Specific heat capacity is the heat energy transferred per unit mass for a stated temperature change. For a material that does not change phase and whose specific heat is approximately constant over the interval, use Q = mcΔT. Rearranging gives c = Q/(mΔT). Here Q is heat transferred in joules, m is mass in kilograms, c is specific heat in J/(kg·K), and ΔT = Tfinal − Tinitial. A temperature difference has the same numeric value in kelvins and degrees Celsius.
Specific heat capacity formula and units
Specific heat capacity, usually written c, is an intensive property: doubling the amount of the same material doubles its heat capacity, but does not double its specific heat. The SI unit is joules per kilogram-kelvin, J/(kg·K), which is numerically equivalent to J/(kg·°C) for a temperature difference. If you use grams, report c in J/(g·°C). Convert consistently: 1 J/(g·°C) = 1,000 J/(kg·K).
The equation is an engineering approximation over a defined temperature range. More precisely, specific heat can vary with temperature and depends on conditions such as phase; for gases, constant-volume and constant-pressure specific heats differ. For high accuracy across a wide range, use material data for the relevant temperature and integrate c(T) rather than assuming one constant value.
How to calculate specific heat
- Record the sample mass and convert it to the units used for c.
- Calculate ΔT as final temperature minus initial temperature.
- Measure or determine the heat transferred to the sample, accounting for the apparatus when needed.
- Substitute into c = Q/(mΔT), then check that the units reduce to energy per mass per temperature.
Worked example: heat water from 20°C to 30°C
Suppose 100 g of water receives 4,180 J of heat and warms by 10°C. Using c = Q/(mΔT), c = 4,180 J ÷ (100 g × 10°C) = 4.18 J/(g·°C), or 4,180 J/(kg·K). The example uses the approximate room-temperature value for liquid water; the exact value varies with temperature and conditions.
Worked example: energy to heat aluminum
For a first-pass estimate, heat 0.50 kg of aluminum from 25°C to 100°C. Using c ≈ 900 J/(kg·K), ΔT = 75 K, so Q = mcΔT = 0.50 × 900 × 75 = 33,750 J, or about 33.8 kJ. This estimate assumes no heat is lost to the surroundings or container and uses one approximate specific-heat value for the interval.
What the equation does not include
Q = mcΔT estimates sensible heat: energy associated with a temperature change within one phase. During melting, boiling, or another phase transition, temperature may remain nearly constant while energy changes the phase; use latent heat, Q = mL, for that portion. In a real calorimeter, energy can also warm the container or escape to the environment. Include those terms in an energy balance when they are material to the result.
Specific heat is also different from thermal conductivity. Specific heat describes energy stored per unit mass per degree; conductivity describes how readily heat flows through a material. A material can store substantial heat without conducting it quickly. See our guide to materials used for heat dissipation for the separate conductivity question.
Common mistakes
- Mixing units: pair grams with J/(g·°C), or kilograms with J/(kg·K).
- Using the wrong temperature difference: subtract initial from final; for a cooling sample, ΔT and Q for the sample are negative under this sign convention.
- Confusing heat capacity and specific heat: heat capacity C is for the whole object and equals mc when c is constant; specific heat c is per unit mass.
- Applying one value outside its range: values depend on material, phase, temperature and, for gases, the process condition.
- Ignoring the apparatus: experimental calorimetry may require the container’s heat capacity and a heat-loss correction.
Frequently asked questions
What is the formula for specific heat capacity?
For approximately constant c and no phase change, Q = mcΔT. Solve for specific heat with c = Q/(mΔT).
Is ΔT in Celsius or kelvins?
Either is valid for a temperature difference: a change of 10°C equals a change of 10 K. Do not use an absolute Celsius temperature where the formula requires a difference.
Why did my calculated value differ from a reference table?
Check mass and energy units, the sign and size of ΔT, heat absorbed by the container, heat lost to the surroundings, sample purity, phase, and the temperature range represented by the table value.
For approximate specific-heat reference values for selected metals and nonmetals, see our comparison guide.
Sources and further reading
- OpenStax, University Physics Volume 2: heat transfer, specific heat, and calorimetry (equation, assumptions, and tabulated reference values).
- OpenStax, College Physics 2e: temperature change and heat capacity (units, material and phase dependence).
