Science & Engineering
Specific Heat Calculator - Q = mcΔT Solver
Solve the specific heat equation Q = m × c × ΔT for any of the four variables: heat energy (Q), mass (m), specific heat capacity (c), or temperature change (ΔT). Includes a table of common materials.
heat
41860.0000 J
Specific Heat Formula
The relationship between heat energy, mass, specific heat capacity, and temperature change is: Q = m × c × ΔT. Rearrange for any unknown: m = Q/(c·ΔT), c = Q/(m·ΔT), ΔT = Q/(m·c).
What is specific heat capacity?
Specific heat capacity (symbol: c) is the amount of energy required to raise the temperature of 1 gram of a substance by 1 °C (or 1 K). Its SI unit is J/(g·°C) or equivalently J/(g·K). Water has an exceptionally high specific heat of 4.186 J/(g·°C), meaning it absorbs a large amount of heat with only a small temperature change - which is why the oceans moderate coastal climates and why water is an excellent coolant.
Specific heat reference table
| Substance | Specific heat (J/g·°C) |
|---|---|
| Water (liquid) | 4.186 |
| Ice | 2.09 |
| Steam | 2.01 |
| Ethanol | 2.44 |
| Glycerol | 2.43 |
| Aluminum | 0.897 |
| Iron / steel | 0.449 |
| Copper | 0.385 |
| Gold | 0.129 |
| Lead | 0.128 |
| Silver | 0.235 |
| Glass (window) | 0.84 |
| Granite | 0.79 |
| Wood (oak) | 1.76 |
| Air (dry) | 1.005 |
Calorimetry worked example
How much energy (in joules) is required to heat 250 g of water from 20 °C to 100 °C?
Q = m × c × ΔT = 250 g × 4.186 J/(g·°C) × (100 − 20)°C = 250 × 4.186 × 80 = 83,720 J ≈ 83.7 kJ
This is roughly the energy in one food calorie (kcal), which is defined as the energy to heat 1 kg of water by 1 °C. Our 250 g example therefore requires approximately 20 food calories.