Specific Heat Calculator (Q = mcΔT)
Find the heat needed to warm something, or solve Q = mcΔT for mass, specific heat or temperature change. Mix two substances to find the final temperature, and take water through melting and boiling with latent heat.
How to Use
- Pick what to find: heat Q, mass, specific heat or temperature change; Mix for two substances meeting at one temperature; or Phase change for ice, water and steam.
- Choose a material to fill in its specific heat, or type your own in J/(kg·K), J/(g·K), cal/(g·°C) or BTU/(lb·°F).
- Enter the known values with their units. Add a start temperature to get the final temperature as well.
- Read the answer, the graph of temperature against heat added, and the table with the heat in kJ, kcal, BTU and Wh.
- Press a preset for a ready-made example, and check Show Work for each step and conversion.
Worked Example
Boiling a kettle. 1 kg of water from 20 °C to 100 °C: ΔT = 80 K (a change of 1 °C is the same as 1 K). Q = m × c × ΔT = 1 × 4,186 × 80 = 334,880 J = 334.88 kJ. A 2 kW kettle supplies 2,000 J each second, so it needs at least 167.44 s, just under 3 minutes.
Ice to steam. 1 kg of ice at −20 °C becomes steam at 120 °C in five stages: warm the ice 1 × 2,090 × 20 = 41,800 J; melt it 1 × 334,000 = 334,000 J; warm the water 1 × 4,186 × 100 = 418,600 J; boil it 1 × 2,257,000 = 2,257,000 J; heat the steam 1 × 2,010 × 20 = 40,200 J. Total 3,091,600 J, and boiling alone is 73% of it.
The common mistake: averaging the two temperatures. Drop 200 g of copper at 100 °C into 500 g of water at 20 °C and the answer is not (100 + 20) ÷ 2 = 60 °C. Weight each temperature by its heat capacity m × c: copper 0.2 × 385 = 77 J/K, water 0.5 × 4,186 = 2,093 J/K, so T_f = (77 × 100 + 2,093 × 20) ÷ 2,170 = 22.84 °C. The water barely changes.
Show Work
Formulas
Joseph Black and the Calorie
The Scottish chemist Joseph Black, working in Glasgow around 1760, was the first to separate the quantity of heat from temperature. He noticed that ice at its melting point absorbs a great deal of heat without getting any warmer, which he called latent heat, and that equal masses of different substances need different amounts of heat for the same rise, the idea of specific heat. James Prescott Joule’s paddle-wheel experiments in the 1840s then showed that heat is a form of energy, with a fixed exchange rate between mechanical work and heat.
The calorie was defined as the heat to warm 1 g of water by 1 °C; because that depends on the starting temperature, several versions existed (the 15 °C calorie is 4.1855 J). The thermochemical calorie is now fixed at exactly 4.184 J, and the “Calories” on food labels are kilocalories. Values used here are typical figures near room temperature from the CRC Handbook of Chemistry and Physics; ice (2,090) and steam (2,010) are textbook values, and the latent heats are for water at 1 atm.
About This Calculator
This calculator solves Q = mcΔT for any of heat, mass, specific heat or temperature change, with a start temperature giving the final one. Mix two finds where two substances settle when they share heat, and Phase change takes water through melting and boiling with the latent heats, drawing the full heating curve. It refuses temperatures below absolute zero and warns when liquid water would freeze or boil. Every value has its own unit, from joules to kilowatt-hours and BTU, and from °C to °F and kelvin.
Everything runs in your browser; nothing you enter is sent anywhere.
Related tools: BTU Calculator, Water Heater Calculator, and Unit Converter.
Frequently Asked Questions
What is the formula for specific heat?
Q = m × c × ΔT: heat equals mass times specific heat times the temperature change. Warming 1 kg of water (c = 4,186 J/(kg·K)) from 20 °C to 100 °C takes 1 × 4,186 × 80 = 334,880 J, or 334.88 kJ. A 1 kW heater with no losses would take 334.88 s, about 5.58 minutes.
Why is water’s specific heat given as 4,186 or 4,184?
Water’s specific heat changes slightly with temperature: about 4,186 J/(kg·K) near 15 °C (the 15 °C calorie is 4.1855 J) and 4,181 at 25 °C. 4,184 is exactly one thermochemical calorie per gram per degree, the calorie used on food labels. This calculator uses 4,186; the three differ by about 0.1%.
Why does water take so long to heat up?
Its specific heat is very high. 10 kJ warms 1 kg of water by only 2.39 K, but 1 kg of aluminium by 11.15 K and 1 kg of copper by 25.97 K. That is why seas warm slowly and water is used to carry heat in radiators.
How do I find the final temperature when I mix two things?
Heat lost by the hot one equals heat gained by the cold one, so T_f = (m₁c₁T₁ + m₂c₂T₂) ÷ (m₁c₁ + m₂c₂). 200 g of copper at 100 °C dropped into 500 g of water at 20 °C gives (77 × 100 + 2,093 × 20) ÷ 2,170 = 22.84 °C, assuming no heat escapes.
Why does a steam burn hurt more than boiling water?
Steam releases its latent heat of vaporisation, 2,257 kJ/kg, as it condenses on the skin. 10 g of water cooling from 100 °C to 37 °C gives up 2.64 kJ; 10 g of steam gives up 22.57 kJ condensing first, 25.21 kJ in all, about 9.6 times as much.
How do I use the Specific Heat Calculator (Q = mcΔT)?
Just type your numbers. The answer shows up right away — there is no button to press. Change anything and it updates by itself.
Do I need to install or sign up for anything?
Not at all — it runs in the browser with nothing to install and no account. After it loads once, it even works without an internet connection.
Is my information private?
Yes. Everything happens in your browser. Nothing you type is sent to a server or saved anywhere.
Common Use Cases
Kettles and hot water
A 2 kW kettle brings 1 kg of water from 20 °C to the boil in 334.88 kJ ÷ 2 kW = 2.79 minutes, ignoring losses.
Home energy
A 150 L bath heated from 15 °C to 40 °C needs 150 × 4,186 × 25 = 15.7 MJ, or 4.36 kWh. Turned round, 1 kWh warms 21.5 kg of water by 40 K.
Cooking and metalwork
A 1.5 kg aluminium pan heated from 20 °C to 180 °C absorbs 1.5 × 897 × 160 = 215.28 kJ before any food goes in.
Identifying a metal
A 500 g block that takes 9,625 J to warm by 50 °C has c = 9,625 ÷ (0.5 × 50) = 385 J/(kg·K), which is copper.
Ice and steam
Taking 1 kg of ice at −20 °C to steam at 120 °C needs 3,091.6 kJ, and 73% of that is boiling. Melting 1 kg of ice (334 kJ) takes about as much as heating the water from 0 °C to 80 °C (334.88 kJ).
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