Heat Energy (Q = mcΔT)Free online science calculator — instant results, no sign-up
Calculate the heat needed to change a substance's temperature.
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About the Heat Energy (Q = mcΔT)
The Heat Energy (Q = mcΔT) calculator estimates the heat transferred when a substance changes temperature without changing phase. Enter its mass, specific heat, and temperature change to obtain energy in joules, kilojoules, and kilowatt-hours. The defaults use water's specific heat of 4186 J/kg·K to calculate the heat needed for a temperature increase of 80 °C in 1 kg of water.
- Mass (kg)
- The mass of the substance being heated or cooled, measured in kilograms, defaults to 1 kg.
- Specific heat (J/kg·K)
- The energy needed to raise the temperature of one kilogram of the substance by one kelvin, measured in J/kg·K, defaults to 4186 for water.
- Temperature change (°C)
- The final temperature minus the initial temperature, measured in degrees Celsius, defaults to an increase of 80 °C.
Formula and method
ΔT = T_final − T_initial; Q = m × c × ΔT; Q_kJ = Q / 1000; Q_kWh = Q / 3,600,000.
- QHeat energy transferred to the substance, in joules (J).
- mMass of the substance, in kilograms (kg).
- cSpecific heat capacity of the substance, in joules per kilogram per kelvin (J/kg·K).
- ΔTTemperature change in °C or K; a temperature difference has the same numerical value in either unit.
- T_finalFinal temperature of the substance, in °C.
- T_initialInitial temperature of the substance, in °C.
- Q_kJHeat energy expressed in kilojoules (kJ).
- Q_kWhHeat energy expressed in kilowatt-hours (kWh).
Multiply mass by specific heat and temperature change to obtain heat energy in joules. With the default inputs, Q = 1 × 4186 × 80 = 334,880 J. Divide by 1000 to obtain 334.88 kJ, and divide the joule result by 3,600,000 to obtain 0.09302 kWh, rounded to five decimal places.
Worked example
These are the calculator's starting values and the results it actually produces for them.
Inputs
- Mass1 kg
- Specific heat4186 J/kg·K
- Temperature change80 °C
Results
- Heat energy334,880 J
- In kilojoules334.88 kJ
- In kWh0.09302
- NoteWater's specific heat is 4186 J/kg·K
How to use the Heat Energy (Q = mcΔT)
- 1Enter the substance's mass in kilograms; keep 1 kg for the default example.
- 2Enter its specific heat in J/kg·K; keep 4186 for the water example.
- 3Subtract the initial temperature from the final temperature and enter that change in °C; the default is 80 °C.
- 4Calculate the heat energy using Q = mcΔT.
- 5Read the equivalent energy outputs: 334,880 J, 334.88 kJ, and 0.09302 kWh for the defaults, and check the water-specific-heat note.
Reading your result
The 'Heat energy' row gives the heat transferred in joules: 334,880 J for the defaults. 'In kilojoules' shows the same energy as 334.88 kJ. 'In kWh' expresses that energy as 0.09302 kWh; this is energy, not power or operating time. The note, 'Water's specific heat is 4186 J/kg·K', identifies the reference value used by the default example, not a value suitable for every substance. Positive Q means heat is absorbed; negative Q means heat is released.
Limitations
This model assumes a constant specific heat over the temperature interval, an unchanged mass, and no phase transition. It does not include latent heat for melting or boiling, heat absorbed by a container, heat loss to the surroundings, or heater inefficiency. Specific heat depends on the material, temperature, phase, and sometimes pressure, so 4186 J/kg·K is an approximate value for liquid water rather than a universal constant. Actual electrical or fuel consumption may exceed the calculated heat requirement.
Common questions
Why can I enter a temperature change in °C when specific heat uses kelvins?+
Celsius and kelvin have equal-sized increments, so an increase of 80 °C is also an increase of 80 K. Do not add 273.15 to a temperature difference.
Does the default 80 °C mean the water ends at 80 °C?+
No. It means the final temperature is 80 °C higher than the initial temperature. The calculator uses the difference, not either absolute temperature.
Can I use 4186 J/kg·K for substances other than water?+
Use a specific heat value appropriate to the actual substance and its phase. The default 4186 J/kg·K is for liquid water and should not be assumed valid for other materials, ice, or steam.
Does 0.09302 kWh tell me the electricity needed to heat the default water sample?+
It is the ideal heat transferred to the water for the specified temperature increase. Electrical consumption can be higher because of heat losses and energy used to warm the container; predicting consumption requires additional efficiency and system information.
Does the calculator save my information?+
Calculations run in your browser. While signed in, inputs and results are automatically recorded in account history; saving a result creates a separate saved record.
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Results are estimates for informational purposes and are not financial, medical or legal advice.