Specific Heat Calculator — Q = mcΔT
Calculate heat energy using Q = mcΔT. Find temperature change, mass, or specific heat capacity for any material with practical thermal physics examples.
Why does a cast iron pan hold heat so well after you turn off the burner, while a thin aluminum pan cools quickly? Specific heat capacity. Different materials require different amounts of energy to raise their temperature — and that property affects everything from cooking to climate science to engine design.
The CalcHub specific heat calculator solves for heat energy, temperature change, mass, or specific heat capacity using the fundamental thermal equation.
The Formula
Q = m × c × ΔT- Q = heat energy (Joules, J)
- m = mass (kg)
- c = specific heat capacity (J/kg·°C or J/kg·K)
- ΔT = temperature change (°C or K)
Specific Heat Capacities
| Material | c (J/kg·°C) |
|---|---|
| Water | 4186 |
| Ice | 2090 |
| Alcohol (ethanol) | 2440 |
| Aluminum | 900 |
| Steel/iron | 490 |
| Copper | 390 |
| Lead | 128 |
| Air | 1005 |
Worked Example
How much energy does it take to heat 2 liters (2 kg) of water from 20°C to 100°C for tea?
Q = 2 × 4186 × (100 − 20) = 2 × 4186 × 80 = 669,760 J ≈ 670 kJ
An electric kettle rated at 2000 W (2000 J/s) would take:
t = 670,000 / 2000 = 335 s ≈ 5.6 minutes
That matches real-world kettle boiling times well.
What's the difference between specific heat and heat capacity?
Specific heat capacity is per unit mass — how much energy per kg per degree. Heat capacity (without "specific") is for a specific object: C = mc. A large object has a high heat capacity even if its material has a low specific heat. A cooking pot has more total heat capacity than a coin of the same metal.
Why does water have such high specific heat?
Water molecules form hydrogen bonds that require extra energy to disrupt when heating. These bonds absorb energy without immediately raising temperature, giving water a large "thermal buffer" capacity. It's one reason water is used as coolant in engines and nuclear reactors.
How do I calculate heat loss for a room?
You need specific heat of air and mass of air in the room. For a 3m × 4m × 2.5m room: volume = 30 m³, mass ≈ 36 kg. Cooling 10°C: Q = 36 × 1005 × 10 = 361,800 J. Real heat loss calculations are more complex (conduction through walls, infiltration), but this estimates the energy stored in the air volume.