Specific Heat Calculator
The specific heat equation Q = m × c × ΔT links heat energy, mass, material, and temperature change. Solve for any of the four, with real specific heat values built in.
Example: with Solve for Heat energy Q = m·c·ΔT · Material (sets c, J/kg·K) Water — 4186 · Custom c (J/kg·K) 4186 · Mass (kg) 2 · Temperature change ΔT (°C or K) 30 · Heat energy Q in joules (solve modes) 251160 → Result: 251,160 J.
Computed by the calculator below using its default values. Change any input to see your own numbers.
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Check it outThe specific heat formula, worked through
Q = m × c × ΔT: heat energy equals mass times specific heat capacity times temperature change. Heating 2 kg of water (c = 4,186 J/kg·K) by 30 °C takes Q = 2 × 4,186 × 30 = 251,160 J — about 251 kJ, or 70 watt-hours. Specific heat is the energy needed to warm 1 kg of a material by 1 K, so low-c materials like copper (385) heat up far faster than water on the same burner.
To find specific heat itself, rearrange the equation: c = Q ÷ (m × ΔT). If 10,000 J warms a 0.5 kg metal block by 25 °C, then c = 10,000 ÷ (0.5 × 25) = 800 J/kg·K — close to aluminum’s 900. Because ΔT is a difference, Celsius and kelvin give identical answers.
How it’s calculated
Q = m × c × ΔT, rearranged as c = Q ÷ (mΔT), m = Q ÷ (cΔT), and ΔT = Q ÷ (mc). Built-in specific heat capacities (J/kg·K): water 4186, aluminum 900, copper 385, iron 449 — standard reference values near room temperature. Energy conversions: 1 kJ = 1,000 J; 1 kcal = 4,184 J; 1 Wh = 3,600 J.
Results update as you type and are estimates, not professional advice — verify important decisions with a qualified professional.
Common mistakes
- Using a total temperature instead of the change — ΔT is final minus initial, so 20 °C to 50 °C means ΔT = 30.
- Entering grams where the formula expects kilograms — with c in J/kg·K, 500 g must go in as 0.5 kg.
- Applying Q = mcΔT across a phase change — melting and boiling need latent heat on top.
Frequently asked questions
What is the specific heat formula?
Q = mcΔT: heat energy (J) = mass (kg) × specific heat capacity (J/kg·K) × temperature change (K or °C). Rearrange to c = Q ÷ (mΔT) to find the specific heat itself.
How do you calculate specific heat?
Divide the heat added by mass times temperature change: c = Q ÷ (mΔT). If 4,500 J warms 1 kg of a metal by 10 K, c = 450 J/kg·K — essentially iron (449).
What is the specific heat of water?
About 4,186 J/kg·K, i.e. 1 calorie per gram per °C — unusually high, which is why water heats slowly, cools slowly, and makes an excellent coolant.
Is specific heat the same as heat capacity?
No. Specific heat c is per kilogram of material; heat capacity C = m × c belongs to a whole object. A 2 kg iron pan has C = 2 × 449 = 898 J/K.
Do I use Celsius or kelvin for ΔT?
Either — the degree size is identical, so a change of 30 °C is a change of 30 K. Only the change matters in Q = mcΔT, not the absolute temperature.