Specific Heat Capacity Calculator

Solve for heat energy (q), mass (m), specific heat capacity (c), or temperature changes (ΔT). Switch to the equilibrium tab to solve multi-substance calorimetry mixing problems instantly.

Specific Heat Transfer Solver

Solve For:
Joules (J)
grams (g)
J/(g·°C)
°C

Calculated Missing Parameter

15690.000 Joules (J)

Workspace Expression Trace

Formula: q = m · c · ΔT
q = 250g × 4.184J/(g·°C) × 15°C = 15690.00 Joules (J)

The Heat Transfer Equation (q = mcΔT)

The quantity of heat energy (\(q\)) gained or lost by a substance during a temperature change is directly proportional to its mass and its unique physical molecular structure. This foundational relationship is defined by the classical equation:

q = m · c · ΔT

Where the individual variables stand for:

  • q (Heat Energy Change): Measured in Joules (J). A positive value indicates heat absorbed (endothermic), while a negative value signifies heat released (exothermic).
  • m (Mass): The total physical weight of the substance quantified in grams (g).
  • c (Specific Heat Capacity): The constant amount of energy required to raise the temperature of exactly one gram of substance by one degree Celsius, measured in J/(g·°C).
  • ΔT (Temperature Change): Calculated as Final Temperature minus Initial Temperature (\(T_f - T_i\)).

Calorimetry and Thermal Equilibrium

According to the Law of Conservation of Energy, in an isolated system, heat cannot simply disappear. When a warm object is mixed with a cooler medium, the heat lost by the warmer material must perfectly equal the heat gained by the cooler material:

−q_lost = q_gained  →  m₁·c₁·(T₁ − Tf) = m₂·c₂·(Tf − T₂)