Specific Heat Equation (q = mcΔT)
Calculates the heat absorbed or released when a substance's temperature changes, from its mass, specific heat, and ΔT.
What it means
This is the formula for sensible heat — heat that changes temperature rather than causing a phase change. It says the heat required grows with three things: how much substance you heat (m), how stubborn the substance is about warming up (c, its specific heat), and how far you push the temperature (ΔT). Water's famously high specific heat is why oceans moderate climate: heating a gram of water takes over four joules per degree. When to use it: any time a known mass of a single substance warms or cools without boiling or freezing — the building block of almost every calorimetry problem.
Variables
| Symbol | Name | Unit | Meaning |
|---|---|---|---|
| Heat | J (or kJ) | Heat transferred; positive when the substance absorbs heat, negative when it releases it. | |
| Mass | g | Mass of the substance being heated or cooled. | |
| Specific heat capacity | J/(g·°C) | Energy needed to raise one gram of the substance by one degree — a material property. | |
| Temperature change | °C (or K) | Final temperature minus initial temperature (T_final − T_initial). |
Assumptions
- The specific heat c is constant over the temperature range (valid for small ranges).
- No phase change occurs; the substance stays solid, liquid, or gas throughout heating.
- No heat is lost to the surroundings.
Worked example
How much heat is needed to warm 200. g of water from 20.0°C to 80.0°C? (c = 4.184 J/(g·°C))
Given:
- m = 200. g of water
- c = 4.184 J/(g·°C) for liquid water
- ΔT = 80.0 − 20.0 = 60.0 °C
Write q = mcΔT.
Find ΔT: 80.0°C − 20.0°C = 60.0°C.
Substitute: q = (200. g)(4.184 J/(g·°C))(60.0°C).
Multiply: 200. × 4.184 = 836.8; 836.8 × 60.0 = 50,208 J.
Convert: 50,208 J = 50.208 kJ, which rounds to 50.2 kJ (positive — heat absorbed).
Warming a beaker of water (200 g) by 60 degrees requires about 50 kJ — heat flows INTO the water, so q is positive.