Specific Heat Calculator: Solve for Any Variable
Solve q = mcΔT for heat energy, mass, specific heat, or temperature change, with joule and calorie units and a cited table of common specific heat capacities.
Heat one gram of water by one degree and it takes 4.184 joules. Heat a gram of copper the same amount and it takes 0.385. Specific heat is why. Enter any three of heat, mass, specific heat, and temperature change, then leave the fourth blank, and this solver returns it from q = m·c·ΔT with every step shown.
Solved value
Enter any three values
Leave the fourth field blank and press Solve.
Endothermic Exothermic No temperature change
| Quantity | Value |
|---|
| Substance | c, J/(g·°C) |
|---|---|
| Helium | 5.193 |
| Water (liquid) | 4.184 |
| Ethanol | 2.376 |
| Ice | 2.093 |
| Water vapor | 1.864 |
| Nitrogen | 1.040 |
| Air | 1.007 |
| Oxygen | 0.918 |
| Aluminum | 0.897 |
| Carbon dioxide | 0.853 |
| Silicon | 0.712 |
| Argon | 0.522 |
| Iron | 0.449 |
| Copper | 0.385 |
| Lead | 0.130 |
| Gold | 0.129 |
q = m·c·ΔT
How?
How this is calculated
Method. The tool solves q = m·c·ΔT, with ΔT = T_final - T_initial, rearranged for whichever field you leave blank: c = q / (m·ΔT), m = q / (c·ΔT), ΔT = q / (m·c). Every entry is converted to canonical units (q in J, m in g, c in J/(g·°C), ΔT in °C) before the algebra runs, then the answer is converted back to the unit you chose. Results show 4 significant figures by default.
Convention shown, water. Water's specific heat is taken as 4.184 J/(g·°C) (OpenStax Chemistry 2e, Table 5.1), the value chosen over 4.18 or 4.186. Water is the dominant real input, so this constant is load-bearing.
Convention shown, the calorie. One calorie is 4.184 J exactly, the thermochemical calorie (IUPAC), so 1 kcal equals one food Calorie equals 4184 J. The IT calorie (4.1868 J) and the 15 °C calorie (4.1855 J) are not used.
Convention shown, temperature intervals. A Celsius interval and a kelvin interval are the same size, so ΔT in °C equals ΔT in K. A Fahrenheit interval converts by a factor of 5/9, never by subtracting 32, because a difference carries no offset. The default specific-heat unit is the chemistry J/(g·°C), not the engineering J/(kg·K).
Sign convention. q > 0 means heat absorbed by the sample, so the temperature rises and the process is endothermic. q < 0 means heat released, the temperature falls, and the process is exothermic. The signs of q and ΔT are preserved, never taken as absolute values.
Scope. One substance, one phase, with c constant across the interval. Phase changes (q = m·L at melting and boiling) and calorimetry mixing of two bodies are separate calculations, not covered here.
Formula: q = m·c·ΔT
Sources
- OpenStax Chemistry 2e, 5.1 Energy Basics (Table 5.1, Specific Heats of Common Substances). OpenStax. Retrieved .
- Introductory Chemistry (LibreTexts), 3.12: Energy and Heat Capacity Calculations. LibreTexts. Retrieved .