Calorimetry Calculator

Find the final equilibrium temperature when two substances mix, or the specific heat of a sample from a calorimeter experiment, with heat flow.

Find Final temperature Specific heat of object 1

Object 1 (e.g. hot sample)Ignored when finding the specific heat of object 1

Object 1 specific heat (custom) (J/(g·°C))

Object 1 mass (g)

Object 1 starting temperature (°C)

Object 2 (e.g. water in the calorimeter)

Object 2 specific heat (custom) (J/(g·°C))

Object 2 mass (g)

Object 2 starting temperature (°C)

Calorimeter heat capacity (optional) (J/°C)The cup itself; it starts at object 2's temperature. Leave 0 for an ideal insulated cup.

Measured final temperature (°C)Only used when finding the specific heat

When a hot object meets a cooler one in an insulated container, heat flows until both reach the same temperature, and the heat lost by one equals the heat gained by the other. This calorimetry calculator uses that energy balance two ways: to predict the final temperature of a mixture, or to work out the specific heat of an unknown sample from a coffee-cup calorimeter experiment.

How to use this tool

Choose whether you want the final temperature or the specific heat of object 1.
Pick each material from the list, or choose Custom and type its specific heat in J/(g·°C).
Enter each mass in grams and each starting temperature in °C.
Add the calorimeter's own heat capacity if your lab gives one; otherwise leave 0.
To find a specific heat, enter the final temperature you measured.

Formula

m₁c₁(T₁ − T_f) = (m₂c₂ + C_cal)(T_f − T₂) T_f = (m₁c₁T₁ + (m₂c₂ + C_cal)T₂) / (m₁c₁ + m₂c₂ + C_cal)

m is mass in grams, c specific heat in J/(g·°C), T the starting temperatures and T_f the shared final temperature. C_cal is the heat capacity of the calorimeter cup, which starts at the water's temperature. Rearranging the first equation for c₁ gives the specific heat of a sample.

Example

Drop 200 g of aluminium at 100 °C into 250 g of water at 20 °C. The aluminium holds 200 × 0.897 = 179.4 J/°C and the water 250 × 4.184 = 1,046 J/°C, so T_f = (179.4 × 100 + 1,046 × 20) ÷ 1,225.4 ≈ 31.71 °C. About 12,251 J moves from the metal to the water: the aluminium cools by 68.29 °C while the water warms by only 11.71 °C.

Frequently asked questions

Why does the water temperature change so little?

Water stores far more heat per gram per degree than metals. In the example the water can absorb almost six times as much heat per degree as the aluminium, so it moves much less.

What if ice melts or water boils?

This calculator only covers temperature changes. If the mixture crosses 0 °C or 100 °C, part of the heat goes into melting or boiling instead; work that part out with the latent heat calculator.

Why is my lab result different from the textbook value?

Heat escapes to the air and the cup, the sample cools while being transferred, and thermometers lag. Including the calorimeter heat capacity and moving the sample quickly both bring the result closer.

Sources

Last updated: 2026-10-30