Example 1
c = 4186, m = 1 kg, ΔT = 10 → Q = 41860 J.
Heat Calculator
Enter specific heat, mass, and how much the temperature changes. You get the energy you have to add or take away. A liter of water at c = 4186 J/(kg·K) and a 10 K rise is 41860 J, roughly what a small kettle delivers in a few seconds.
Melting ice: Q = m·Lₜ. Boiling water: Q = m·Lₚ.
Result
Specific heat c (J/(kg·K)), Mass m (kg) and ΔT (K). The result shows up here.
How it works
Q = c·m·ΔT is the energy you add or remove to change the temperature of mass m by ΔT. For water, school tables use c ≈ 4186 J/(kg·K), sometimes rounded to 4200. One kilogram warmed by 10 K needs 41860 J; with 4200 you get a round 42000 J. The difference is small. The formula is the same.
ΔT is a change, not the number on a thermometer. A ten degree Celsius difference is also 10 K. Do not add 273. If the water cools by 20 K, type −20. The result is negative because the body gives heat away.
Type specific heat in joules per kilogram-kelvin and mass in kilograms. The result is in joules. A comma and a period are the same, so 0.5 kg can be 0,5.
Use this formula while temperature actually changes. During melting or boiling the temperature holds still and the energy goes into the phase change. Then you want Q = m·Lₜ or Q = m·Lₚ on the neighboring pages, not this one.
Fill c, m, and ΔT, then Calculate. The calculator example 4186, 1 kg and 10 gives 41860 J. For aluminum, c is about 900, so the same kilogram and the same 10 K is only 9000 J.
Joule heat from current in a resistor is another route to joules. Here the energy comes from mass, specific heat, and a temperature change, not from I²Rt.
How to use
- Type specific heat c. For water, 4186 J/(kg·K) is the usual value; school 4200 is fine too.
- Type mass in kilograms, for example 1, or 0.5 if you are heating half a liter of water.
- Type ΔT as a change: 10 when the temperature rises by 10 °C. Do not add 273.
- Click Calculate. 4186, 1 kg and 10 K give 41860 J. A negative ΔT means heat leaving the body.
- If ice is melting or water is boiling, open fusion or vaporization. Temperature does not rise there.
Formula
Q = c · m · ΔT
SI unit: J. ΔT is a temperature difference (K), not an absolute temperature.
Letters in Q = c·m·ΔT
The calculator multiplies c, m and ΔT. A liter of water at 4186 J/(kg·K) and a 10 K rise is 41860 J, energy for the temperature climb only.
- Q
- Energy in joules. With c = 4186, m = 1 kg and ΔT = 10 K the result is 41860 J.
- c
- Specific heat in J/(kg·K). Water here is 4186; rounding to 4200 would give exactly 42000 J.
- m
- Mass in kilograms. One kilogram at those 4186 and 10 K builds 41860 J.
- ΔT
- Temperature rise in kelvin. A ten degree Celsius gap is 10 K, not a 20 °C thermometer reading.
Real-life examples
Example 2
m = 0.5 kg, ΔT = 80, water.
Example 3
m = 0.2 kg, ΔT = 60.
Example 4
c ≈ 900, m = 1 kg, ΔT = 20.
Example 5
c ≈ 450, m = 2 kg, ΔT = 50.
Example 6
c ≈ 2100, m = 1 kg, ΔT = 5 (before melting).
Example 7
c ≈ 2000, m = 1 kg, ΔT = 15.
Example 8
water, m = 0.1 kg, ΔT = 30.
Example 9
m = 1000 kg, ΔT = 1, water.
Example 10
m = 1 kg, ΔT = −20: negative Q (heat released).
Ways to use this calculator
- You estimate how much energy a kettle needs to warm a liter of water by 10 K: about 42 kJ.
- You compare water with a metal at the same ΔT: aluminum at c ≈ 900 uses less energy.
- You build a balance: warm the ice with c m ΔT first, then melt it on the fusion page.
Frequently asked questions
How many joules for 1 kg of water, c = 4186 and ΔT = 10 K?
Q = 41860 J, about 41.9 kJ. Two kilograms at the same rise is 83720 J.
Do I add 273 to Celsius?
Not in this field. ΔT is a difference. 20 °C minus 10 °C is 10, not 283.
Should I type 4186 or 4200 for water?
School work often uses 4200. Then 1 kg and 10 K give 42000 J, 140 J more than 4186.
When is this the wrong formula?
While melting or boiling, temperature holds still. Energy is Q = m L on the fusion and vaporization pages.
What does a negative ΔT mean?
The body is cooling and giving heat away. At 1 kg, 4186 and −20 K you get −83720 J.
What unit is the result?
Joules. With the US switch the calculator also shows foot-pounds. The formula stays the same.
Does c depend on temperature?
A little. In school problems you take a constant c from a table and leave it there.
How do I handle ice starting at −10 °C?
Warm the ice to 0 °C with this formula first (ice c ≈ 2100), then Q = m Lₜ. The sum is larger than melting alone.
How is this different from Joule heat?
Here Q comes from mass and ΔT. There Q = I² R t, energy from current in a resistor.
Can I type 0.5 kg with a comma?
Yes. 0,5 and 0.5 are the same mass. A comma and a period mean the same value.
Knowledge sources
The formula is the school one. Units follow SI; NIST SP 330 and BIPM define the measures, not your result.
Page updated in 2026.