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Specific Heat Calculator

Free specific heat calculator — solve Q = mcΔT for heat, mass, c, or temperature change. SI units, worked water example, NIST-cited cal/J conversion.

Specific Heat Calculator

Solve for
Mass of the substance being heated or cooled, in kilograms. 1 kg = 1000 g = 2.2046 lb. Required unless solveFor = mass.
kg
Specific heat capacity of the substance in joules per kilogram per kelvin. Liquid water = 4186, ice = 2090, steam = 2010, aluminium = 897, copper = 385, iron = 449, gold = 129, dry air = 1005, ethanol = 2440. Required unless solveFor = specificHeat.
J/(kg·K)
Temperature change in kelvin (or equivalently degrees Celsius — a 1 K change equals a 1 °C change). Use a positive value when heating, negative when cooling. Required unless solveFor = deltaT.
K
Heat energy transferred, in joules. 1 kJ = 1000 J; 1 calorie (IT) = 4.184 J; 1 food Calorie (kcal) = 4184 J; 1 BTU ≈ 1055 J. Required when solving for mass, specific heat, or ΔT.
J
Heat energy (Q)
41,860
Heat energy transferred Q, computed as m·c·ΔT when solving for heat and echoed back otherwise. Reported in joules; a positive value means energy entering the substance (heating), a negative value means energy leaving (cooling).
Heat energy (kJ)
41.86 kJ
Heat energy (calories)
10,004.7801 cal
Mass
1 kg
Specific heat capacity (c)
4,186 J/(kg·K)
Temperature change (ΔT)
10 K

Background.

This specific heat calculator solves Q = m·c·ΔT — the single equation that links the energy you put into a substance to the temperature rise you observe — for whichever of the four quantities you do not already know. Pick a 'solve for' mode (heat energy, mass, specific heat capacity, or temperature change), enter the three known values in SI units, and the tool returns the missing quantity together with energy unit conversions into kilojoules and IT calories.

The equation Q = m·c·ΔT is the working equation of calorimetry, the experimental technique by which most of nineteenth-century thermal physics was built. James Prescott Joule's paddle-wheel experiments of 1843–1845, published in his landmark Philosophical Magazine paper 'On the Mechanical Equivalent of Heat' (1845, volume 27, page 205), measured exactly the constant of proportionality between mechanical work and the temperature rise of water in a stirred bath. Joule's number — 772 foot-pounds per BTU, equivalent to 4.158 joules per IT calorie in modern units — was the first quantitative bridge between mechanics and heat, the empirical foundation of the first law of thermodynamics, and the reason the SI unit of energy is named after him. The modern value of the IT calorie, fixed by NIST in Special Publication 811 as exactly 1 cal = 4.184 J, is essentially Joule's measurement refined by a hundred and fifty years of better thermometry and adopted by international convention so that engineering tables and chemistry textbooks could finally agree on a single number.

The 'specific' in specific heat refers to the per-unit-mass normalisation: total heat capacity C (capital) is an extensive property that grows with the size of the sample, while specific heat capacity c (lowercase) is the intensive material property C/m that depends only on what the substance is, not how much of it you have. Two textbooks of the same material have the same c but twice the C; a cup of water and an Olympic swimming pool both have c ≈ 4186 J/(kg·K), but the pool has many millions of times the total heat capacity.

The numerical value of water's specific heat — 4186 J/(kg·K), or equivalently 1.000 cal/(g·°C) by historical definition — is anomalously high compared to almost every other liquid or solid on the periodic table. Mercury sits at 140, ethanol at 2440, aluminium at 897, copper at 385, iron at 449, gold at 129, dry air at 1005. Water's value is roughly three times that of typical organic liquids, ten times that of typical metals, and thirty times that of mercury. The physical reason is hydrogen bonding: each water molecule donates and accepts hydrogen bonds with its neighbours, and a substantial fraction of any added thermal energy goes into stretching and breaking those bonds rather than into translational kinetic energy that the thermometer can read. The consequence — for climate, biology, cooking, and engineering — is that water resists temperature change far more strongly than its mass would suggest.

Coastal climates are mild because the ocean buffers temperature swings the land cannot. Mammalian thermoregulation works because blood (which is mostly water) carries enormous amounts of heat without overheating the tissues it passes through. Boiling pasta takes longer than frying it because every gram of water you put in the pot is a gram you have to push up sixty or seventy degrees at 4186 joules per kelvin per kilogram. Steam radiators heat buildings efficiently because the steam carries not only sensible heat but also the enormous latent heat of vaporisation that it releases when condensing. Calorimetry — the experimental measurement of heats of reaction, heats of dissolution, and heat capacities — exploits exactly this fact: a known mass of water, with its precisely known c, makes an excellent thermal 'ruler' for absorbing the heat released or consumed by a chemical or physical process. You drop a measured-mass sample of unknown specific heat into a water bath, you read the equilibrium temperature change of the water, you solve Q = m_water · c_water · ΔT_water for the heat absorbed, and by conservation that equals the heat lost by the sample, which lets you back out c_sample. The same arithmetic underlies bomb calorimetry for enthalpies of combustion, differential scanning calorimetry for phase transitions, and the calibration of every heat capacity number in the CRC Handbook.

Below the widget you will find the derivation of Q = m·c·ΔT from first principles, the precise definitions of heat capacity versus specific heat versus molar specific heat, why c varies with temperature (and how much it matters), why the equation breaks down through a phase change (where latent heat takes over), the cal–joule conversion arithmetic and which calorie definition to use, and a discussion of the Dulong–Petit law that predicts c for metals from first principles. The solver beneath the calculator is registered at specificHeat.calculate in the Quanta engine and routes through pure functions that throw on division-by-zero edge cases.

What is specific heat calculator?

Specific heat capacity is the amount of heat energy required to raise the temperature of one kilogram of a substance by one kelvin. Its symbol is the lowercase c, its SI units are joules per kilogram per kelvin J/(kg·K), and its operational definition is the constant of proportionality in the equation Q = m·c·ΔT, where Q is heat energy in joules, m is mass in kilograms, and ΔT is the temperature change in kelvin (equivalently in degrees Celsius, since a kelvin interval and a Celsius interval are identical in magnitude). Specific heat is an intensive material property: it depends on what the substance is and on the conditions (pressure for gases, phase for any substance) but not on how much of the substance you have. Pure liquid water at 25 °C and atmospheric pressure has c = 4186 J/(kg·K) — equivalently 4.186 kJ/(kg·K) or 1.000 cal/(g·°C) by the historical definition that one calorie raises one gram of water by one Celsius degree. Other common values: ice 2090, steam 2010, ethanol 2440, mercury 140, aluminium 897, copper 385, iron 449, gold 129, lead 130, glass ≈ 840, wood ≈ 1700, dry air at constant pressure 1005, dry air at constant volume 718. Distinguish three closely related quantities. Total heat capacity C (capital) is c × m and has units of J/K — it is the extensive version of c and tells you how much energy the specific sample in front of you needs per kelvin. Molar specific heat capacity is c × M (where M is the molar mass) and has units of J/(mol·K); the Dulong–Petit law states that for most monatomic solids at room temperature, molar specific heat is approximately 3R ≈ 25 J/(mol·K) regardless of the element. Gases have two distinct specific heats — c_p at constant pressure and c_v at constant volume — that differ by R (the universal gas constant) per mole because heating at constant pressure does work pushing back the surroundings while heating at constant volume does not. The Q = m·c·ΔT equation in this calculator handles only sensible heat — energy that changes the temperature of a single phase. It does not handle latent heat — the energy absorbed or released at constant temperature during a phase change (melting, boiling, sublimation). To melt ice without warming the meltwater you supply 334 kJ/kg of latent heat of fusion at 0 °C; to boil water without superheating the steam you supply 2257 kJ/kg of latent heat of vaporisation at 100 °C. Those calculations live outside this widget; pair Q = m·c·ΔT with a separate latent-heat term when a process crosses a phase boundary.

How to use this calculator.

  1. Pick the quantity you want to compute from the 'Solve for' menu — heat energy Q, mass m, specific heat c, or temperature change ΔT. The calculator computes that quantity from the other three.
  2. Enter the three known values in SI units. Mass in kilograms (1 kg = 1000 g), specific heat in J/(kg·K), temperature change in kelvin or Celsius degrees (they are interchangeable for ΔT), heat in joules (1 kJ = 1000 J, 1 cal = 4.184 J).
  3. Convert your inputs before entering them if your data is in other units. For mass: 1 g = 0.001 kg, 1 lb = 0.4536 kg. For specific heat: cal/(g·°C) × 4184 = J/(kg·K); BTU/(lb·°F) × 4186.8 = J/(kg·K). For heat: 1 cal = 4.184 J, 1 kcal (food Calorie) = 4184 J, 1 BTU = 1055.06 J, 1 kWh = 3.6 × 10⁶ J.
  4. Use a positive ΔT when the substance is being heated and a negative ΔT when it is being cooled. The sign of the computed Q follows: positive Q means energy entering the substance, negative Q means energy leaving it.
  5. Look up c in a reference table when the substance is not water. The Engineering Toolbox, the CRC Handbook of Chemistry and Physics (Table 2-170), and NIST WebBook all tabulate specific heats; built-in defaults in the input hint cover the dozen most common materials.
  6. Do not use this calculator across a phase change. If your process melts, boils, sublimates, condenses, or freezes anything, split it into separate sensible-heat steps (Q = mcΔT inside each phase) plus latent-heat steps (Q = mL at the phase boundary).
  7. Read the three energy outputs — joules, kilojoules, and IT calories — to cross-check against textbook tables that may use any of the three. The cal value uses the NIST exact conversion 1 cal = 4.184 J.

The formula.

Q = m × c × ΔT

The equation Q = m·c·ΔT is the operational definition of specific heat capacity rearranged to its most useful form. The four explicit rearrangements are:

Q = m × c × ΔT (solve for heat) m = Q / (c × ΔT) (solve for mass) c = Q / (m × ΔT) (solve for specific heat) ΔT = Q / (m × c) (solve for temperature change)

The calculator implements all four and routes to the correct one based on your 'solve for' selection. Each rearrangement is a single algebraic step; the only constraint is that the denominator must be non-zero, which the calculator enforces by throwing an InvalidInputError when m, c, or ΔT is zero in a position where it would divide.

The physical content of the equation is the first law of thermodynamics applied to a single phase of a single substance with no work done other than the thermal energy transfer itself. When you add heat Q to a sample, its internal energy rises by Q (no work, no phase change); the temperature rise per joule of internal energy added is set by the molecular degrees of freedom available to absorb the energy — translation, rotation, vibration, hydrogen-bond stretching for water — and that ratio is exactly what c measures. The specific heat at constant pressure c_p that this calculator uses for solids and liquids differs negligibly from the specific heat at constant volume c_v because solids and liquids barely expand on heating; for gases the distinction matters and you must specify which c you mean (c_p − c_v = R for an ideal gas, per mole).

The IT calorie conversion factor built into the calculator's heat output is exactly 1 cal = 4.184 J, as defined by NIST Special Publication 811 (Guide for the Use of the International System of Units, Appendix B). This is the 'International Table' calorie used in steam tables and modern engineering; the older 'thermochemical calorie' (1 cal_th = 4.184 J also by adoption) and the '15 °C calorie' (1 cal_15 = 4.1858 J) survive in some chemistry literature but the IT value is the modern standard. Food labels report energy in kilocalories (kcal), often misleadingly called 'Calories' with a capital C: 1 food Calorie = 1 kcal = 4184 J. A 2000 Calorie/day diet is therefore 8.37 megajoules per day.

The equation breaks down in two regimes you must watch for. First, across a phase change. When water reaches 100 °C and starts boiling, every joule you add goes into vaporising water at constant temperature, not into warming what is still liquid; the heat absorbed per kilogram is the latent heat of vaporisation L_v = 2257 kJ/kg, computed by Q = m·L not Q = m·c·ΔT. The same is true at melting (latent heat of fusion L_f = 334 kJ/kg for water at 0 °C). Second, when c varies significantly with temperature. The values tabulated in handbooks are typically reported at 25 °C, and for water c rises by only about 0.6% between 25 °C and 100 °C — negligible for almost any practical calculation. For metals and many solids c falls toward zero as T approaches absolute zero (Debye T³ law); the constant-c approximation is excellent for engineering work above about 100 K and breaks down for cryogenic problems.

A worked example.

Example

The canonical introductory-physics question: how much heat does it take to warm one kilogram of liquid water by ten degrees Celsius? Pick solveFor = 'heat', enter mass = 1 kg, specific heat = 4186 J/(kg·K) (the standard tabulated value for liquid water at room temperature), and ΔT = 10 K (a 10 °C rise — kelvin and Celsius differences are identical). The calculator computes Q = m·c·ΔT = 1 × 4186 × 10 = 41,860 J, which it also reports as 41.86 kJ and 10,005 cal (equivalently about 10.0 kilocalories, or roughly the food energy of a single Brazil nut). Reading the result back: warming a litre of water from room temperature (20 °C) to 30 °C requires about 42 kilojoules. For comparison, warming the same kilogram of aluminium by the same 10 °C requires only 1 × 897 × 10 = 8,970 J — about a fifth as much, because aluminium's specific heat is about a fifth of water's. Warming a kilogram of copper takes 3,850 J, about a tenth as much. This is why a copper pot heats up so much faster than the water in it, and why pouring boiling water onto a cold ceramic mug barely cools the water at all (the mug is light and ceramic's c is low; the water has to dump enough heat to warm the mug but most of its thermal energy is still 'inside' the water). The same arithmetic, run in reverse, tells you the temperature rise from a known heat input: a 1500-watt kettle delivers 1500 J/s, so 41,860 J takes 27.9 seconds — provided no heat leaks to the surroundings, which is why real kettles take longer. Switching solveFor = 'specificHeat' with Q = 41,860 J, m = 1 kg, ΔT = 10 K returns c = 4,186 J/(kg·K), demonstrating the round-trip identity the equation satisfies by construction.

heat0
mass1
specific Heat4,186
delta T10
solve Forheat

Frequently asked questions.

What is the difference between specific heat capacity and heat capacity?
Heat capacity (capital C) is the total thermal energy a specific sample needs per kelvin of temperature rise, with units J/K. It is extensive — it grows with the size of the sample. Specific heat capacity (lowercase c) is heat capacity per unit mass, c = C/m, with units J/(kg·K). It is intensive — it depends only on what the substance is, not how much. Two kilograms of water have twice the heat capacity of one kilogram, but the same specific heat capacity. Q = m·c·ΔT in this calculator uses the specific (per-mass) form because tabulated values in handbooks are reported per kilogram or per gram. Multiply c by your sample mass to get the heat capacity C, then C·ΔT gives the energy directly — the two forms are algebraically equivalent.
Why does water have such a high specific heat capacity?
Water's c = 4186 J/(kg·K) is roughly three times that of typical organic liquids, ten times that of typical metals, and thirty times that of mercury. The physical reason is hydrogen bonding. Each water molecule donates two hydrogen bonds and accepts two more from its neighbours, knitting the liquid into a transient three-dimensional network. When you add thermal energy, a substantial fraction goes into stretching, bending, and partially breaking those hydrogen bonds rather than into translational kinetic energy that the thermometer reads as temperature. Mercury, by contrast, is a monatomic metal with only translational and electronic degrees of freedom and no inter-atomic bonds to absorb energy — so its c is tiny (140 J/(kg·K)). The same hydrogen-bond network explains water's anomalously high boiling point, surface tension, heat of vaporisation, and the density maximum at 4 °C.
Why does water resist temperature change so strongly?
Because of that high c. To raise one kilogram of water by one kelvin you must inject 4186 joules; to raise one kilogram of aluminium by the same kelvin you need only 897 joules; for copper, 385 joules; for mercury, 140 joules. The planet's oceans hold roughly 1.4 × 10²¹ kg of water, which means about 5.9 × 10²⁴ J/K of heat capacity — enough that even a one-degree rise in average ocean temperature represents trillions of times the world's annual energy production. This is why coastal climates are milder than continental ones (the ocean buffers seasonal temperature swings), why hot springs cool slowly, why mammalian thermoregulation works (blood is mostly water and carries large amounts of heat without overheating tissues), and why boiling pasta takes so much longer than frying it. Water is, by mass, the best practical thermal ballast in nature.
Does Q = m·c·ΔT work during a phase change like melting or boiling?
No. The equation handles only sensible heat — energy that changes the temperature of a single phase. During a phase change the temperature stays constant while large amounts of energy are absorbed or released, so ΔT = 0 and the equation collapses. Use latent heat instead: Q = m·L where L is the specific latent heat of the phase transition. For water at standard pressure, the latent heat of fusion (melting/freezing at 0 °C) is L_f = 334 kJ/kg, and the latent heat of vaporisation (boiling/condensing at 100 °C) is L_v = 2257 kJ/kg. To heat 1 kg of ice at −10 °C into steam at 110 °C you need four sequential calculations: warm the ice (Q = 1 × 2090 × 10), melt it (Q = 1 × 334,000), warm the water to 100 °C (Q = 1 × 4186 × 100), boil it (Q = 1 × 2,257,000), warm the steam (Q = 1 × 2010 × 10). Sum: about 3.05 megajoules, dominated by the vaporisation step.
How do I convert between J/(kg·K) and cal/(g·°C)?
Multiply cal/(g·°C) by exactly 4184 to obtain J/(kg·K), or divide J/(kg·K) by 4184 to go the other way. The factor decomposes into two pieces: 1 cal = 4.184 J (the IT calorie, fixed by NIST), and 1 g·°C = 0.001 kg·K (1 g is 0.001 kg and 1 °C is 1 K for differences). So 1 cal/(g·°C) = 4.184 J / (0.001 kg × 1 K) = 4184 J/(kg·K). Water's specific heat is 4186 J/(kg·K) = 1.000 cal/(g·°C) — the historical definition, which is why the calorie was originally named to give water exactly unit specific heat. For BTU/(lb·°F), the engineering unit common in US HVAC work, multiply by 4186.8 to get J/(kg·K); water is 1.000 BTU/(lb·°F) by the same historical coincidence.
What is molar specific heat capacity and when do I use it?
Molar specific heat (also called molar heat capacity) is heat capacity per mole rather than per kilogram, with units J/(mol·K). It equals c × M where M is the molar mass in kg/mol. For copper, M = 0.0635 kg/mol and c = 385 J/(kg·K), so molar specific heat = 24.4 J/(mol·K) — almost exactly 3R, the prediction of the Dulong–Petit law for monatomic solids. The molar form is the natural one when you care about chemistry (where stoichiometry is in moles) or fundamental physics (where the Dulong–Petit law, Debye model, and equipartition theorem all predict values per mole). The mass-specific form this calculator uses is the natural one for engineering (where you weigh materials in kilograms, not count molecules). Convert with multiplication by M; both forms describe the same physical quantity.
What is the Dulong–Petit law and does this calculator use it?
The Dulong–Petit law, formulated by Pierre Dulong and Alexis Petit in 1819, states that the molar specific heat of most monatomic solids at room temperature is approximately 3R ≈ 25 J/(mol·K), independent of the element. It works because at room temperature each atom in a crystalline solid contributes six classical degrees of freedom (three kinetic, three potential from oscillation against its neighbours) and each contributes ½kB of energy per kelvin via equipartition, giving 3kB per atom or 3R per mole. The law explains why iron (M = 0.0558 kg/mol, c × M = 25.1 J/(mol·K)), copper (24.4), aluminium (24.2), gold (25.4), and lead (26.9) all cluster around 25 J/(mol·K) despite vastly different masses per mole. This calculator does not invoke Dulong–Petit; it takes c as a user input. But you can use Dulong–Petit as a sanity check: if you are looking up c for an unfamiliar metal, expect molar specific heat near 25 J/(mol·K) and divide by molar mass to estimate c.
What is the difference between c_p and c_v for gases?
For a gas, the specific heat at constant pressure (c_p) is larger than the specific heat at constant volume (c_v) because heating at constant pressure does mechanical work pushing back the surroundings while heating at constant volume does not. For an ideal gas the molar difference is exactly R: c_p − c_v = R = 8.314 J/(mol·K). For dry air at room temperature, c_p ≈ 1005 J/(kg·K) and c_v ≈ 718 J/(kg·K), giving the ratio γ = c_p/c_v ≈ 1.40 that appears in adiabatic-process calculations and the speed of sound. For solids and liquids the distinction barely matters because thermal expansion is tiny (a one-kelvin rise in water expands its volume by only about 0.02%); the c_p and c_v values differ by less than a percent and tables almost always tabulate c_p. This calculator does not distinguish c_p from c_v; for gas problems, use c_p when the process is at constant pressure (a balloon expanding in open air) and c_v when it is at constant volume (a sealed rigid container).
Does specific heat capacity change with temperature?
Yes, but for most practical engineering calculations the variation is small enough to ignore. For liquid water, c rises from 4178 J/(kg·K) at 35 °C to 4218 J/(kg·K) at 100 °C — about a 1% variation, negligible for kitchen and HVAC arithmetic. For solids at moderate temperatures the Debye model predicts c ∝ T³ as T → 0 K (cryogenic regime) and asymptotes to the classical 3R per mole at high temperature (the Dulong–Petit limit). For a 25 °C calculation you can treat c as constant; for cryogenic problems below about 100 K or for thousand-degree problems where atomic vibrations approach their high-temperature limit you must integrate c(T) over the temperature range: Q = m·∫c(T)dT from T₁ to T₂. The NIST WebBook tabulates c(T) for water and dozens of other substances if you need the temperature-dependent value.
How does calorimetry use Q = m·c·ΔT to measure unknown specific heats?
Drop a sample of unknown specific heat c_s and known mass m_s, pre-heated to a known temperature T_s, into a calorimeter containing a known mass m_w of water at known initial temperature T_w. Wait for thermal equilibrium and measure the final temperature T_f. By conservation of energy, the heat lost by the sample equals the heat gained by the water (plus the calorimeter, which has its own heat capacity you measure separately): m_s · c_s · (T_s − T_f) = m_w · c_w · (T_f − T_w). Solve for c_s = [m_w · c_w · (T_f − T_w)] / [m_s · (T_s − T_f)]. This 'method of mixtures' is the oldest calorimetric technique, used by Joseph Black in the 1760s to measure the first heat capacities and the first latent heats, and it still works in undergraduate physics labs today. More accurate variants — adiabatic calorimetry, differential scanning calorimetry, bomb calorimetry for combustion enthalpies — wrap better thermal isolation and better temperature measurement around the same underlying Q = m·c·ΔT arithmetic.

References& sources.

  1. [1]Çengel, Y. A., Boles, M. A., & Kanoğlu, M. (2019). Thermodynamics: An Engineering Approach, 9th ed., §4-1, 'Specific Heats'. McGraw-Hill Education. The canonical undergraduate engineering text; derives Q = m·c·ΔT from the first law, distinguishes c_p from c_v, and tabulates values for common solids, liquids, and ideal gases.
  2. [2]NIST Chemistry WebBook (Standard Reference Database 69). Thermophysical Properties of Fluid Systems — water (H₂O). The primary US reference for temperature- and pressure-dependent specific heat of water and ~7,000 other substances. Used to verify c_p,water ≈ 4186 J/(kg·K) at 25 °C, 1 atm.
  3. [3]Rumble, J. R. (ed.) (2019). CRC Handbook of Chemistry and Physics, 100th ed., Table 2-170 'Heat Capacity of the Elements'. CRC Press. The standard tabulation of c for solid elements, the source for most undergraduate problem-set values (aluminium 897, copper 385, iron 449, gold 129, lead 130 J/(kg·K)).
  4. [4]Atkins, P. W. & de Paula, J. (2018). Atkins' Physical Chemistry, 11th ed., §2A, 'Internal Energy'. Oxford University Press. Defines specific, molar, and total heat capacity; derives c_p − c_v = R for the ideal gas; introduces the Dulong–Petit and Debye models for solids.
  5. [5]Joule, J. P. (1845). 'On the Existence of an Equivalent Relation between Heat and the ordinary Forms of Mechanical Power'. The London, Edinburgh, and Dublin Philosophical Magazine and Journal of Science, 3rd series, 27 (179): 205–207. The foundational paper establishing the mechanical equivalent of heat — Joule's paddle-wheel measurement of 4.158 J per IT calorie, from which the modern definition 1 cal = 4.184 J descends.
  6. [6]Taylor, B. N. & Thompson, A. (2008). NIST Special Publication 811, 'Guide for the Use of the International System of Units (SI)', Appendix B.8. National Institute of Standards and Technology. Defines the exact conversion 1 cal_IT = 4.184 J used in the heatCal output of this calculator.

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