Thermochemistry Calculator
What this calculator solves
Two thermochemistry workhorses share this widget.
Heat transfer: q = m × c × ΔT. Four variables, three given, one unknown — the calculator solves for the missing one. The relation assumes c is constant over the temperature range and that no phase changes occur (those need separate latent heat terms).
Reaction enthalpy via Hess’s law:
ΔH_rxn = Σ ν · ΔH_f(products) − Σ ν · ΔH_f(reactants)
where ν is the stoichiometric coefficient. This works because enthalpy is a state function — the path does not matter, only the start and end. Element standard states (O₂(g), C(graphite), H₂(g)) have ΔH_f = 0 by definition, so they fall out of the sum.
How to use it
For heat transfer, enter any three of {q, m, c, ΔT} and the fourth comes out. Pick a substance from the preset list (water, aluminum, iron, copper, ethanol) or type a custom c. The temperature-difference field accepts T_initial and T_final separately.
For enthalpy, enter the balanced equation’s coefficients and the ΔH_f values for each species (kJ/mol). The calculator multiplies, sums products, sums reactants, subtracts, and reports ΔH_rxn for the reaction as written.
Common specific heat values
| Substance | c (J/(g·K)) |
|---|---|
| Water (liquid) | 4.184 |
| Ice | 2.09 |
| Steam | 2.01 |
| Aluminum | 0.897 |
| Iron | 0.449 |
| Copper | 0.385 |
| Gold | 0.129 |
| Ethanol | 2.44 |
Water’s value is anomalously high. That single fact explains a startling amount of climate, biology, and engineering.
Worked examples
Heating water. Heat needed to bring 250 g of water from 20.0 °C to 80.0 °C. ΔT = 60.0 K. q = 250 × 4.184 × 60.0 = 62,760 J = 62.8 kJ.
Identifying a metal. A 50.0 g sample absorbs 1,500 J and rises from 25.0 to 95.0 °C. ΔT = 70.0 K. c = q/(m·ΔT) = 1500/(50.0 × 70.0) = 0.429 J/(g·K) — close to iron at 0.449.
Bomb-style calorimetry. Burning 1.00 g of fuel raises 200.0 g of water from 22.0 to 35.5 °C. q_water = 200.0 × 4.184 × 13.5 = 11,297 J. The fuel released −11.3 kJ per gram, with the negative sign reflecting heat leaving the system.
Hess’s law: methane combustion. CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l). ΔH_f values: CH₄ = −74.8, O₂ = 0, CO₂ = −393.5, H₂O(l) = −285.8 kJ/mol. ΔH = [(−393.5) + 2(−285.8)] − [(−74.8) + 2(0)] = −965.1 + 74.8 = −890.3 kJ/mol. Strongly exothermic, which is why methane is a fuel.
The sign convention trap
Heat released by a reaction has a negative sign from the system’s perspective and a positive sign from the surroundings’. In a calorimetry problem, the water gets warmer (q_water > 0), but the reaction itself has q_reaction < 0 of equal magnitude. Mixing up which side of the energy ledger you are on is the most common stumbling block in introductory thermochemistry.