Specific Heat Formula
The specific heat formula Q = mcΔT calculates the heat energy needed to change the temperature of a substance.
Includes worked examples with common materials.
The Formula
The heat energy absorbed or released by a substance depends on its mass, its specific heat capacity, and the temperature change. Different materials require different amounts of energy to change temperature.
Variables
| Symbol | Meaning |
|---|---|
| Q | Heat energy (measured in joules, J) |
| m | Mass of the substance (measured in grams, g, or kilograms, kg) |
| c | Specific heat capacity (measured in J/(g·°C) or J/(kg·K)) |
| ΔT | Change in temperature (final temperature minus initial temperature, in °C or K) |
Example 1
How much energy is needed to heat 500 g of water from 20°C to 80°C? The specific heat of water is 4.186 J/(g·°C).
Identify the values: m = 500 g, c = 4.186 J/(g·°C), ΔT = 80 - 20 = 60°C
Apply the formula: Q = mcΔT = 500 × 4.186 × 60
Q = 125,580 J (approximately 125.6 kJ)
Example 2
A 200 g piece of iron absorbs 4,500 J of heat. If iron's specific heat is 0.449 J/(g·°C), what is the temperature change?
Rearrange: ΔT = Q / (mc)
ΔT = 4,500 / (200 × 0.449)
ΔT = 4,500 / 89.8
ΔT ≈ 50.1°C
When to Use It
Use the specific heat formula for any problem involving heat and temperature change.
- Calculating the energy to heat or cool a substance
- Calorimetry experiments in the lab
- Comparing how quickly different materials heat up
- Cooking, heating systems, and industrial thermal processes
Key Notes
- Formula: q = mcΔT: q is heat energy (J), m is mass (g or kg), c is specific heat capacity, and ΔT is the temperature change. Specific heat must be in units consistent with m — c for water is 4.18 J/(g·°C) or 4,180 J/(kg·K).
- Water's exceptionally high specific heat: Water's c = 4.18 J/(g·°C) is unusually high compared to metals (aluminum: 0.897, iron: 0.449 J/(g·°C)). This is why coastal climates are moderate, why water is used as engine coolant, and why large bodies of water slow seasonal temperature swings.
- Constant pressure vs constant volume: At constant pressure (cp), some energy goes into expansion work (PΔV). At constant volume (cv), all heat goes into temperature change. For solids and liquids, cp ≈ cv. For gases, cp > cv and cp − cv = R (per mole).
- Calorimetry — measuring specific heat: In a calorimeter, q_lost by hot object = q_gained by cold water. Solving q = mcΔT for both sides and equating them yields the unknown specific heat. This technique assumes no heat loss to the environment.
- Applications: Specific heat calculations govern HVAC system design, cooking time estimation, industrial heating/cooling processes, thermal storage systems, and climate modeling where heat exchange between ocean and atmosphere is critical.