The formula
q = m × c × ΔT
- q: heat added (positive) or released (negative), in joules
- m: mass, in grams
- c: specific heat capacity, the heat needed to warm 1 g by 1 °C, in J/(g·°C)
- ΔT: final temperature minus initial temperature. A change of 1 °C is the same as a change of 1 K, so either unit works for ΔT.
Rearranged: c = q ÷ (m × ΔT), m = q ÷ (c × ΔT) and ΔT = q ÷ (m × c).
Example: heating water
How much heat does it take to warm 250 g of water from 20 °C to 80 °C? q = 250 × 4.184 × (80 − 20) = 62,760 J, or 62.8 kJ.
Example: identifying a metal
A 100 g piece of metal gives off 1,800 J as it cools by 40 °C. c = 1,800 ÷ (100 × 40) = 0.45 J/(g·°C), which matches iron.
Example: a calorimetry problem
A 50.0 g metal sample at 100.0 °C is dropped into 100.0 g of water at 20.0 °C, and both end at 23.5 °C. The heat the metal loses is the heat the water gains:
- Water: q = 100.0 × 4.184 × (23.5 − 20.0) = 1,464 J.
- Metal: ΔT = 100.0 − 23.5 = 76.5 °C.
- c = 1,464 ÷ (50.0 × 76.5) = 0.383 J/(g·°C), close to copper.
Specific heats of common materials
| Material | c, J/(g·°C) |
|---|---|
| Water (liquid) | 4.184 |
| Ice | 2.09 |
| Steam | 2.01 |
| Ethanol | 2.44 |
| Aluminum | 0.897 |
| Glass | 0.84 |
| Iron | 0.449 |
| Copper | 0.385 |
| Lead | 0.129 |
When q = mcΔT does not apply
The formula only covers a temperature change within one phase. While ice melts or water boils, heat goes in with no change in temperature, so use the latent heat instead: about 334 J/g to melt ice and 2,260 J/g to boil water. Warming ice at −10 °C to steam at 110 °C takes five steps: warm the ice, melt it, warm the water, boil it, warm the steam. Specific heat also changes a little with temperature, which matters only for large temperature ranges.
Calculators
The Specific Heat Calculator solves q = mcΔT for any one of the four quantities, with common materials built in.