Specific Heat (Q = mcΔT) Calculator
Specific heat capacity c is the energy needed to raise 1 kg of a substance by 1 K (units: J/(kg·K)). With the formula Q = m × c × ΔT you only need three of the four variables — mass m, specific heat c, temperature change ΔT or heat energy Q — to solve for the missing one. The widget bundles c for 22 common materials, fills it in automatically and converts the answer into kJ, kcal, Wh and BTU.
Please enter valid positive numbers (ΔT may be negative)
Heat Q
—
Derived
Mass m
—
Given
Specific heat c
—
Given
Temperature change ΔT
—
Given
Equivalent energy
- kJ—
- kcal—
- Wh—
- BTU—
Formula
Q = m × c × ΔT
Specific heat reference — c in J/(kg·K)
- Water (liquid, 25 °C) 4186 J/(kg·K)
- Ice (−10 °C) 2090 J/(kg·K)
- Steam (100 °C) 2010 J/(kg·K)
- Ethanol 2440 J/(kg·K)
- Olive oil 1970 J/(kg·K)
- Milk (whole) 3930 J/(kg·K)
- Air (constant pressure, 300 K) 1005 J/(kg·K)
- Wood (oak) 1700 J/(kg·K)
- Concrete 880 J/(kg·K)
- Glass 840 J/(kg·K)
- Sand (dry) 830 J/(kg·K)
- Granite 790 J/(kg·K)
- Brick 840 J/(kg·K)
- Aluminium 897 J/(kg·K)
- Steel 466 J/(kg·K)
- Iron 449 J/(kg·K)
- Brass 380 J/(kg·K)
- Copper 385 J/(kg·K)
- Silver 235 J/(kg·K)
- Mercury 140 J/(kg·K)
- Gold 129 J/(kg·K)
- Lead 128 J/(kg·K)
Values from the CRC Handbook of Chemistry and Physics (97th ed., 2016–17) at ~20–25 °C and 1 atm; air is at constant pressure c_p.
Formula
Q = m × c × ΔT // heat (J) = mass (kg) × specific heat (J/(kg·K)) × temperature change (K or °C) m = Q ÷ (c × ΔT) c = Q ÷ (m × ΔT) ΔT = Q ÷ (m × c)
- · Temperature interval: 1 °C and 1 K are the same magnitude, so ΔT can be entered in either; for °F first convert with ΔT_K = ΔT_F × 5/9.
- · The substance must stay in one phase (liquid / solid / gas). If a phase change happens (e.g. water → steam) you also need the latent heat term Q = m × L — this calculator does not include it.
- · Air and other gases use the constant-pressure specific heat c_p; under constant volume use c_v, which is smaller, depending on the problem.
- · Q > 0 means heat absorbed (warming); Q < 0 means heat released (cooling). Negative ΔT and Q are accepted so the tool works for cooling problems too.
- · Specific heat varies slightly with temperature and pressure. The table holds typical values near room conditions — for precise lab work consult NIST or the CRC Handbook directly.
- · Energy unit equivalents: 1 cal = 4.184 J (exact), 1 kcal (food Calorie) = 4184 J, 1 Wh = 3600 J (exact), 1 BTU (IT) = 1055.056 J.
- · Sources: CRC Handbook of Chemistry and Physics (97th ed., 2016–17); NIST Special Publication 811 (unit conversions); HKEAA HKDSE Physics — "Heat and Gases" strand.
Frequently asked
Why is ΔT the same in K and °C?
Kelvin and Celsius use identical scale spacing — only the zero point differs (Kelvin starts at absolute zero, −273.15 °C). So "30 °C → 80 °C" is the same interval as "303.15 K → 353.15 K"; both equal 50. Since Q = mcΔT depends on a temperature difference, °C and K give the same number. If the problem hands you Fahrenheit, convert first with ΔT_K = ΔT_F × 5 ÷ 9.
How much energy does it take to boil 1.5 L of water from room temperature?
One litre of water weighs about 1 kg. Assuming room temperature 20 °C and boiling point 100 °C, ΔT = 80 K. Q = 1.5 × 4186 × 80 = 502,320 J ≈ 502 kJ ≈ 120 kcal ≈ 139 Wh. A real kettle is roughly 80–90 % efficient, so it draws around 150–170 Wh from the wall — about 4.5 minutes on a 2 kW kettle. In the widget set Solve for = Q, m = 1.5, c = 4186 (auto-filled when you pick "Water"), ΔT = 80 to see all of this.
Why is water's specific heat so high?
Water's c = 4186 J/(kg·K) is one of the highest of any everyday substance — roughly 4.7× aluminium, 11× copper, and 4.2× the constant-pressure value for air. The reason is hydrogen bonding: a large share of incoming energy first goes into breaking and re-forming hydrogen bonds between molecules rather than into raw kinetic energy. That high heat capacity is why oceans, lakes and even bodies buffer temperature so well — coastal climates are milder than continental ones, and the property is one of the reasons liquid water makes life possible.
What if I cross a melting or boiling point?
Phase changes (solid↔liquid, liquid↔gas) require an additional "latent heat" L, which is not part of Q = mcΔT. For example, 1 kg of 0 °C ice → 1 kg of 0 °C water needs 334 kJ (latent heat of fusion L_f); 1 kg of 100 °C water → 1 kg of 100 °C steam needs 2,260 kJ (latent heat of vaporisation L_v). Problems crossing a phase change need to be split into stages: (1) heat the ice with c_ice up to 0 °C, (2) melt it with m × L_f, (3) heat the water with c_water up to 100 °C, (4) vaporise it with m × L_v, (5) superheat the steam with c_steam. This tool handles one phase at a time — run it once per stage and add the Q values.
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