Chemistry: Atoms First 2e · Thermochemistry

Calorimetry

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On this page 9 sections

In 30 seconds
Why this matters
The college version
Eli explains
Worked example
Key takeaway
Check yourself
Study tools
Sources & references

In 30 seconds

Heat is invisible — you cannot weigh or count it directly. measures heat indirectly, by watching how it changes the temperature of a known amount of material: if you know how much heat a substance needs to warm up by 1 °C (its heat capacity), and you measure how many degrees it warmed, you can calculate exactly how much heat flowed.

This topic develops the two key equations, q = mcΔT and q = CΔT, introduces the two workhorse instruments — the (constant pressure) and the (constant volume) — and shows how "heat gained = heat lost" bookkeeping finds specific heats and food energy content. Calorimetry is also the experimental bridge to the next topic: at constant pressure, the measured heat is the enthalpy change.

Why this matters

Nutrition labels come from calorimetry. Food Calories are measured with bomb calorimeters — burn the food, measure the heat.
Specific heat is a practical number. It explains why water cools engines efficiently, why coastal climates are milder, and why metal pan handles get hot while wooden ones stay cool.
Identifying unknowns. Measuring an unknown metal's specific heat and comparing it with a table of values is a classic identification method and exam scenario.
Industrial and safety applications. Calorimetry measures the energy released by fuels and explosives — information engineers need for safe reactor design.
It defines the chapter's central measurements. Calorimetry is how chemists measure ΔH (constant pressure) and ΔE (constant volume).

The college version

Core Concepts

Heat capacity, specific heat, and molar heat capacity

Three related quantities describe how a material responds to heat:

Heat capacity (C) is the heat required to raise the temperature of a whole object by 1 °C: units J/^°C. An engine block has a large heat capacity; a paperclip has a tiny one.
Specific heat (c) is the heat required to raise 1 gram of a substance by 1 °C: units J g^-1^°C^-1. It is an intensive property — it describes the material, not the sample size.
Molar heat capacity (C_m) is the heat required to raise 1 mole by 1 °C: units J mol^-1K^-1.

Water's specific heat is the benchmark: c_water = 4.184 J g^-1^°C^-1 — the same number as the calorie (1 cal = 4.184 J). Metals are far lower (e.g., ~0.385 J g⁻¹ °C⁻¹ for copper), which is why a metal spoon heats up far faster than the soup it sits in.

The calorimetry equations

For a substance of mass m, specific heat c, and temperature change ΔT = T_final - T_initial:

$q = mc Δ T$

For a whole object or instrument with heat capacity C:

$q = C Δ T$

The sign follows the first-law convention: q > 0 when the object gains heat, q < 0 when it loses heat.

Coffee-cup calorimetry: constant pressure

The coffee-cup calorimeter is a Styrofoam cup (a good insulator) holding a known mass of solution, with a thermometer and stirrer. A reaction or mixing occurs in the cup; the thermometer records the temperature change. Because the cup is open to the atmosphere, pressure stays constant — so the heat measured equals the enthalpy change:

$q reaction = -q solution = -mc Δ T$

The negative sign implements "heat lost by the reaction = heat gained by the solution" (or vice versa). This is the device behind most general-chemistry labs: dissolving salts, neutralizing acids and bases, and measuring heats of solution.

Bomb calorimetry: constant volume

The bomb calorimeter is a heavy steel vessel (the "bomb") containing a sample and high-pressure oxygen, submerged in a known mass of water inside an insulated jacket. The sample is ignited electrically and burns completely; the heat released warms the water and all calorimeter parts. Instead of tracking the water alone, chemists calibrate the entire assembly — water, bomb, stirrer, thermometer — as one object with a measured heat capacity C_cal:

$q reaction = -q calorimeter = -C cal Δ T$

Because the bomb is sealed, volume — not pressure — is constant. At constant volume, no pressure–volume work is done (w = 0), so by the first law the heat measured equals the change in internal energy:

$q V = Δ E$

Bomb calorimetry is the standard method for measuring the energy content of foods and fuels.

The golden rule: heat gained = heat lost

Every calorimetry problem is a conservation-of-energy statement:

$q hot + q cold = 0 or q gained = -q lost$

The workflow: identify all objects exchanging heat, write q = mcΔT (or q = CΔT) for each, set the sum to zero, and solve. The final temperature is always between the initial temperatures — a built-in sanity check.

Common Confusions

Do Not Confuse	With	Difference
Heat capacity C	Specific heat c	C is for a whole object (J/°C); c is per gram (J g⁻¹ °C⁻¹)
Coffee-cup calorimeter	Bomb calorimeter	Coffee cup = constant pressure, measures ΔH; bomb = constant volume, measures ΔE
ΔT sign	Absolute temperature	ΔT = T_f - T_i; a cooling object has negative ΔT and negative q
Heat gained by water	Heat released by reaction	Equal in magnitude, opposite in sign: q_rxn = -q_water
Calorimeter heat capacity	Water's specific heat	C_cal includes water + bomb + stirrer as one object; don't use c_water alone for a bomb
°C and K in ΔT	°C and K as absolute values	A temperature difference is numerically identical in °C and K; absolute values are not
Temperature change	Heat	Big ΔT doesn't mean big heat — mass and specific heat matter too

Eli explains

The same idea, in plain words

Explain it like I’m 10

Imagine measuring how much money a fire gives off by watching how much a piggy bank warms up. First you learn the piggy bank's "warm-up rule" — how much heat it takes to get 1° warmer. Then you watch a fire and see the bank warm up 5°, and you can figure out exactly how much heat the fire gave. A calorimeter is that piggy bank for chemistry: when a reaction warms the water inside, you can count the reaction's heat. A bomb calorimeter is the same idea inside a strong steel box, used to burn food and fuel completely.

Worked example

Example 1: Finding the specific heat of an unknown metal

A 30.0 g sample of an unknown metal is heated to 100.0 °C and dropped into a coffee-cup calorimeter containing 100.0 g of water at 20.0 °C. The final temperature reaches 23.5 °C. Find the metal's specific heat (c_water = 4.184 J g^-1^°C^-1).

Step 1 — Write the conservation statement:

$q metal + q water = 0$

Step 2 — Write the heat equation for each object:

$mc Δ T metal + mc Δ T water = 0$

Step 3 — Compute the temperature changes:

$Δ T water = 23.5 - 20.0 = +3.5 ° C$

$Δ T metal = 23.5 - 100.0 = -76.5 ° C$

Step 4 — Substitute and solve:

$(30.0 g)(c metal)(-76.5 ° C) + (100.0 g)(4.184 J g -1 ° C -1)(+3.5 ° C) = 0$

$-2295 c metal + 1464 J = 0 \Rightarrow c metal = 14642295 = 0.638 J g -1 ° C -1$

Sanity check: 0.638 J g⁻¹ °C⁻¹ lies in the metallic range (typical metals: 0.2–0.9), and the final temperature 23.5 °C lies between 20.0 and 100.0 °C as required.

Example 2: Bomb calorimetry of a food sample

A 0.500 g sample of a snack food is burned in a bomb calorimeter with calibrated heat capacity C_cal = 2.50 kJ/^°C. The water temperature rises from 21.00 °C to 24.35 °C. Calculate the heat released per gram of food.

Step 1 — Write the bomb-calorimeter equation:

$q sample = -C cal Δ T$

Step 2 — Compute ΔT:

$Δ T = 24.35 - 21.00 = 3.35 ° C$

Step 3 — Substitute:

$q sample = -(2.50 kJ / ° C)(3.35 ° C) = -8.38 kJ$

The negative sign means the sample released 8.38 kJ.

Step 4 — Divide by sample mass:

$-8.38 kJ 0.500 g = -16.8 kJ g -1$

This food releases 16.8 kJ per gram when completely oxidized — about 4.0 nutritional Calories per gram, in the range expected for a food of mostly carbohydrates and proteins. (Dimensional check: kJ/°C × °C = kJ; kJ ÷ g = kJ/g.)

Example 3: How much heat does water absorb?

How much heat is required to warm 250.0 g of water from 22.0 °C to 40.0 °C?

Step 1 — Write the equation:

$q = mc Δ T$

Step 2 — Compute ΔT and substitute:

$Δ T = 40.0 - 22.0 = 18.0 ° C$

$q = (250.0 g)(4.184 J g -1 ° C -1)(18.0 ° C) = 1.88 \times 104 J = 18.8 kJ$

Note how units cancel: g × J g^-1^°C^-1 × ^°C = J. Water's high specific heat is why 18.8 kJ — the energy in ~4.5 food Calories — only warms a cup of water by 18 °C.

Key takeaways

Core equation: q = mcΔT for a substance; q = CΔT for an object/instrument.
Specific heat of water: 4.184 J g^-1^°C^-1 — the benchmark; metals are far lower.
Coffee-cup (constant pressure): q_rxn = -mcΔT; the heat measured equals ΔH.
Bomb (constant volume): q_rxn = -C_calΔT; the heat measured equals ΔE.
Conservation rule: q_gained + q_lost = 0; final temperature lies between initial temperatures.
ΔT = T_final - T_initial, in °C or K (differences are identical in both scales).
Write the formula before substituting; check units: mass in g, c in J g⁻¹ °C⁻¹, ΔT in °C.

Check yourself

6 review questions from the chapter. Try each one, then open the answer.

Write q = mcΔT, defining every symbol and its units.
Show answer
q = heat (J), m = mass (g), c = specific heat (J g⁻¹ °C⁻¹), ΔT = T_f - T_i (°C).
Why does a coffee-cup calorimeter measure ΔH, while a bomb calorimeter measures ΔE?
Show answer
The coffee-cup calorimeter is open to the atmosphere, so pressure is constant — at constant pressure, q_p = ΔH. The sealed bomb keeps volume constant, so no PΔV work is done and q_V = ΔE.
A 50.0 g piece of aluminum (c = 0.900 J g^-1^°C^-1) cools from 80.0 °C to 25.0 °C. How much heat does it release?
Show answer
ΔT = 25.0 - 80.0 = -55.0 ^°C; q = (50.0)(0.900)(-55.0) = -2.48 × 10³ J; the metal releases ~2.5 kJ.
In Example 2, if the same 0.500 g food were burned in a calorimeter with C_cal = 5.00 kJ/^°C, would ΔT be larger or smaller, and why?
Show answer
Smaller. Doubling C_cal means the same heat produces half the temperature rise (ΔT = q/C); the heat measured is the same, but the thermometer readout changes less.
Which requires more heat: warming 10 g of water or 10 g of copper by the same amount? Explain using specific heats.
Show answer
Water — its specific heat (4.184 J g⁻¹ °C⁻¹) is about 11× copper's (0.385 J g⁻¹ °C⁻¹), so warming 10 g of water by the same amount needs ~11× more heat.
A student forgets the stirrer and thermometer in the bomb calorimeter's heat capacity. Will the measured food energy be too high or too low? Why?
Show answer
Too low. The omitted parts absorb heat that is not counted, so the calorimeter's true heat capacity is larger than the value used, and the computed heat (-C_calΔT) underestimates the actual energy released.

Keep learning

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Study tools & related lessonsKey vocabulary · Related

Key vocabulary

Calorimetry: Measuring heat by observing temperature changes
Heat capacity (C): Heat to raise a whole object by 1 °C (J/^°C)
Specific heat (c): Heat to raise 1 g of a substance by 1 °C (J g^-1^°C^-1)
Molar heat capacity (Cm): Heat to raise 1 mol by 1 °C (J mol^-1K^-1)
Coffee-cup calorimeter: Insulated cup measuring heat at constant pressure
Bomb calorimeter: Sealed steel vessel measuring heat at constant volume
Calorimeter heat capacity (Ccal): Heat capacity of the entire bomb-calorimeter assembly
Thermal equilibrium: State where all objects share the same temperature

Sources & references

openstax.org — Chemistry Atoms First 2e

This lesson was adapted from the open educational references above; their licenses and attributions are preserved. See Copyright & Licensing.

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