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Specific Heat & Heat Energy Calculator

q = mcΔT for any variable, phase changes with latent heat, and calorimetry.

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Heat to change the temperature of one substance.

Temperature
°C
°C
Heater power optional — for the time
The time assumes every joule goes into the substance; real heaters lose some.
Heat —

    How it was worked out

    Specific heats of common substances

    SubstanceStateJ/(kg·K)J/(g·°C)ConditionsSource
    Water, liquid liquid 4,184 4.184 25 °C Chemistry 2e
    Water, liquid liquid 4,186 4.186 15 °C Univ. Physics 2
    Ice solid 2,093 2.093 −10 °C Chemistry 2e
    Ice solid 2,090 2.090 average, −50 °C to 0 °C Univ. Physics 2
    Water vapour gas 1,864 1.864 25 °C Chemistry 2e
    Steam gas 2,020 2.020 100 °C, constant pressure, 1 atm Univ. Physics 2
    Aluminium solid 897 0.897 25 °C Chemistry 2e
    Copper solid 385 0.385 25 °C Chemistry 2e
    Iron solid 449 0.449 25 °C Chemistry 2e
    Iron or steel solid 452 0.452 25 °C Univ. Physics 2
    Gold solid 129 0.129 25 °C Chemistry 2e
    Silver solid 235 0.235 25 °C Univ. Physics 2
    Lead solid 130 0.130 25 °C Chemistry 2e
    Silicon solid 712 0.712 25 °C Chemistry 2e
    Glass solid 840 0.840 25 °C Univ. Physics 2
    Concrete or granite (average) solid 840 0.840 25 °C Univ. Physics 2
    Wood solid 1,700 1.700 25 °C Univ. Physics 2
    Human body (average) solid 3,500 3.500 37 °C Univ. Physics 2
    Ethanol liquid 2,376 2.376 25 °C Chemistry 2e
    Glycerin liquid 2,410 2.410 25 °C Univ. Physics 2
    Benzene liquid 1,740 1.740 25 °C Univ. Physics 2
    Mercury liquid 139 0.139 25 °C Univ. Physics 2
    Air (dry) gas 1,007 1.007 25 °C, 1 bar Chemistry 2e
    Nitrogen gas 1,040 1.040 25 °C, 1 bar Chemistry 2e
    Oxygen gas 918 0.918 25 °C, 1 bar Chemistry 2e
    Carbon dioxide gas 853 0.853 25 °C, 1 bar Chemistry 2e
    Argon gas 522 0.522 25 °C, 1 bar Chemistry 2e
    Helium gas 5,193 5.193 25 °C, 1 bar Chemistry 2e
    Ammonia gas 2,190 2.190 constant pressure, 1 atm Univ. Physics 2

    Next steps

    About the Specific Heat & Heat Energy Calculator

    The heat needed to warm something is q = m c ΔT: mass times specific heat times the temperature change. This calculator solves that formula for any one unknown — the heat, the mass, the specific heat or the temperature — in joules, calories, kilocalories, Btu or kilowatt-hours, with temperatures in °C, K or °F, and gives the time a heater of a given power would take. A table of about thirty sourced specific heats (water, ice, steam, metals, glass, wood, liquids and gases) fills in c for you.

    Two more modes go beyond the simple formula. Heating curve adds the latent heat of melting and boiling, so it can take ice at −20 °C all the way to steam at 120 °C stage by stage, with the classic temperature–heat graph. Mixing finds the final temperature when substances at different temperatures meet in a calorimeter — including ice that melts, or steam that condenses — or works backwards from a measured final temperature to an unknown specific heat, a starting temperature or the heat released by a reaction.

    How to use it

    1. Choose q = mcΔT, Heating curve or Mixing.
    2. For q = mcΔT, pick what to solve for, choose a material (or type c) and enter the other values. Use “Change ΔT” if you only know the temperature difference. Add a heater power to see how long it takes.
    3. For a heating curve, enter the mass and the start and end temperatures. Water’s properties are filled in; open Property values to change them or to use another substance.
    4. For mixing, add each substance with its mass and temperature (and the calorimeter’s heat capacity if it matters). To work backwards, choose what is unknown and type the measured final temperature.
    5. Read the answer, the stages or the heat flows, and the full working; copy the result or download the heating-curve stages as CSV.

    Examples

    Heating water
    Input
    800 g of water from 21 °C to 85 °C, c = 4.184 J/(g·°C)
    Result
    q = 214,221 J ≈ 2.1 × 10² kJ

    OpenStax Chemistry 2e, Example 5.1.

    Identify a metal
    Input
    348 g absorbs 6.64 kJ and warms from 22.4 °C to 43.6 °C
    Result
    c = 0.900 J/(g·°C) — aluminium

    Chemistry 2e, Example 5.2.

    Kettle
    Input
    1.5 kg of water from 20 °C to 100 °C with a 2 kW element
    Result
    q = 502 kJ; at least 4 min 11 s (no losses)
    Ice to steam
    Input
    1 kg of ice at −20 °C to steam at 120 °C
    Result
    41.8 kJ + 334 kJ + 418.6 kJ + 2,256 kJ + 40.4 kJ = 3.09 MJ

    OpenStax University Physics Vol. 2 §1.5 values.

    Ice in a drink
    Input
    18 g of ice at 0 °C in 250 g of soda (as water) at 20 °C
    Result
    Final temperature 13.3 °C

    University Physics Vol. 2, Example 1.9.

    Bomb calorimeter
    Input
    775 g of water + 893 J/°C bomb, 23.8 °C → 35.6 °C
    Result
    The reaction released 48.8 kJ

    Chemistry 2e, Example 5.7.

    Common uses

    • Chemistry and physics homework on specific heat, latent heat and calorimetry, with every substitution shown.
    • Lab work: finding the specific heat of an unknown metal from a calorimetry experiment.
    • Kitchen and home: the energy and time to heat water in a kettle, geyser or pot.
    • Estimating the heat to melt ice or raise steam for a process.
    • Checking coffee-cup and bomb-calorimeter results for reaction heats.

    The formulas

    • Temperature change: q = m c ΔT with ΔT = T_final − T_initial; a temperature fall makes q negative (heat given out). A change of 1 K equals 1 °C; 1 °F is 5/9 K.
    • Phase change: q = m L, with L_f (fusion) for melting or freezing and L_v (vaporisation) for boiling or condensing; the temperature stays put while it happens.
    • Mixing: with no heat lost, Σ m c (T_final − T_initial) = 0, so without phase changes T_final = Σ m c T_initial ÷ Σ m c. A calorimeter of heat capacity C adds C (T_final − T_cal). Ice, water and steam are handled through their total energy (enthalpy), so the answer can be “0 °C with some ice left”.
    • Heater time: t = q ÷ P, ignoring heat lost to the surroundings and the heater itself — real kettles take longer.

    Water’s heating curve

    At normal pressure, heating 1 kg of ice from −20 °C to steam at 120 °C takes (OpenStax University Physics Vol. 2, Tables 1.3 and 1.4):

    • warm the ice: 1 kg × 2,090 J/(kg·K) × 20 K = 41.8 kJ
    • melt it at 0 °C: 1 kg × 334 kJ/kg = 334 kJ
    • warm the water: 1 kg × 4,186 J/(kg·K) × 100 K = 418.6 kJ
    • boil it at 100 °C: 1 kg × 2,256 kJ/kg = 2,256 kJ
    • warm the steam: 1 kg × 2,020 J/(kg·K) × 20 K = 40.4 kJ

    The total is 3.09 MJ, and 83.8 % of it goes into melting and boiling. These are textbook constants: the specific heat of water actually drifts between about 4.18 and 4.22 kJ/(kg·K) from 0 to 100 °C (IAPWS-IF97), and other references give slightly different latent heats, so the property values can be edited.

    Where the specific heats come from

    • OpenStax Chemistry 2e, Table 5.1 — “Specific Heats of Common Substances at 25 °C and 1 bar”: helium, water, ethanol, ice (at −10 °C), water vapour, nitrogen, air, oxygen, aluminium, carbon dioxide, argon, iron, copper, lead, gold and silicon.
    • OpenStax University Physics Vol. 2, Table 1.3 — “Specific Heats of Various Substances”: water at 15 °C, ice averaged over −50 °C to 0 °C, steam at 100 °C (at constant pressure), iron or steel, silver, glass, concrete or granite, wood, the human body, benzene, glycerin, mercury and ammonia.

    For gases the values are at constant pressure, which is what applies when a gas is heated in the open. Specific heats change with temperature, so treat them as averages near the stated conditions.

    Sources

    Limitations

    • Specific heats are treated as constant over the temperature range; for wide ranges or high accuracy, use temperature-dependent data.
    • Latent heats and transition temperatures are for normal atmospheric pressure (1 atm); water boils at lower temperatures at altitude.
    • Mixing assumes a perfectly insulated system: no heat to or from the surroundings, no evaporation, no chemical reaction (except in the reaction-heat solve).
    • Only water, ice and steam change phase in the mixing mode; other substances keep their specific heat (a warning appears if they would cross a melting or boiling point that is known).
    • Heater times are minimums: real appliances lose heat and take longer.

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    Frequently asked questions

    What is the formula for heat energy?

    q = m × c × ΔT. For 800 g of water warmed from 21 °C to 85 °C with c = 4.184 J/(g·°C): q = 800 × 4.184 × 64 = 214,221 J, about 214 kJ.

    How much energy does it take to boil water?

    To bring 1.5 kg (1.5 L) of water from 20 °C to 100 °C takes 1.5 × 4,184 × 80 = 502 kJ — at least 4 minutes 11 seconds in a 2 kW kettle. Boiling it all away takes another 1.5 × 2,256 = 3,384 kJ.

    What is the specific heat of water?

    4.184 J/(g·°C) at 25 °C (OpenStax Chemistry 2e), or 4,186 J/(kg·°C) at 15 °C (OpenStax University Physics) — about 1 calorie per gram per degree. Ice is about 2.09 J/(g·°C) and steam about 2.02 J/(g·°C).

    What is latent heat?

    The heat to change phase without changing temperature: q = mL. For water at 1 atm, melting takes 334 kJ/kg and boiling 2,256 kJ/kg — boiling 1 kg of water away takes more than five times the heat needed to bring it from 0 °C to 100 °C.

    How do I find the final temperature when two things are mixed?

    Heat lost by the hot one equals heat gained by the cold one, so without phase changes T_final = (m₁c₁T₁ + m₂c₂T₂) ÷ (m₁c₁ + m₂c₂). A 0.500 kg aluminium pan at 150 °C with 0.250 kg of water at 20 °C ends at 59.0 °C with the table’s 897 J/(kg·K) for aluminium and 4,186 J/(kg·K) for water; OpenStax University Physics Vol. 2 Example 1.7 gets 59.1 °C with 900 J/(kg·°C) for aluminium.

    How do I identify a metal from a calorimetry experiment?

    Measure the final temperature, then solve the energy balance for the metal’s c. In OpenStax Chemistry 2e Example 5.4, 59.7 g of metal from boiling water warms 60.0 g of water from 22.0 °C to 28.5 °C, giving c = 0.38 J/(g·°C), close to copper (0.385).

    Why can’t I use q = mcΔT across 0 °C or 100 °C for water?

    Because melting and boiling absorb latent heat without any temperature change, which q = mcΔT leaves out. Use the heating-curve mode: it adds q = mL at each phase change.

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