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🔥 Thermodynamics

Specific Heat Calculator

Solve q = mcΔT for heat energy, mass, specific heat capacity or temperature change — with common material values and the full working.

Solve for any term
Heat in J & kJ
Absorbed or released
Material c values
100% Free
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Specific heat — Quick answer

Heat equals mass times specific heat times the temperature change.

q = m · c · ΔT  →  m = q/(cΔT) · c = q/(mΔT) · ΔT = q/(mc)

Worked example: 100 g water × 4.184 × 25 °C = 10,460 J.

Specific heats (J/g·°C)

Materialc
Water4.184
Aluminium0.897
Copper0.385

Covers temperature change only — not melting or boiling.

🔥 Specific Heat Calculator

Choose what to solve for, then fill the other three.

Result
Heat in kJ
Direction
Equation

ℹ️ q = mcΔT. Leave the "solve for" quantity blank or any value — it's computed from the other three. Positive q = absorbed, negative = released.

The heat equation q = mcΔT links the energy q transferred to a substance with its mass m, specific heat capacity c, and temperature change ΔT. This calculator solves for whichever quantity you choose, shows the result in joules and kilojoules, and notes whether heat is absorbed or released.

Reviewed: June 20, 2026 · Author: Naveen P N, Founder — AI Calculator · Verified against: the calorimetry equation, recomputed in code.

The formula, rearranged

Four forms
q = mcΔT · m = q ÷ (cΔT) · c = q ÷ (mΔT) · ΔT = q ÷ (mc)

Specific heat capacity is the energy to raise 1 gram by 1 °C. Water's is unusually high (4.184 J/g·°C), so it stores a lot of heat and resists temperature swings; metals are far lower. A positive ΔT (warming) gives a positive q — heat absorbed; a negative ΔT (cooling) gives a negative q — heat released. The equation holds only within one phase, with no melting or boiling.

Worked examples

Heat 100 g of water by 25 °C:

10,460 J
q = 100 × 4.184 × 25 = 10,460 J = 10.46 kJ

Heat 50 g of copper by 30 °C:

577.5 J
q = 50 × 0.385 × 30 = 577.5 J

Find c from q = 577.5 J, m = 50 g, ΔT = 30 °C:

0.385 J/g·°C
c = 577.5 ÷ (50 × 30) = 0.385 — that's copper

Because water's specific heat is more than ten times copper's, the same 50 g warmed 30 °C would take about 6,276 J for water versus 578 J for copper — which is why a metal pan heats up so much faster than the water in it.

Frequently Asked Questions

What is the specific heat formula?

q = mcΔT. Rearrange to solve for m, c or ΔT.

Heat to warm 100 g water by 25 °C?

≈ 10,460 J (10.46 kJ), using c = 4.184.

What is specific heat capacity?

Energy to raise 1 g by 1 °C. Water 4.184, copper 0.385 J/g·°C.

What does negative q mean?

Heat is released (the substance cools, ΔT < 0).

Does it cover melting or boiling?

No — phase changes use latent heat, q = mL, at constant temperature.

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