Example 1 — Find heat \(q\)
Aluminum sample: \(m = 120\,\text{g}\), \(c = 0.900\,\mathrm{J\,g^{-1}\,^\circ C^{-1}}\), \(\Delta T = 15.0\,^\circ\text{C}\).
Specific Heat Calculator computes c, q, m, ΔT with clear equations, unit-aware guidance, and stepwise examples for labs and reports.
c = Q / (m·ΔT)
A Specific Heat Calculator is a tool for experiments and learning that relates heat energy \(q\), mass \(m\), temperature change \(\Delta T\), and specific heat capacity \(c\). In thermochemistry, the core relationship
quantifies energy absorbed or released when a substance changes temperature without a phase change. The calculator solves for any one variable given the other three, checks unit consistency, and emphasizes correct sign conventions and significant figures. It is ideal for classroom exercises, lab planning, and homework verification.The calculator applies the constant-pressure calorimetry model. Input values for heat energy \(q\), mass \(m\), temperature change \(\Delta T\), and specific heat capacity \(c\). The missing variable is computed using
.Supported units include: \(q\) in J or kJ, \(m\) in g or kg, and \(\Delta T\) in °C or K (numerically identical for differences). Specific heat \(c\) can be in \(\mathrm{J\,g^{-1}\,^\circ C^{-1}}\) or \(\mathrm{J\,kg^{-1}\,K^{-1}}\).
For thermal equilibrium problems, the energy balance equation
computes unknown temperatures or specific heats. Optionally, a calorimeter constant \(C_{\text{cal}}\) accounts for heat absorbed by the container:Sensible heating/cooling:
Solve for specific heat:
Solve for mass or temperature change:
Mixture at equilibrium (no losses):
With calorimeter constant \(C_{\text{cal}}\) (optional):
Aluminum sample: \(m = 120\,\text{g}\), \(c = 0.900\,\mathrm{J\,g^{-1}\,^\circ C^{-1}}\), \(\Delta T = 15.0\,^\circ\text{C}\).
Metal sample: \(m = 250\,\text{g}\), \(q = 3.15\,\text{kJ}\), \(\Delta T = 10.0\,^\circ\text{C}\).
Mix hot water: \(m_h = 100.0\,\text{g}, T_{h,i} = 60.0\,^\circ\text{C}\) with cold water: \(m_c = 150.0\,\text{g}, T_{c,i} = 20.0\,^\circ\text{C}\), \(c = 4.184\,\mathrm{J\,g^{-1}\,^\circ C^{-1}}\).
The energy required to raise 1 gram (or kilogram) of a substance by \(1^\circ\text{C}\) (or 1 K).
For temperature differences, \(\Delta^\circ\text{C} = \Delta \text{K}\). Ensure consistent units in calculations.
Exothermic cooling gives \(q < 0\); endothermic heating gives \(q > 0\). Magnitude is often reported.
No. Use latent heat \(q = mL\) for phase transitions and sum with sensible heating segments separately.
Typically, \(q\) in J, \(m\) in g, \(c\) in \(\mathrm{J\,g^{-1}\,^\circ C^{-1}}\). Convert consistently for kg/K units.
Include a calorimeter constant or perform calibration runs to account for system heat capacity.
Coffee-cup calorimetry assumes approximately constant pressure; \(q\) approximates solution enthalpy change.
Heat losses, evaporation, or measurement errors may shift \(T_f\). Stirring and insulation reduce discrepancies.
Yes. Compute \(c\) experimentally and compare with reference tables, accounting for uncertainties.
Use the least precise measurement, typically 2–3 significant figures in introductory labs.
For small temperature ranges, \(c\) is treated as constant; larger ranges may require temperature-dependent values.
Heat \(q\) is energy transfer; temperature reflects average kinetic energy. They are related but distinct concepts.