Specific Heat Calculator
Find the energy needed to change the temperature of a material, with metric and imperial inputs and outputs.
What it does
Specific heat says how much energy it takes to raise the temperature of one kilogram of a material by one degree. Water has the highest value of anything you meet in daily life at 4186 joules per kilogram per kelvin, which is why a bathtub takes so long to heat and why a hot pan cools quickly compared with a pot of water.
The arithmetic follows from the definition. Multiply the mass by the specific heat and by the size of the temperature change and you have the heat that has to be added or removed. The OpenStax physics text gives the same relation, Q equals m times c times delta T, with the temperature change in kelvins when the specific heat is in joules per kilogram per kelvin.
The numbers on this page come from a standard table of specific heats. Water is 4186 joules per kilogram per kelvin, aluminum 900, copper 387, iron and steel 452, and concrete and granite about 840. Any material you can name has a value in a handbook, and you can type it in when the table here does not cover it.
A temperature difference of one degree Celsius is the same size as one degree Fahrenheit, so the change does not need converting between the two scales. Only the mass unit and the answer unit change, and this page handles both.
Keep the ideal case in mind. The result assumes the specific heat does not change with temperature, that nothing is lost to the container or the room, and that no phase change happens along the way. A real kettle has to make up for all three.
How to use
- Pick metric or imperial units.
- Choose the material, or pick your own value and type in its specific heat.
- Enter the mass of the material.
- Enter the size of the temperature change you need.
- Read the heat needed in kilojoules or BTU, plus watt hours and food calories.
- Add a margin for heat lost to the container and the room if you are sizing a heater.
Formula
- Q
- Heat needed, in joules.
- m
- Mass of the material.
- c
- Specific heat of the material in J/(kg K).
- delta T
- Size of the temperature change, the same number in °C and °F.
Specific heat depends on temperature, so the tabulated values are representative rather than exact. The NIST reference data for water gives a molar heat capacity of 75.38 joules per mole per kelvin at 298 K, which is 4186 joules per kilogram per kelvin at that point. NIST also fixes the thermochemical calorie at exactly 4.184 joules, which is the conversion used for the kilocalorie figure on this page.
Worked example
Example: 2.5 kg of water heated by 40 degrees
Defaults: metric units, water, 2.5 kg, a temperature change of 40 degrees.
- Specific heat of water: 4186 J/(kg K).
- Q = 2.5 kg x 4186 J/(kg K) x 40 K = 418,600 J.
- In kilojoules: 418,600 / 1000 = 418.6 kJ.
- In watt hours: 418,600 / 3600 = 116.3 Wh.
- In food calories: 418,600 / 4184 = 100.0 kcal.
- The same energy would raise 2.50 kg of water by 40 degrees, since water is the reference.
- The tool shows 418.6 kJ, 116.3 Wh, 100.0 kcal and 2.50 kg of water.
Reference table
| Material | J/(kg K) | kJ per kg per degree | BTU per lb per °F |
|---|---|---|---|
| Water | 4186 | 4.186 | 1.000 |
| Aluminum | 900 | 0.900 | 0.215 |
| Iron and steel | 452 | 0.452 | 0.108 |
| Concrete and granite | 840 | 0.840 | 0.201 |
| Copper | 387 | 0.387 | 0.092 |
Limits
- Assumes the specific heat does not change across the temperature range.
- Ignores heat absorbed by the container, which matters most for small batches in metal pans.
- Leaves out heat lost to the surroundings, so a real system needs more energy than the result.
- Does not cover phase changes, where the temperature stops moving while the energy goes into melting or boiling.
Common mistakes
- Using the starting temperature instead of the size of the change, which gives a much larger answer.
- Entering a temperature difference in Fahrenheit and a specific heat per degree Celsius, which is fine, but then treating the mass in pounds with a metric specific heat.
- Forgetting that a degree difference is the same size in Celsius and Fahrenheit, so no conversion is needed.
- Using the heat capacity of the whole object instead of the specific heat per kilogram.
- Leaving the heat lost to the kettle out of a heating estimate and sizing a heater that is too small.
Try it yourself
Quick checks. Answers sit tucked under each question.
How much energy does it take to heat 0.75 kg of aluminum by 60 degrees?
Q = 0.75 x 900 x 60 = 40,500 J, so 40.5 kJ, 11.3 Wh and 9.7 kcal. The same heat would raise 0.16 kg of water by 60 degrees.
Heating 5 kg of water by 10 degrees needs how much energy?
Q = 5 x 4186 x 10 = 209,300 J, so 209.3 kJ, 58.1 Wh and 50.0 kcal.
An imperial user wants the same job as the example: how many BTU does it take to heat 5.51 lb of water by 40 degrees?
5.51 lb is 2.5 kg, so the answer is the same energy as the metric example: 418.6 kJ, which is 396.7 BTU.
Questions
What does specific heat mean?
It is the energy needed to raise one kilogram of a material by one kelvin. Water is 4186 joules per kilogram per kelvin, aluminum 900 and copper 387.
Why does a degree Celsius equal a degree Fahrenheit for this calculation?
A temperature change of one degree Celsius is exactly as large as a change of one degree Fahrenheit, only the zero points differ. Since the formula uses a difference, no conversion is needed.
How much energy does it take to heat a bathtub of water?
For 150 kg of water raised by 40 degrees, Q = 150 x 4186 x 40 = 25.1 MJ, which is about 7 kWh. A typical immersion heater runs at 3 kW, so the water needs close to two and a half hours.
Why is water used as the reference?
Its specific heat is high and it is easy to measure, so the calorie was originally defined from water. A kilocalorie is the energy to raise one kilogram of water by one degree, which is why the tool shows a water equivalent for every result.
Does the tool work for cooling?
Yes. Enter the size of the temperature drop instead of the rise. The energy is the same size, it just leaves the material rather than entering it.
What about melting ice or boiling water?
That is a different calculation, because the temperature does not change while the phase does. The energy goes into the latent heat of fusion or vaporization instead, and this page does not cover it.
Sources
- OpenStax University Physics Volume 2: heat transfer, specific heat and calorimetry
- NIST Chemistry WebBook: thermochemical data for liquid water
- NIST Guide to the SI: notes on the energy unit and the calorie