CalculateY
Quick Answer
Heating 100 g of water by 10°C requires 4,186 J (4.186 kJ). Heating 100 g of aluminum by 10°C requires only 897 J because aluminum has a lower specific heat capacity.
Common Examples
| Input | Result |
|---|---|
| m = 100 g water, ΔT = 10°C, c = 4.186 J/g°C | Q = 4,186 J (4.19 kJ) |
| Q = 5,000 J, c = 4.186 J/g°C (water), ΔT = 25°C | m = 47.78 g |
| Q = 2,000 J, m = 200 g, ΔT = 15°C | c = 0.667 J/g°C |
| Q = 10,000 J, m = 500 g, c = 0.897 J/g°C (aluminum) | ΔT = 22.30°C |
| m = 250 g copper, ΔT = 50°C, c = 0.385 J/g°C | Q = 4,813 J (4.81 kJ) |
How It Works
The formula
The heat transfer equation (also called the calorimetry equation) relates the heat energy exchanged by a substance to its mass, specific heat capacity, and change in temperature:
Q = m × c × ΔT
Where:
- Q = heat energy transferred, in joules (J)
- m = mass of the substance, in grams (g)
- c = specific heat capacity, in joules per gram per degree Celsius (J/g°C)
- ΔT = change in temperature, in degrees Celsius (°C)
Rearranged to solve for any variable:
- Heat energy: Q = m × c × ΔT
- Mass: m = Q / (c × ΔT)
- Specific heat: c = Q / (m × ΔT)
- Temperature change: ΔT = Q / (m × c)
Common specific heat values (J/g°C):
| Material | Specific heat (J/g°C) |
|---|---|
| Water | 4.186 |
| Aluminum | 0.897 |
| Iron | 0.449 |
| Copper | 0.385 |
| Gold | 0.129 |
Why water’s high specific heat matters
Water’s specific heat of 4.186 J/g°C is exceptionally high compared to most materials. This means water absorbs a large amount of energy before its temperature rises noticeably. Coastal climates are more moderate than inland climates partly because oceans act as thermal buffers. Industrial cooling systems and car radiators use water for the same reason.
Worked example
To find the heat required to warm 500 g of iron from 20°C to 100°C: ΔT = 100 − 20 = 80°C. Q = 500 × 0.449 × 80 = 17,960 J = 17.96 kJ. The same temperature rise in 500 g of water would require Q = 500 × 4.186 × 80 = 167,440 J = 167.44 kJ, nearly ten times more energy.