Interactive Enthalpy Change Calculator
Use this tool to calculate heat transfer (q) and, if needed, molar enthalpy change (ΔH) from calorimetry data.
What is enthalpy change?
Enthalpy change, written as ΔH, measures the heat absorbed or released by a process at constant pressure. In practical chemistry, this is one of the most useful quantities for understanding whether a process is energetically favorable and how much thermal energy moves between a system and its surroundings.
If a process absorbs heat, enthalpy increases (ΔH > 0, endothermic). If it releases heat, enthalpy decreases (ΔH < 0, exothermic).
How this calculator works
1) Heat change mode
The calculator uses:
- q = m × c × ΔT
- ΔT = Tf - Ti
Where:
- m = mass in grams (g)
- c = specific heat capacity in J/g°C
- ΔT = temperature change in °C
The output is heat transfer in J and kJ. Positive q means the sample absorbed heat; negative q means it released heat.
2) Reaction enthalpy mode (calorimetry)
In many lab setups, you measure the temperature change of the solution (surroundings), then infer reaction enthalpy with opposite sign:
- qsolution = m × c × ΔT
- qrxn = -qsolution
- ΔHrxn = qrxn / n = -qsolution / n
That is why this calculator asks for moles in reaction mode and returns kJ/mol.
Step-by-step usage
- Select your calculation mode.
- Enter mass and specific heat capacity.
- Enter initial and final temperatures.
- If using reaction mode, enter moles of the limiting reactant.
- Click Calculate to see q, ΔT, and ΔH (if applicable).
Sign convention quick guide
- q > 0: sample/surroundings gain heat.
- q < 0: sample/surroundings lose heat.
- ΔHrxn < 0: reaction is exothermic.
- ΔHrxn > 0: reaction is endothermic.
Common mistakes to avoid
- Mixing units (for example, using kg with J/g°C).
- Forgetting the negative sign when converting from measured solution heat to reaction enthalpy.
- Using the wrong specific heat value for the material.
- Rounding too early in multi-step calculations.
Worked example
Suppose 150 g of water warms from 20°C to 28°C. Use c = 4.184 J/g°C.
- ΔT = 28 - 20 = 8°C
- q = (150)(4.184)(8) = 5020.8 J = 5.0208 kJ
So the water absorbed 5.0208 kJ. If this came from a reaction and n = 0.0400 mol, then:
- ΔHrxn = -5.0208 / 0.0400 = -125.52 kJ/mol
Negative means the reaction is exothermic.