Heating a pot of water takes far longer than heating an empty pan of the same mass, because water stores much more energy per degree. That property is the specific heat capacity: the energy needed to raise one kilogram of a material by one kelvin. This calculator uses it in the basic calorimetry equation, Q = mcΔT, to find the heat energy, the mass, the specific heat or the temperature change, with presets for twenty common materials and an optional heater power that turns the energy into a heating time.
How to use the specific heat calculator
- Under Solve for, choose Heat (Q), Mass, Specific heat or Temperature change.
- Pick a Material or choose “Custom value” and enter the specific heat in J/(kg·K), J/(g·°C), kJ/(kg·K), cal/(g·°C) or BTU/(lb·°F).
- Enter the Mass in kg, g, lb, oz or metric tons.
- Give the temperatures as Start and end values in °C, °F or K, or as a Change only. Use a negative change for cooling.
- When solving for anything other than heat, enter the Heat energy: positive when heat is added, negative when removed. When solving for the temperature change, add an optional starting temperature to get the final temperature.
- Optionally enter a Heater power to see how long the heating would take.
Specific heat formula
Q is the heat in joules, m the mass in kilograms, c the specific heat in J/(kg·K) and ΔT = Tfinal − Tinitial. The product mc is the heat capacity of the whole sample, in joules per kelvin. With a heater of power P, the ideal heating time is t = Q ÷ P.
Worked example
Boiling water for tea
Heat 1.5 kg (about 1.5 L) of water from 20 °C to 100 °C. Water's specific heat is 4,184 J/(kg·K) and ΔT = 80 K.
Q = 1.5 × 4,184 × 80 = 502,080 J = 502.08 kJ
That equals 120 kcal, 475.9 BTU or 0.139 kWh. A 1,500 W kettle needs 502,080 ÷ 1,500 = 334.7 s, about 5.6 minutes, with no losses.
Identifying a metal. A 0.25 kg sample absorbs 5.0 kJ and warms from 20 °C to 64.5 °C. Solve for specific heat: c = 5,000 ÷ (0.25 × 44.5) = 449.4 J/(kg·K), which matches iron (449 J/(kg·K)).
The same energy in copper. Heating 200 g of copper from 20 °C to 220 °C takes only 0.2 × 385 × 200 = 15.4 kJ, about 3% of what the kettle needed, even though the temperature rise is more than twice as large.
Specific heats of common materials
| Material | c (J/(kg·K)) |
|---|---|
| Water | 4,184 |
| Ethanol | 2,440 |
| Ice | 2,090 |
| Wood (typical) | 1,700 |
| Air (constant pressure) | 1,005 |
| Aluminum | 897 |
| Glass | 840 |
| Iron | 449 |
| Copper | 385 |
| Lead and gold | 129 |
Values are typical figures near room temperature from standard references such as the CRC Handbook; real samples vary with composition and temperature.
Practical notes
High specific heat makes water ideal for radiators, hot-water bottles and thermal storage tanks; it also explains why coastal climates swing less than inland ones. Metals, with low specific heats, warm and cool quickly, which is why a metal spoon in hot soup heats up so fast. In real heating, some energy always escapes to the container and the air, so measured heating times are longer than the ideal figure.
To convert the energy into kilowatt-hours or calories, use the energy converter; for heat flowing through a wall rather than into a material, use the heat conduction calculator; and to price the electricity, try the electricity cost calculator.
Frequently asked questions
What is the specific heat of water?
About 4,184 J/(kg·K), which is 4.184 J/(g·°C) or 1 cal/(g·°C). It varies slightly with temperature, from about 4,218 at 0 °C to 4,180 near 25 °C. Water's value is unusually high, which is why oceans moderate climate and why water makes a good coolant.
Can I use °C or °F for the temperature change?
Yes. A change of 1 °C equals a change of 1 K, so either works directly in J/(kg·K). A change of 1 °F is 5/9 K; the calculator converts for you. Only absolute temperatures need kelvin, and those are not used in Q = mcΔT.
How long will it take to boil water in my kettle?
Heating 1.5 L of water from 20 °C to 100 °C takes 502 kJ. A 1,500 W kettle delivers 1,500 J each second, so about 335 seconds, or 5.6 minutes, if no heat were lost. Real kettles take a little longer.
Why does the calculator warn about phase changes?
Q = mcΔT only covers warming or cooling within one phase. Melting ice at 0 °C absorbs about 334 J/g and boiling water at 100 °C absorbs about 2,257 J/g without any temperature change, so those latent heats must be added separately.
What does a negative heat value mean?
Heat leaving the material. Cooling 1.5 kg of water from 100 °C to 20 °C gives Q = −502 kJ: the same amount of energy that heating it took, now released to the surroundings.