Choose energy transferred, mass, specific heat capacity, or temperature change to solve for. Enter the known values in the supported units; the default specific heat is approximately 4.18 J/(g·°C) for liquid water.
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Energy to Heat Water Formula
The following formula estimates sensible heat transferred to liquid water with an approximately constant specific heat. It excludes phase changes, vessel heating and heat loss; negative energy or temperature change represents cooling.
- Where E is the total energy required (J)
- m is the nonnegative water mass; use matching mass and specific-heat units (the calculator converts supported units to SI)
- c is the specific heat capacity (J/(kg·°C) or J/(g·°C))
- ΔT is final minus initial temperature; 1 °C difference = 1 K difference = 1.8 °F difference, without an absolute-temperature offset
| Water Mass (kg) | ΔT = 10°C (kJ) | ΔT = 40°C (kJ) | ΔT = 80°C (kJ) | ΔT = 100°C (kJ) |
|---|---|---|---|---|
| 0.25 | 10.5 | 41.8 | 83.6 | 104.5 |
| 0.50 | 20.9 | 83.6 | 167.2 | 209.0 |
| 0.75 | 31.4 | 125.4 | 250.8 | 313.5 |
| 1.00 | 41.8 | 167.2 | 334.4 | 418.0 |
| 1.25 | 52.3 | 209.0 | 418.0 | 522.5 |
| 1.50 | 62.7 | 250.8 | 501.6 | 627.0 |
| 2.00 | 83.6 | 334.4 | 668.8 | 836.0 |
| 2.50 | 104.5 | 418.0 | 836.0 | 1045.0 |
| 3.00 | 125.4 | 501.6 | 1003.2 | 1254.0 |
| 3.50 | 146.3 | 585.2 | 1170.4 | 1463.0 |
| 4.00 | 167.2 | 668.8 | 1337.6 | 1672.0 |
| 4.50 | 188.1 | 752.4 | 1504.8 | 1881.0 |
| 5.00 | 209.0 | 836.0 | 1672.0 | 2090.0 |
| 6.00 | 250.8 | 1003.2 | 2006.4 | 2508.0 |
| 7.00 | 292.6 | 1170.4 | 2340.8 | 2926.0 |
| 8.00 | 334.4 | 1337.6 | 2675.2 | 3344.0 |
| 9.00 | 376.2 | 1504.8 | 3009.6 | 3762.0 |
| 10.00 | 418.0 | 1672.0 | 3344.0 | 4180.0 |
| 12.00 | 501.6 | 2006.4 | 4012.8 | 5016.0 |
| 15.00 | 627.0 | 2508.0 | 5016.0 | 6270.0 |
| Assumes same-phase liquid water with approximate constant c = 4.18 kJ/kg°C. E (kJ) = mass (kg) × 4.18 × ΔT (°C). No latent heat, vessel heating or heat loss is included; a listed temperature rise does not establish that water remains liquid at the operating pressure. Approximation: 1 liter ≈ 1 kg. | ||||
How much energy is required to heat water?
The energy required to heat water is determined by three factors.
First, the total mass of the water. The greater the mass of water, the more energy it will take to heat it.
Next, the specific heat of the water. This value is mostly a constant. Specific heat varies with temperature, pressure and composition. A constant room-temperature approximation may be suitable for an estimate; phase changes require a different model.
Lastly, the amount of temperature change. The greater the change in temperature the greater the energy.
Example Problem
How to calculate energy to heat water?
First, determine the total mass of the water and the change in temperature. For this example, the mass of the water is 400 grams and the temperature change is 20°C.
Next, we will use the specific heat of fresh water at room temperature which is approximately 4.18 J/(g·°C).
Finally, calculate the energy required to heat the water using the formula above:
E = m * c * ΔT
E = 400 * 4.18 * 20
E = 33,440 joules
