Calculators
Specific Heat Calculator
Solve Q = mcΔT for heat, mass, specific heat, or temperature change.
Use Q = mcΔT for sensible heating or cooling with no phase change. Solve heat energy, mass, specific heat capacity, or temperature change.
Calculated result
41840 J
Q = mcΔT
Positive Q/ΔT represents heating and negative Q/ΔT cooling. The formula assumes approximately constant specific heat and no melting, boiling, chemical reaction, or other phase change.
About This Tool
Specific heat capacity describes how much energy is needed to change the temperature of a unit mass of material. For sensible heating or cooling over a range where the material stays in the same phase, the standard calorimetry relationship is Q = mcΔT. This calculator can solve that equation for heat energy, mass, specific heat capacity, or temperature change and converts common units before applying the formula. Calculations stay in your browser.
How To Use It
- Choose the quantity you want to solve for.
- Enter the three known quantities and select their units. Use a positive temperature change for heating and a negative change for cooling.
- Choose the result unit. Celsius and kelvin temperature differences have the same numeric size; Fahrenheit differences are converted by a factor of 5/9.
- Check that the process has no phase change and that treating specific heat as constant is reasonable for the temperature range.
Examples
Heating 1 kg of water by 10 °C
Using c ≈ 4184 J/(kg·°C), Q = 1 × 4184 × 10 = 41,840 J, or 41.84 kJ.
100 g with c = 1 cal/(g·°C)
A 10 °C rise needs 100 × 1 × 10 = 1000 cal, equivalent to 4184 J.
Find specific heat capacity
If 0.2 kg absorbs 8368 J while warming 10 °C, c = 8368 ÷ (0.2 × 10) = 4184 J/(kg·°C).
Cooling
For 1 kg with c = 1000 J/(kg·°C), a −5 °C change gives Q = −5000 J. The negative sign indicates heat leaves the object under this sign convention.
Useful Notes
The Q = mcΔT relationship
Q is heat transferred, m is mass, c is specific heat capacity, and ΔT is final temperature minus initial temperature. In SI units, kilograms, J/(kg·°C), and °C or K differences produce joules.
Temperature differences are not absolute temperatures
The equation uses a temperature change, not an absolute temperature. A change of 1 °C equals a change of 1 K. A change of 1 °F equals 5/9 °C, so Fahrenheit differences require scaling rather than the offset used when converting absolute temperatures.
Heating and cooling signs
This calculator allows signed Q and ΔT. With the stated convention, positive values describe energy entering the object and warming it; negative values describe energy leaving the object and cooling it. When solving for mass or specific heat, inconsistent signs would imply a nonphysical negative result and are rejected.
Specific heat varies in real materials
Published specific heat values are often approximate and can vary with temperature, composition, pressure, moisture, and material state. Use a value appropriate to the substance and conditions when accuracy matters.
Phase changes need another model
Q = mcΔT describes sensible heat within one phase. Melting, freezing, boiling, condensation, and similar transitions require latent heat terms. A process spanning a phase change normally needs separate heating and latent-heat steps.
FAQ
What is the SI unit of specific heat capacity?
The SI unit is joules per kilogram kelvin, J/(kg·K). For temperature differences, J/(kg·°C) has the same numeric scale.
Is specific heat the same as heat capacity?
No. Specific heat capacity is heat capacity per unit mass. An object's total heat capacity is m × c and is commonly expressed in J/K.
Can Q be negative?
Yes. Under the sign convention used here, negative Q represents heat leaving the object. A negative temperature change likewise represents cooling.
Can I use Fahrenheit with Q = mcΔT?
Yes for temperature differences when converted correctly. A Fahrenheit temperature difference is multiplied by 5/9 to obtain the equivalent Celsius or kelvin difference.
Can this calculate melting or boiling energy?
Not by itself. Phase changes require latent heat, and a full process may need separate Q = mcΔT segments before or after the phase change.
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