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Chemistry

How To Calculate Specific Heat

Specific heat is the price of a temperature rise. Every material charges a different number of joules per kilogram per degree, and that number explains why a beach cools faster than the sea beside it.

Quick Answer

Q = mass x specific heat x temperature change

Q
Heat energy added or removed, in joules
m
Mass being heated, in kilograms
c
Specific heat capacity of the material, in joules per kilogram per degree
dT
Temperature change, end temperature minus start temperature

Multiply the mass in kilograms by the specific heat of the material and by the number of degrees it warms up. Water needs 4186 joules per kilogram per degree, so heating two kilograms of it through sixty degrees takes about five hundred and two thousand joules, or a little under half a kilowatt hour.

What Is Specific Heat?

Specific heat capacity is the amount of energy required to raise the temperature of one kilogram of a substance by one degree Celsius. It is a property of the material rather than of the object, so a thimble of water and a bathtub of water have exactly the same specific heat, even though heating them takes wildly different amounts of energy.

Water is the reference point that surprises people. Its specific heat is about 4186 joules per kilogram per degree, which is unusually high for a common substance. Metals sit far lower: aluminium is around 900, copper about 385 and lead only 128. That is why a metal spoon heats up instantly in soup while the soup itself takes minutes.

The formula is a chain of three multiplications. Mass tells you how much material there is, specific heat tells you what the material charges per kilogram per degree, and the temperature change tells you how many degrees you are asking for. Multiply all three and you have the energy in joules.

Temperature change is always end temperature minus start temperature. It does not matter whether the material is heating up or cooling down in terms of the size of the number, only in its sign. A substance cooling from eighty to twenty degrees releases exactly the same energy that heating it from twenty to eighty absorbed.

Because the formula uses a difference, the Celsius and Kelvin scales are interchangeable. A change of one degree Celsius is a change of one kelvin, so there is no offset to worry about. You would only need to be careful if the formula used an absolute temperature, which this one does not.

The energies involved are often larger than people expect. Two kilograms of water warmed through sixty degrees takes just over five hundred thousand joules, which is about a tenth of a kilowatt hour. That is why heating water for a bath is one of the largest single uses of energy in a typical home.

The same calculation runs the other way, which is how calorimetry works. If you drop a hot metal block into cold water and measure how much the water warms, you can work out the specific heat of the metal. This is exactly how the values in tables were measured, and it is a standard school experiment for good reason.

Water's high specific heat has consequences well beyond the laboratory. Oceans absorb enormous quantities of solar energy with only a small temperature rise, which moderates coastal climates and makes sea breezes blow. It is also why sweating works so well: evaporating water carries away a large amount of heat for a modest amount of liquid.

Specific heat changes slightly with temperature and pressure, so the values in tables are quoted at a stated condition, usually twenty five degrees Celsius and standard pressure. For everyday calculations the variation is small enough to ignore, but precise engineering work uses tabulated values at the actual operating temperature.

There is an important distinction between specific heat capacity and heat capacity. Specific heat is per kilogram, so it is a material property. Heat capacity is per object and is simply the specific heat multiplied by the mass. A ten kilogram block of aluminium has ten times the heat capacity of a one kilogram block, but the same specific heat.

Finally, be aware that a phase change breaks the formula. Melting ice or boiling water absorbs a large amount of energy with no temperature change at all, governed by the latent heat of fusion or vaporisation rather than by specific heat. A calculation that crosses a phase boundary needs both terms, and using only Q equals mc delta T will badly understate the energy.

Formula

Q = m x c x dT

Mass in kilograms, specific heat in joules per kilogram per degree, and the temperature change in degrees.

SymbolMeaning
QHeat energy
mMass
cSpecific heat
dTTemperature change

dT = end temperature - start temperature

A difference, so Celsius and kelvin give the same answer with no offset correction.

SymbolMeaning
dTTemperature change
T2End temperature
T1Start temperature

c = Q / (m x dT)

The calorimetry form: measure the energy, the mass and the temperature rise, and solve for the material property.

SymbolMeaning
cSpecific heat
QHeat energy
mMass
dTTemperature change

How To Calculate Specific Heat

  1. 1

    Find the mass in kilograms

    Convert grams to kilograms first, because specific heat is defined per kilogram. Two kilograms is the default, and using grams here makes the answer a thousand times too large.

  2. 2

    Look up the specific heat

    Water is about 4186 joules per kilogram per degree, aluminium about 900 and copper about 385. The value is a property of the material, not of the size of the object.

  3. 3

    Work out the temperature change

    Subtract the start temperature from the end temperature. Twenty to eighty degrees is a change of sixty, and the sign only matters if you care whether heat was absorbed or released.

  4. 4

    Multiply the three together

    Two kilograms times 4186 times sixty is five hundred and two thousand three hundred and twenty joules.

  5. 5

    Convert to a familiar unit if useful

    Divide joules by a thousand for kilojoules, or by three million six hundred thousand for kilowatt hours. The default example is 502.32 kilojoules, or about 0.1395 kilowatt hours.

Examples

Example 1: Heating two litres of water

Mass
2 kg
Specific heat
4186 J/kg/degC
Start temperature
20 degC
End temperature
80 degC
StepCalculationResult
Temperature change80 - 2060
Heat energy in joules2 x 4186 x 60502320
Energy in kilojoules502320 / 1000502.32
Energy in kilowatt hours502320 / 36000000.1395

Result: Two kilograms of water rising 60 degrees needs 502320 joules, which is 502.32 kilojoules, or about 0.1395 kilowatt hours of energy.

Example 2: A small aluminium block

Mass
0.5 kg
Specific heat
900 J/kg/degC
Start temperature
25 degC
End temperature
150 degC
StepCalculationResult
Temperature change150 - 25125
Heat energy in joules0.5 x 900 x 12556250
Energy in kilojoules56250 / 100056.25
Energy in kilowatt hours56250 / 36000000.015625

Result: Half a kilogram of aluminium warmed through 125 degrees absorbs 56250 joules, which is 56.25 kilojoules and only 0.015625 kilowatt hours, because aluminium stores far less heat per kilogram than water does.

Example 3: Bringing five litres of water to the boil

Mass
5 kg
Specific heat
4186 J/kg/degC
Start temperature
15 degC
End temperature
100 degC
StepCalculationResult
Temperature change100 - 1585
Heat energy in joules5 x 4186 x 851779050
Energy in kilojoules1779050 / 10001779.05
Energy in kilowatt hours1779050 / 36000000.4942

Result: Five kilograms of water heated through 85 degrees needs 1779050 joules, which is 1779.05 kilojoules and about 0.4942 kilowatt hours, close to half a unit of electricity.

Calculator

Heat energy

502,320

Energy in kilojoules
502.32
Energy in kilowatt hours
0.1395
Temperature change
60

Values update as you type. This calculator covers the single scenario its formula assumes — see Common Mistakes for what it leaves out.

Prefer a full-width tool? Open the Specific Heat calculator page.

Common Mistakes

  • Using grams instead of kilograms

    Specific heat is defined per kilogram, so a mass entered in grams makes the answer a thousand times too large. Converting to kilograms before multiplying is the single most common error in this calculation.

  • Adding the temperatures instead of subtracting

    The formula needs the change, not the sum or the final value. Twenty to eighty degrees is a change of sixty, not a hundred, and using the wrong quantity throws the energy out by a wide margin.

  • Using the specific heat of the container

    If you heat water in a pan, the pan also warms up and absorbs energy. Ignoring it understates the total, which is why precise calorimetry uses insulated vessels with a known and small heat capacity.

  • Forgetting about latent heat at a phase change

    Melting or boiling absorbs a large amount of energy with no temperature change. A calculation that crosses zero or a hundred degrees needs the latent heat term as well, or it will badly underestimate the energy required.

  • Treating specific heat as constant at every temperature

    Tabulated values are quoted at a reference temperature and drift slightly as the substance warms. For everyday work the difference is negligible, but for engineering calculations across a wide range it matters.

  • Confusing specific heat with heat capacity

    Specific heat is per kilogram and is a material property. Heat capacity is per object and scales with mass. A large block and a small block of the same metal have identical specific heats but very different heat capacities.

FAQ

What is the formula for specific heat?

Q equals m times c times delta T, where Q is energy in joules, m is mass in kilograms, c is the specific heat in joules per kilogram per degree and delta T is the temperature change. Rearranged, c equals Q divided by m times delta T.

What is the specific heat of water?

About 4186 joules per kilogram per degree Celsius. It is unusually high for a common material, which is why water is so effective at storing heat and why oceans moderate the climate of coastal regions.

Why does metal heat up faster than water?

Metals have much lower specific heat capacities, so the same energy produces a much larger temperature rise. Aluminium needs about 900 joules per kilogram per degree against 4186 for water, roughly a fifth as much.

Does the temperature scale matter?

Not for this formula, because it uses a temperature difference. One degree Celsius equals one kelvin, so a change of sixty degrees is a change of sixty kelvin. Only formulas using absolute temperature need the conversion.

Can specific heat be negative?

No. Specific heat is the energy needed to raise a unit mass by one degree, and adding energy always raises the temperature or changes the phase. A negative value would mean a material that cools as you heat it.

How do you measure specific heat experimentally?

By calorimetry. Heat a sample to a known temperature, drop it into a measured mass of water, and record how much the water warms. Then rearrange Q equals mc delta T to solve for the specific heat of the sample.

References

  1. [1]Wikipedia, Specific heat capacity — https://en.wikipedia.org/wiki/Specific_heat_capacity
  2. [2]Wikipedia, Heat capacity — https://en.wikipedia.org/wiki/Heat_capacity
  3. [3]Wikipedia, Calorimetry — https://en.wikipedia.org/wiki/Calorimetry
  4. [4]Wikipedia, Latent heat — https://en.wikipedia.org/wiki/Latent_heat