Specific Heat Calculator
Calculate the heat energy needed to change a substance's temperature from Q = mcΔT, solving for any variable.
How to use this calculator
- 1Choose which quantity you want to find.
- 2Pick a substance from the list, or select Custom and enter its specific heat capacity.
- 3Enter the remaining values in SI units — kilograms, kelvin (or Celsius degrees) and joules.
How the calculation works
Q = mcΔT m = Q ÷ (cΔT) ΔT = Q ÷ (mc) c = Q ÷ (mΔT)- Q
- Heat energy transferred, in joules
- m
- Mass in kilograms
- c
- Specific heat capacity, in joules per kilogram per kelvin
- ΔT
- Temperature change — identical in kelvin or degrees Celsius
Specific heat capacity is how much energy one kilogram of a substance needs to warm by one kelvin. Water's value of about 4,186 J/kg·K is unusually high, which is why it is such an effective coolant and why oceans moderate climate so strongly.
ΔT is a difference, so kelvin and Celsius are interchangeable — the scales have different zero points but identical degree size. Fahrenheit is not interchangeable, since its degrees are 5/9 the size.
A negative Q means heat is released rather than absorbed, which is what a negative ΔT produces.
Worked example
Heating 2 kg of water by 50 K
- 1.Water's specific heat capacity is about 4,186 J/kg·K.
- 2.Q = mcΔT = 2 × 4,186 × 50.
- 3.= 8,372 × 50 = 418,600 J.
- 4.That is 418.6 kJ, or about 0.116 kWh — roughly what a 2 kW kettle draws in three and a half minutes.
Result: 418,600 J (418.6 kJ, 0.116 kWh)
Why water is the outlier
Water has a specific heat capacity of about 4,186 J/kg·K, which is remarkably high — roughly ten times iron's and four times air's. Heating a kilogram of water by one degree takes about as much energy as heating ten kilograms of iron by the same amount. The cause is hydrogen bonding: much of the added energy goes into breaking and reforming bonds between molecules rather than into raising their kinetic energy, which is what temperature actually measures.
The consequences are large and everywhere. Oceans absorb enormous quantities of heat with small temperature change, which moderates coastal climates. Water works well as an engine coolant and in central heating. And it is why boiling a kettle takes noticeably longer than heating the same mass of almost anything else.
What this formula deliberately excludes
Q = mcΔT covers only temperature change within a single phase. It says nothing about melting or boiling, and those transitions absorb striking amounts of energy at constant temperature — latent heat.
The numbers are not small. Melting one kilogram of ice at 0 °C takes about 334,000 J while the temperature stays at 0 °C throughout. Boiling one kilogram of water at 100 °C takes about 2,260,000 J — more than five times the energy needed to heat that same kilogram from freezing to boiling in the first place. Any calculation spanning a phase change has to add these separately, which is why a kettle spends most of its time at a rolling boil rather than getting there.
Reading the comparison table
The relative column shows each substance against water. Metals sit far below: copper at 385 J/kg·K heats and cools roughly eleven times faster than water per unit mass, which is exactly why a metal spoon in hot soup becomes uncomfortable long before the soup cools noticeably.
Note also that ice and steam have quite different specific heats from liquid water, despite being the same compound. Specific heat is a property of a substance in a particular phase, not of the substance alone — which is another reason phase changes have to be handled separately.
What this assumes, and where it stops
Assumptions
- Specific heat capacity is constant over the temperature range. In reality it varies somewhat with temperature, though the variation is small for modest ranges.
- No phase change occurs. Melting or boiling requires latent heat, which this formula does not include.
- All the heat goes into the substance — no losses to the surroundings.
Limitations
- Does not handle phase changes. Spanning a melt or boil needs the latent heat added separately.
- Values in the table are typical figures near room temperature and vary with temperature, pressure and purity.
- Assumes perfect efficiency. A real kettle, heater or engine loses substantial heat to its surroundings, so actual energy consumed will be higher than Q.
Common questions
Do I use kelvin or Celsius for the temperature change?
Either — they give the same answer. A change of 1 K and a change of 1 °C are exactly the same size, since the scales differ only in where zero sits and that offset cancels in a subtraction. Fahrenheit is different: its degrees are 5/9 the size, so a Fahrenheit change must be converted first.
Why does it take so long to boil water?
Two reasons compounding. Water has an unusually high specific heat capacity, so raising its temperature at all takes a lot of energy. Then the phase change itself takes about 2,260,000 J per kilogram at constant temperature — more than five times what it took to heat that water from freezing to boiling. This calculator covers only the first part.
Formula and content last reviewed on .
Results are estimates for information only, not professional advice.
Related calculators
Tools people commonly use alongside the specific heat calculator.