Specific Heat Calculator
The specific heat equation Q = m·c·ΔT links the heat energy added to a substance to its mass, its specific heat capacity, and how much its temperature changes. Pick what you want to solve for, choose a preset material or enter your own specific heat, switch between metric and imperial units, work from a temperature change or from initial and final temperatures, and optionally price the energy used.
Formula
Worked example
Heat 0.5 kg of water (c = 4186 J/(kg·K)) by 10 K: Q = 0.5 × 4186 × 10 = 20,930 J ≈ 20.9 kJ (about 5.8 Wh). Rearranged, that same 20,930 J spread over a 5 K rise would require 1.0 kg of water, and a calorimetry test that measured 20,930 J into 0.5 kg over 10 K would back out c = 4186 J/(kg·K).
What the specific heat equation tells you
Specific heat capacity, written c, measures how much energy a material needs to change its temperature. Formally it is the amount of heat required to raise the temperature of one kilogram of a substance by one kelvin (identical in size to one degree Celsius). The full relationship Q = m·c·ΔT says the total heat Q transferred equals the mass m multiplied by the specific heat c multiplied by the temperature change ΔT. A large specific heat means the material resists temperature change: you must pour in a lot of energy for only a small rise, which is exactly why water, with one of the highest specific heats of any common substance, is so effective at storing and moderating heat.
Solving for any variable, in any units
Because Q = m·c·ΔT is a simple product, you can solve for whichever quantity is unknown as long as you know the other three. Pick the unknown from the selector: to find mass use m = Q / (c·ΔT), to find an unknown material’s specific heat use c = Q / (m·ΔT), the basis of calorimetry, and to find the temperature change a given energy produces use ΔT = Q / (m·c). Every input carries its own unit switch, so you can enter energy in joules, kilojoules, calories, kilocalories, BTU or watt-hours, mass in kilograms, grams, pounds or ounces, specific heat in metric or BTU/(lb·°F) form, and temperature in Celsius, kelvin or Fahrenheit. Internally the tool converts everything to SI base units, solves, and converts the answer back to your chosen display units, so you can mix and match freely.
Substance presets, temperatures, and energy cost
Rather than looking up a value, choose a substance from the dropdown and its specific heat fills in automatically: water, ice, steam, ethanol, oils, air, and metals such as aluminium, iron, copper, silver, gold and lead. Select Custom to type your own. If you know the starting and ending temperatures rather than the change, turn on the initial and final temperature mode and the calculator works out ΔT for you in your chosen scale. Finally, switch on the energy cost estimate to see how much it costs to supply the heat: enter your electricity price per kilowatt-hour and a heater efficiency, and the tool converts joules to watt-hours and prices the input energy, since real heaters waste part of what they draw.
Units, sign conventions, and limits
A positive Q and ΔT mean heat is absorbed and the substance warms; a negative value means heat is released as it cools. The formula assumes c stays roughly constant over the temperature range and that no phase change occurs. Crossing a melting or boiling point requires latent heat, Q = m·L, which this calculator does not include, so keep the substance in a single phase. Specific heat also varies with temperature and, for gases, with whether pressure or volume is held constant; the preset values are room-temperature, constant-pressure figures and are accurate enough for everyday physics, chemistry and HVAC estimates.
Specific heat of common substances
| Substance | Specific heat c [J/(kg·K)] | In BTU/(lb·°F) |
|---|---|---|
| Water (liquid) | 4186 | 1.000 |
| Ice | 2090 | 0.499 |
| Ethanol | 2440 | 0.583 |
| Air (dry, const. pressure) | 1005 | 0.240 |
| Aluminium | 900 | 0.215 |
| Glass | 840 | 0.201 |
| Iron / steel | 450 | 0.108 |
| Copper | 385 | 0.092 |
| Silver | 235 | 0.056 |
| Lead / gold | 129 | 0.031 |
Approximate specific heat capacities near room temperature, in J/(kg·K). Values vary slightly with temperature and source. 1 J/(kg·K) = 1 J/(kg·°C).
Frequently asked questions
What is the difference between specific heat and heat capacity?
Specific heat (c) is per unit mass, the energy to raise one kilogram by one kelvin, so it is a property of the material. Heat capacity (often C) is for a whole object and equals mass times specific heat (C = m·c), so it depends on how much of the substance you have.
Should ΔT be in Celsius, Kelvin, or Fahrenheit?
A temperature difference of one degree Celsius is exactly one kelvin, so Celsius and Kelvin are interchangeable for ΔT. A Fahrenheit degree is smaller, 5/9 of a kelvin, so the tool scales it for you. You only need absolute kelvin for thermodynamic temperatures, not for a temperature change. Switch the temperature unit and the calculator handles the conversion.
How do I find the specific heat of an unknown material?
Measure the heat you add (Q), the mass (m), and the temperature rise (ΔT), then set the calculator to solve for c. It applies c = Q / (m·ΔT), which is exactly what a calorimetry experiment does. Make sure no phase change happens during the test, or the result will be distorted by latent heat.
Why does Q = m·c·ΔT fail during boiling or melting?
During a phase change the temperature stays constant while energy goes into breaking or forming molecular bonds, so ΔT is zero even though heat is flowing. That energy is the latent heat, calculated separately as Q = m·L. Use Q = m·c·ΔT only while the substance stays in a single phase.
How is the energy cost estimated?
The tool converts the heat Q from joules to kilowatt-hours (1 kWh = 3,600,000 J), divides by your heater efficiency to account for losses, then multiplies by your price per kWh. It is a planning figure: real costs depend on standby losses, insulation, and how your energy is billed.