Calorimetry

You’ll be able to

Use q = mcΔT to relate heat, mass, specific heat, and temperature change
Apply conservation of energy between system and surroundings in a calorimeter
Reason about how specific heat controls a substance's temperature response

Measuring heat by watching temperature

We cannot see enthalpy directly, but we can watch a thermometer. Calorimetry measures the heat of a process by trapping it in a known mass of material — usually water — and recording how much the temperature changes. The workhorse is a coffee-cup calorimeter: an insulated cup where a reaction dumps its heat into the surrounding water at constant pressure.

Heat and temperature change

q = m·c·ΔT

q = heat (J), m = mass (g), c = specific heat capacity (J·g⁻¹·°C⁻¹), ΔT = T(final) − T(initial). For water, c = 4.18 J·g⁻¹·°C⁻¹.

Specific heat: how stubborn a substance is

Specific heat capacity (c) is the energy needed to raise 1 gram of a substance by 1 °C. Water's value, 4.18 J·g⁻¹·°C⁻¹, is unusually large — it resists temperature change, which is why oceans moderate climate and why water is the standard calorimeter fluid. A metal with a small c heats and cools quickly for the same amount of heat.

Conservation ties the system to the surroundings

In an isolated calorimeter, energy is conserved: the heat lost by one part is gained by the other, so q(system) = −q(surroundings). If an exothermic reaction (the system) releases heat, that heat is exactly the heat absorbed by the water (the surroundings), which warms up. The minus sign is just the two viewpoints of the same energy.

Worked example

A 50.0 g sample of water is heated from 25.0 °C to 75.0 °C. How much heat does it absorb? (c = 4.18 J·g⁻¹·°C⁻¹)

1.Identify the pieces: m = 50.0 g, c = 4.18 J·g⁻¹·°C⁻¹, ΔT = 75.0 − 25.0 = 50.0 °C.
2.Substitute into q = m·c·ΔT: q = 50.0 × 4.18 × 50.0.
3.Multiply step by step: 50.0 × 4.18 = 209; 209 × 50.0 = 10450 J.
4.Convert to kilojoules: 10450 J ÷ 1000 = 10.45 kJ.

Answer: q = +10450 J ≈ +10.5 kJ (positive because the water absorbs energy)

Watch out

Keep your units consistent. Specific heat is usually J·g⁻¹·°C⁻¹, so mass must be in grams and heat comes out in joules — divide by 1000 only at the end for kJ. And ΔT is final minus initial: a cooling sample has a negative ΔT and therefore a negative q.

Checkpoint

How much heat is required to raise the temperature of 100.0 g of water by 20.0 °C? (c = 4.18 J·g⁻¹·°C⁻¹)

Checkpoint

In a coffee-cup calorimeter, a reaction releases 2090 J of heat into 100.0 g of water. By how much does the water temperature rise? (c = 4.18 J·g⁻¹·°C⁻¹)

Checkpoint

An exothermic reaction runs in a coffee-cup calorimeter and the water warms up. What are the signs of q(system) and q(surroundings)?

On the exam

Free-response calorimetry problems almost always want the heat of the reaction, not just of the water. Compute q(water) with m·c·ΔT, then flip the sign to get q(reaction), and finally divide by moles to report ΔH in kJ·mol⁻¹.

Answer the 3 checkpoints as you read.