Specific Heat Calculator
How much energy heats it, how hot it gets, or what a material’s specific heat is. Plus a calorimetry mixer for the final temperature of hot and cold together.
By Manzoor Ahmad, pharmacist · Updated How do you calculate heat energy with specific heat?
Use Q = m × c × ΔT. Heating 1 kg of water (c = 4.184 J/g·°C) from 20 °C to 100 °C takes 1,000 g × 4.184 × 80 °C = 334,720 J, about 335 kJ or 0.093 kWh. A 2 kW kettle would supply that in roughly 2.8 minutes if no heat were lost.
Same job, other materials
Substance A
Substance B
Specific heat of common substances
Typical values near room temperature and atmospheric pressure. J/g·°C is numerically equal to kJ/kg·K.
| Substance | J/g·°C | J/kg·K | cal/g·°C |
|---|
Specific heat formulas
| To find | Formula | Example |
|---|---|---|
| Heat energy | Q = mcΔT | 500 g aluminium, +30 °C: 13.5 kJ |
| Specific heat | c = Q ÷ (mΔT) | 5,000 J, 100 g, +50 °C: 1.0 J/g·°C |
| Mass | m = Q ÷ (cΔT) | 100 kJ warms 299 g of water by 80 °C |
| Temperature change | ΔT = Q ÷ (mc) | 1,000 J into 100 g copper: +26.0 °C |
| Final temperature (mixing) | (m₁c₁T₁ + m₂c₂T₂) ÷ (m₁c₁ + m₂c₂) | 200 g at 80 °C + 100 g at 20 °C: 60 °C |
Latent heat is separate
Q = mcΔT applies only within one phase. Melting ice takes 334 J/g and boiling water takes 2,257 J/g at constant temperature. To turn 100 g of ice at 0 °C into steam at 100 °C you need 33.4 kJ to melt it, 41.8 kJ to heat it and 225.7 kJ to boil it: 301 kJ in total.
Specific heat FAQs
Q = m × c × ΔT, where Q is heat energy in joules, m is mass in grams, c is specific heat capacity in J/g·°C and ΔT is the temperature change in °C (or K).
Multiply mass by 4.184 J/g·°C by the temperature rise. Heating 1 kg (1,000 g) of water from 20 °C to 100 °C needs 1,000 × 4.184 × 80 = 334,720 J, about 335 kJ or 0.093 kWh.
About 4.184 J/g·°C (4,184 J/kg·K) for liquid water near room temperature, which is why water heats and cools slowly. Ice is about 2.09 J/g·°C and steam about 2.01 J/g·°C.
Measure the heat supplied, the mass and the temperature rise, then use c = Q ÷ (m × ΔT). If 5,000 J raises 100 g of a metal by 50 °C, c = 5,000 ÷ (100 × 50) = 1.0 J/g·°C.
Rearrange to ΔT = Q ÷ (m × c) and add it to the starting temperature. 1,000 J added to 100 g of copper (0.385 J/g·°C) raises it by 26.0 °C.
With no heat loss, heat lost by the hot one equals heat gained by the cold one, so T_final = (m₁c₁T₁ + m₂c₂T₂) ÷ (m₁c₁ + m₂c₂). Mixing 200 g of water at 80 °C with 100 g at 20 °C gives 60 °C.
Yes. A change of 1 °C equals a change of 1 K, so specific heat in J/g·°C equals J/g·K. A change of 1 °F is 5/9 of a degree Celsius.
Specific heat is per unit mass (J/g·°C). Heat capacity is for a whole object (J/°C) and equals mass times specific heat.
Metals have low specific heat and high thermal conductivity, so they heat up quickly and transfer heat to your skin fast. Specific heat sets how much energy a material stores; conductivity sets how fast it flows.
Divide the energy by the power. 1 L of water from 20 °C to 100 °C needs 335 kJ; a 2 kW kettle supplies 2 kJ per second, so about 167 s in theory, or roughly 3 minutes with real-world losses.
No. During a phase change temperature stays constant while latent heat is absorbed or released: 334 J/g to melt ice and 2,257 J/g to boil water. Calculate those steps separately.
Common units are J/g·°C, J/kg·K, kJ/kg·K, cal/g·°C and BTU/lb·°F. 1 cal/g·°C = 1 BTU/lb·°F = 4.1868 J/g·°C.
Sources
- OpenStax University Physics Volume 2, Section 1.5: Heat Transfer, Specific Heat, and Calorimetry
- OpenStax Chemistry 2e, Section 5.2: Calorimetry
- NIST Chemistry WebBook: Water (thermophysical properties)
- USGS Water Science School: Specific Heat Capacity and Water
- NIST SP 811: Unit conversion factors (energy)
Results assume no heat loss and constant specific heat over the temperature range. Real heaters and kettles lose energy to their surroundings, so actual times are longer.