Heat & Specific Heat Calculator
Calculate heat with Q = mcΔT, heat capacity with Q = CΔT, the final temperature when two objects mix, and latent heat with Q = mL. Shows unit conversions (J, kJ, cal, kcal, g, kg), the working and whether heat is absorbed or released.
A difference is the same in K and °C
Try an example
Fill in every value except the one you solve for.
How to use
- 1
Choose the calculation
Select heat (specific heat or heat capacity), thermal equilibrium, or a phase change (latent heat).
- 2
Enter values and units
Enter mass, specific heat, temperature change and heat. Specific heat can be picked from material presets.
- 3
Check the result and working
See the unit conversions, whether heat is absorbed or released, and the heat exchanged in mixing problems.
Features
- Solve Q = mcΔT for heat, mass, specific heat or temperature change
- Heat capacity mode with Q = CΔT
- Find the final temperature and heat exchanged when two objects mix
- Latent heat mode with Q = mL and water fusion/vaporisation presets
- Unit conversion for J, kJ, cal, kcal, g and kg, with absorbed/released heat shown
- Material presets (water, aluminium, iron, copper, ethanol and more) and 1 cal = 4.18, 4.184 or 4.2 J
Use cases
Physics and chemistry homework
Check heat and specific heat problems with the unit conversions and substitution shown.
Lab preparation and reports
Estimate the heat for heating water or cooling a metal, and the temperature after mixing.
Everyday estimates
Work out the energy needed to warm a bathtub by 1 °C, for example.
Details
The heat Q = mcΔT is the energy needed to change the temperature of a mass m by ΔT. The specific heat c of water is about 4.2 J/(g·K) (some textbooks use 4.18). With mass in g, specific heat in J/(g·K) and temperature difference in K, Q comes out in joules. A difference of 20 °C equals a difference of 20 K.
Using the heat capacity C = mc gives Q = CΔT. The heat capacity is the energy to raise the whole object by 1 K, in J/K. For 100 g of water, C is about 420 J/K.
When two objects are mixed, the final temperature equalises the heat lost by the hot object and the heat gained by the cold one (conservation of energy). For an insulated system with no phase change, t = (m₁c₁t₁ + m₂c₂t₂) / (m₁c₁ + m₂c₂). Mixing 100 g of water at 80 °C with 200 g at 20 °C gives 40 °C.
During melting or boiling the temperature stays constant and the heat goes into latent heat (Q = mL). For water, the latent heat of fusion is about 334 J/g and of vaporisation about 2256 J/g. If a calculation crosses 0 °C or 100 °C, add the Q = mcΔT and Q = mL steps separately.
FAQ
Which units should I use in Q = mcΔT?
Mass in g, specific heat in J/(g·K) and temperature difference in K (or °C) give Q in joules. Using kg with J/(kg·K) also works, but mixing g and kg introduces a factor of 1000. This tool converts the units you enter.
What is the difference between specific heat and heat capacity?
Specific heat is the energy needed to raise 1 g of a material by 1 K (J/(g·K)) and depends on the material. Heat capacity is the energy to raise the whole object by 1 K (J/K) and scales with mass: C = m × c.
How many joules is 1 calorie?
By the thermochemical definition, 1 cal = 4.184 J. Japanese textbooks often use 4.18 J, and introductory courses sometimes use 4.2 J. The tool lets you switch between these.
Should I use K or °C for the temperature change?
A temperature difference is the same in K and °C: a change from 20 °C to 40 °C is 20 K. Absolute temperatures are only needed for other equations, such as the ideal gas law.
Is the mixing temperature just an average?
Only when the masses and specific heats are equal. In general it is a weighted average using the heat capacities m × c. Mixing 100 g and 200 g of water gives a temperature closer to the larger mass.
How do I calculate the heat to melt ice?
While ice at 0 °C turns into water at 0 °C the temperature does not change, so use Q = mL, not Q = mcΔT. For water the latent heat of fusion is about 334 J/g, so 100 g of ice needs about 33,400 J. If the ice starts below 0 °C, add the warming step separately.
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Verified: Known values (100 g water × 4.18 × 20 K giving 8,360 J = 2,000 cal, 100 g water × 4.2 × 20 K giving 8,400 J, 100 g iron × 0.45 × 10 K giving 450 J, 840 J/K × 20 K giving 16,800 J, 100 g water at 80 °C mixed with 200 g at 20 °C giving 40 °C, 50 g aluminium at 100 °C with 100 g water at 20 °C giving 27.78 °C, melting 100 g of ice giving 33,400 J and vaporising 100 g of water giving 225,600 J) plus error handling are covered by browser tests
Did you know?
The specific heat of water is about 4.2 J/(g·K). Warming 100 g by 10 K takes about 4,200 J for water but only about 450 J for iron — roughly nine times less, because the specific heat of iron (0.45) is much smaller. That is why metals heat up so quickly.