value as metabolizing it.
- The sign convention is critical: qreaction = −qcalorimeter. A temperature increase in the calorimeter means the reaction is exothermic (ΔH < 0).
- Specific heat capacity (c) is the heat required to raise one gram of a substance by one degree Celsius, and the equation q = mcΔT relates heat to mass, specific heat capacity, and temperature change.
- A substance with a lower specific heat capacity heats up faster — that is why an aluminum park bench in the sun burns you, but the pool water stays pleasant.
- Heating curves show that temperature rises within each phase (use q = mcΔT) but remains constant during phase changes (use q = n × ΔH).
- Multi-step heating curve problems require separate calculations for each of the 5 segments, with careful attention to unit conversion (joules vs. kilojoules) before summing.
- On a heating curve, a shallower slope indicates a higher specific heat capacity, and a longer plateau indicates a larger enthalpy of phase change.

