Specific Heat Calculator
Solve Q = mcΔT for any variable or find the final temperature when two objects mix
Solve Q = mcΔT for heat, mass, specific heat or temperature change with 31 built-in materials, and find the final temperature when two objects exchange heat in a calorimetry problem.
PhysicsUpdated October 21, 2026Checked against 6 reference values
Negative for cooling.
Examples
Heat needed
334.72kJ
80 kcal, 0.09298 kWh
Chart data as a table
| J/(kg·K) | |
|---|---|
| Water | 4,184 |
| Ice | 2,090 |
| Air | 1,005 |
| Aluminum | 897 |
| Steel | 490 |
| Copper | 385 |
| Gold | 129 |
- Heat
- 334,720 J
- Heat capacity mc
- 4,184 J/K
- BTU
- 317.25 BTU
- Time on a 1 kW heater
- 334.7 s
- with no losses
Working
- 1
Heat equation
Q = mcΔT - 2
Substitute
334,720 J = 1 kg × 4,184 J/(kg·K) × 80 K
What this calculator does
Specific heat capacity is the energy needed to warm one kilogram of a substance by one degree. This calculator solves Q = mcΔT for whichever quantity you need: the heat to warm or cool something, the temperature change a given amount of heat produces, the mass that can be heated, or the specific heat of an unknown material from a measurement. Choose from 31 common materials or enter your own value in J/(kg·K), cal/(g·°C) or BTU/(lb·°F).
The mixing mode handles the classic calorimetry problem: two objects at different temperatures placed in contact, such as a hot metal block dropped into water, and finds the temperature where they meet. Results show joules, kilojoules, kilocalories, kilowatt-hours and BTU, plus how long a 1 kW heater would take.
The formulas
Q = mcΔT
Mixing: Tf = (m₁c₁T₁ + m₂c₂T₂) ÷ (m₁c₁ + m₂c₂)
Q is the heat added in joules, m the mass in kilograms, c the specific heat in J/(kg·K) and ΔT the change in temperature. A change of 1 K equals a change of 1 °C, so either can be used for ΔT, while a change of 1 °F is 5/9 of that. A negative ΔT or Q means the object is cooling and giving off heat. The mixing formula comes from setting the heat lost by the hotter object equal to the heat gained by the colder one. Both formulas assume no phase change: melting ice or boiling water needs latent heat in addition.
Worked examples
Boiling water for tea. Heating 1 kg of water from 20 °C to 100 °C takes 334.7 kJ, about 0.093 kWh. A 2 kW kettle with no losses would need 2.79 minutes.
An aluminum pan. 10 kJ into a 0.5 kg aluminum pan raises it by 22.3 °C. The same energy would warm 0.5 kg of water by only 4.78 °C.
Identifying a metal. If 3.85 kJ warms a 100 g sample by 100 °C, its specific heat is 385 J/(kg·K), which matches copper.
Hot copper in water. A 200 g copper block at 100 °C dropped into 500 g of water at 20 °C ends at 22.84 °C. The water barely warms because its heat capacity is about 27 times larger.
A swimming pool. Raising 50 m³ of water by 5 °C takes 291 kWh, which is why pool heaters are sized in tens of kilowatts.
Specific heat of common substances
| Substance | J/(kg·K) | cal/(g·°C) |
|---|---|---|
| Water (liquid) | 4,184 | 1 |
| Seawater | 3,993 | 0.954 |
| Ice (near 0 °C) | 2,090 | 0.5 |
| Steam (100 °C) | 2,010 | 0.48 |
| Ethanol | 2,440 | 0.583 |
| Glycerol | 2,430 | 0.581 |
| Acetone | 2,170 | 0.519 |
| Vegetable oil | 2,000 | 0.478 |
| Air (constant pressure) | 1,005 | 0.24 |
| Nitrogen gas | 1,040 | 0.249 |
| Oxygen gas | 918 | 0.219 |
| Hydrogen gas | 14,300 | 3.42 |
| Helium gas | 5,193 | 1.24 |
| Aluminum | 897 | 0.214 |
| Copper | 385 | 0.092 |
| Iron | 449 | 0.107 |
| Steel (carbon) | 490 | 0.117 |
| Stainless steel | 500 | 0.12 |
| Brass | 380 | 0.0908 |
| Gold | 129 | 0.0308 |
| Silver | 235 | 0.0562 |
| Lead | 129 | 0.0308 |
| Mercury | 140 | 0.0335 |
| Silicon | 705 | 0.168 |
| Glass | 840 | 0.201 |
| Granite | 790 | 0.189 |
| Concrete | 880 | 0.21 |
| Brick | 840 | 0.201 |
| Wood (typical) | 1,700 | 0.406 |
| Dry sand | 835 | 0.2 |
| Human body (average) | 3,470 | 0.829 |
Values near room temperature and atmospheric pressure from the CRC Handbook of Chemistry and Physics, with typical values for building materials and wood. Gases are at constant pressure.
Why water's high specific heat matters
Water stores about 4.2 kJ per kilogram per degree, roughly ten times as much as iron and more than almost any other common substance, because hydrogen bonds between molecules absorb energy that would otherwise raise the temperature. This is why coastal climates are mild, why oceans moderate the planet's temperature, and why water is used in car radiators, home heating loops and industrial cooling. It is also why a pot of water takes much longer to heat than the metal pot holding it, and why your body, which is mostly water, holds a steady temperature.
Phase changes and latent heat
At a melting or boiling point, added heat changes the state rather than the temperature. Melting ice takes 334 kJ/kg and boiling water takes 2,257 kJ/kg, far more than warming it: boiling away 1 kg of water already at 100 °C takes about seven times the energy needed to bring it from 20 °C to 100 °C. For a problem that crosses a phase change, calculate each stage separately, warming the solid, melting it, warming the liquid and so on, and add the energies. Ice and steam have their own specific heats, both listed above.
Specific Heat Calculator: common questions
What is the formula for heat energy?
Q = mcΔT, where Q is heat in joules, m is mass in kilograms, c is specific heat in J/(kg·K) and ΔT is the temperature change. Rearrange it to find any one value from the other three.
What is the specific heat of water?
About 4,184 J/(kg·K), or 1 cal/(g·°C), near room temperature. It varies slightly with temperature, from about 4,218 at 0 °C to 4,216 at 100 °C, with a minimum near 35 °C.
Is a change of 1 °C the same as 1 K?
Yes. The Celsius and Kelvin scales have the same step size and differ only in their zero point, so temperature differences are identical. A change of 1 °F equals 5/9 of a kelvin.
Why does metal feel colder than wood at the same temperature?
Metal conducts heat away from your skin much faster. Specific heat is not the main reason: wood actually has a higher specific heat than most metals. Thermal conductivity controls how cold a surface feels.
Can I use this for melting ice or boiling water?
Not in one step. Q = mcΔT applies within a single phase. For a phase change, add latent heat: 334 kJ/kg to melt ice and 2,257 kJ/kg to boil water, then compute the warming stages separately.
Sources
- CRC Handbook of Chemistry and Physics, Heat capacity of the elements and common substances
- OpenStax University Physics Volume 2, Section 1.4: Heat Transfer, Specific Heat, and Calorimetry
- NIST Special Publication 811, conversion factors for calorie and BTU
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