Specific Heat Calculator (Q = mcΔT)
Solve Q = mcΔT for heat, mass, specific heat or final temperature in J, kJ, kWh, BTU or calories, with ice melting and water boiling added when needed.
The interactive tool loads here when JavaScript runs.
About this toolWhat is the Specific Heat Calculator?
Specific heat is the energy needed to raise one kilogram of a substance by one kelvin. Multiply it by the mass and the temperature change and you have the heat that went in or came out: Q = m·c·ΔT. Two kilograms of water warmed from 20 °C to 95 °C, a large pot on a stove, takes 627.9 kJ, or about 0.17 kWh.
This calculator solves that equation for whichever quantity you do not know: heat, mass, specific heat, or the final temperature reached for a given amount of heat. It has typical specific heats for eighteen materials from water and air to steel, copper and concrete, and it shows the answer in joules, kilojoules, kilowatt-hours, BTU and kilocalories together. When water starts as ice or ends as steam, switch on phase changes and it adds the latent heat of melting and boiling, step by step, with a heating curve.
- Solve Q = mcΔT for heat, mass, specific heat or final temperature
- Eighteen materials with typical specific heats, all editable
- Energy in J, kJ, MJ, kWh, BTU, cal and kcal; mass in kg, g and lb
- Temperatures in °C, °F or K with unit changes converted in place
- Water phase changes with a step table and heating-curve chart
How to use the Specific Heat Calculator
- Choose what to solve for: heat Q, mass, specific heat or final temperature. The result field turns blue and read-only.
- Pick a material to load its typical specific heat, or type a value from a data sheet in J/kg·K, kJ/kg·K, BTU/lb·°F or cal/g·°C.
- Enter the mass and the start and final temperatures in °C, °F or K. For a final-temperature solve, enter the heat instead.
- For water that freezes, melts or boils, turn on phase changes to include the latent heat at 0 °C and 100 °C.
- Read the heat in every energy unit, then copy the results or share a link with your inputs.
Worked examples
Heating a pot of water (page default)
Input
2 kg water, c 4,186 J/kg·K, 20 °C to 95 °C
Output
627.9 kJ = 0.1744 kWh = 595.1 BTU = 150.1 kcal
A 2 kW hob that delivered all of its output to the water would need just over five minutes. Real hobs lose a third or more of their output to the room.
Ice at −10 °C to steam at 120 °C
Input
1 kg, phase changes on
Output
3,069.7 kJ in five steps
Boiling alone takes 2,256 kJ, 73.5% of the total. Warming the liquid from 0 to 100 °C takes only 418.6 kJ.
Aluminium pan on a burner
Input
1.5 kg aluminium, c 897 J/kg·K, 20 °C to 200 °C
Output
242.2 kJ
Aluminium has about a fifth of the specific heat of water per kilogram, which is why an empty pan heats so much faster than a full one.
Background and insight
The relation Q = m·c·ΔT is sensible heat: energy that changes temperature without changing phase. It assumes c stays constant over the range, which is close enough for most solids and liquids over a few tens of degrees. Water at 4,186 J/kg·K is the reference value in most textbooks; its true value dips to about 4,179 near 35 °C and rises to about 4,220 at 0 °C and 4,216 at 100 °C, a spread of under 1%.
The calorie and the BTU were originally defined from water: one calorie raises one gram of water by one degree Celsius, and one BTU raises one pound by one degree Fahrenheit. Modern definitions fix them in joules instead. The calculator uses the thermochemical calorie (4.184 J) and the International Table BTU (1,055.056 J), which makes 1 BTU/lb·°F equal to 4,186.8 J/kg·K.
At a phase change the temperature stops rising while the energy goes into breaking bonds. Melting ice takes 334 kJ/kg and boiling water at atmospheric pressure takes 2,256 kJ/kg, more than five times the energy to heat the same water from 0 to 100 °C. Ignoring these terms is the most common way to get a heating calculation badly wrong.
Common use cases
Process heating and cooling
Estimate the energy to bring a batch of product, oil or water to temperature, or the heat a chiller must remove.
Thermal storage sizing
Find how much heat a water tank, a concrete slab or a bed of rock stores between two temperatures.
Lab and coursework
Check calorimetry problems, solve for an unknown specific heat, or work through a full ice-to-steam heating curve.
Heat-treatment and cooking
Work out the energy to heat a steel part for tempering or a large volume of liquid in a kitchen.
Tips and limitations
- A temperature change of 1 °C is the same as 1 K, but 1 °F is only 5/9 of that. Enter temperatures, not differences, and the calculator handles it.
- For gases, use cp for heating at constant pressure (open systems, air in a duct) and cv for a sealed rigid container. Air cp is about 1,005 J/kg·K.
- For a mixture or an alloy, a mass-weighted average of the component specific heats is a reasonable first estimate.
- Heat lost from the container and to the air is not included. For an equipment estimate, divide the result by an efficiency.
- Specific heat per kilogram favours light materials. Per unit volume, water (4.19 MJ/m³·K) and concrete (about 2.1 MJ/m³·K) store more than they seem to.
Frequently asked questions
What does Q = mcΔT mean?
Q is the heat energy, m the mass, c the specific heat capacity and ΔT the temperature change. Multiply mass by specific heat by the change in temperature to get the energy absorbed (positive) or released (negative).
What is the specific heat of water?
About 4,186 J/kg·K (4.186 kJ/kg·K, 1 BTU/lb·°F, 1 cal/g·°C) for liquid water near room temperature. Ice is about 2,090 and steam about 2,010 J/kg·K.
How much energy does it take to boil water?
To heat 1 kg from 20 °C to 100 °C takes 335 kJ. Turning that water into steam then takes another 2,256 kJ, the latent heat of vaporization at atmospheric pressure.
How do I find the specific heat from an experiment?
Rearrange to c = Q / (m x ΔT). If 8,970 J raises 1 kg of a metal by 10 K, c = 8,970 / 10 = 897 J/kg·K, which matches aluminium.
How do I calculate the final temperature after adding heat?
ΔT = Q / (m x c), then add ΔT to the starting temperature. 3,850 J into 1 kg of copper (c = 385) raises it by 10 K.
How many joules are in a BTU and a calorie?
One International Table BTU is 1,055.056 J. One thermochemical calorie is 4.184 J, and a food Calorie (kcal) is 4,184 J.
Is there an in-depth guide to the Specific Heat Calculator?
Yes. “Specific Heat and Latent Heat Explained: Q = mcΔT and the Heating Curve” covers the background, how it works, common pitfalls and worked examples. Read the full guide.
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