Specific Heat & Heat Energy Calculator
q = mcΔT for any variable, phase changes with latent heat, and calorimetry.
Heating curve
Stages
How it was worked out
Specific heats of common substances
| Substance | State | J/(kg·K) | J/(g·°C) | Conditions | Source |
|---|---|---|---|---|---|
| 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 |
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
- Choose q = mcΔT, Heating curve or Mixing.
- 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.
- 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.
- 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.
- 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
800 g of water from 21 °C to 85 °C, c = 4.184 J/(g·°C)
q = 214,221 J ≈ 2.1 × 10² kJ
OpenStax Chemistry 2e, Example 5.1.
348 g absorbs 6.64 kJ and warms from 22.4 °C to 43.6 °C
c = 0.900 J/(g·°C) — aluminium
Chemistry 2e, Example 5.2.
1.5 kg of water from 20 °C to 100 °C with a 2 kW element
q = 502 kJ; at least 4 min 11 s (no losses)
1 kg of ice at −20 °C to steam at 120 °C
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.
18 g of ice at 0 °C in 250 g of soda (as water) at 20 °C
Final temperature 13.3 °C
University Physics Vol. 2, Example 1.9.
775 g of water + 893 J/°C bomb, 23.8 °C → 35.6 °C
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 ΔTwith 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 changesT_final = Σ m c T_initial ÷ Σ m c. A calorimeter of heat capacity C addsC (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
- OpenStax, University Physics Volume 2 (CC BY 4.0): §1.4 Heat Transfer, Specific Heat, and Calorimetry (Eq. 1.5, Table 1.3, Examples 1.5 and 1.7) and §1.5 Phase Changes (Table 1.4, the −20 °C ice heating curve, Example 1.9).
- OpenStax, Chemistry 2e (CC BY 4.0): §5.1 Energy Basics (Table 5.1, Examples 5.1–5.2) and §5.2 Calorimetry (Examples 5.3–5.7).
- IAPWS, Industrial Formulation 1997 for the Thermodynamic Properties of Water and Steam (IAPWS-IF97), for the temperature dependence of water’s specific heat.
- NIST, SP 811 Appendix B.8: calorie (4.184 J), Btu (IT) and the Btu/(lb·°F) = 4.1868 kJ/(kg·K) relation.
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.