Specific Heat Calculator
Fast, accurate, and free online Specific Heat Calculator tool that runs directly in your browser.
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Specific Heat Calculator
Calculate the thermal energy, mass, temperature or specific heat capacity of a substance in seconds.
Enter the data on the left to see the result.
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Other tools you may find usefulSpecific heat calculator - calculate Q, ΔT, m or c
This tool calculates the thermal energy balance based on the equationQ = m · c · ΔT. The interface allows you to indicate which quantity is unknown (mode: Q, dT, m, c) and automatically converts the data between units:min kg or g,ΔTin K,Qin J or kJ. The additional parameterprecisioncontrols the number of decimal places in the results.
Description of fields and operating modes
- mode– selection of an unknown:Q(heat quantity),dT(temperature difference),m(mass) orc(specific heat).
- m– sample mass. Unitm_unit: kg or g. Conversion to SI is automatic.
- c– specific heat in units ofJ/(kg·K). Reference values for materials can be found below.
- ΔT– temperature difference. UnitdT_unit: K. In practice, ΔT in K is equivalent to the change in °C.
- Q– amount of heat supplied or received. UnitQ_unit: J or kJ.
- precision– number of decimal places in the results (0–6).
Results are returned in the output fieldsout_Q, out_dT, out_m, out_caccording to the selected mode.
Formulas and theory
Basic relationship
Q = m · c · ΔT
whereQis thermal energy [J],m– mass [kg],c– specific heat [J/(kg · K)], andΔT– temperature difference [K]. The relationship assumes no phase changes and constantscin the considered temperature range.
Conversions to calculator modes
Q mode: knownm, c, ΔT
Q = m · c · ΔT
dT mode: knownm, c, Q
ΔT = Q / (m · c)
Mode m: knownQ, c, ΔT
m = Q / (c · ΔT)
Mode c: knownQ, m, ΔT
c = Q / (m · ΔT)
Model assumptions
- Energy-closed system: no heat loss to the environment.
- No phase transformations in the considered temperature range.
- The specific heatcis constant in the considered range ΔT.
Units and conversions
| Size | Symbol | Units available | Default |
|---|---|---|---|
| Weight | m | kg, g (1 kg = 1000 g) | kg |
| Difference temperatures | ΔT | K | K |
| Heat quantity | Q | J, kJ (1 kJ = 1000 J) | J |
| Specific heat | c | J/(kg·K) | J/(kg·K) |
Specific heat reference values
| Material (approx.) | c [J/(kg K)] | Notes |
|---|---|---|
| Water (20°C) | 4180 | High thermal capacity, good for heat storage |
| Air (dry, 25°C) | 1005 | Under constant conditions pressure |
| Aluminum | 900 | Light structural metal |
| Copper | 385 | Very good thermal conductivity |
| Steel | ~460 | Value depending on the alloy |
| Concrete | ~880 | Variable material composition |
| Wood (dry) | 1200–2400 | Strongly dependent on humidity and species |
Calculation examples
Example 1 - calculation of Q
- m = 2.5 kg
- c = 900 J/(kg·K) (aluminum)
- ΔT = 35 K
Q= 2.5 · 900 · 35 = 78,750 J = 78.75 kJ. This is how much energy is needed to heat 2.5 kg of aluminum by 35 K.
Example 2 - determining ΔT
- m = 0.8 kg
- c = 4180 J/(kg·K) (water)
- Q = 100 kJ
ΔT= 100,000 / (0.8 · 4180) ≈ 29.9 K. A portion of water will increase the temperature by approximately 30°C.
Example 3 – mass determination
- c = 385 J/(kg·K) (copper)
- ΔT = 50 K
- Q = 10 kJ
m= 10,000 / (385 · 50) ≈ 0.519 kg. The weight of the copper sample is approximately 0.52 kg.
Example 4 - determination of c
- m = 150 g = 0.15 kg
- ΔT = 20 K
- Q = 2.1 kJ
c= 2100 / (0.15 · 20) = 700 J/(kg · K). Approximate value for polymer composite.
Common pitfalls and good practices
- SI consistent units- make sure mass is in kg, ΔT in K, and Q in J. The calculator performs conversions, but the input data must match the selected unit.
- ΔT instead of absolute temperature– in the case of a temperature difference, ΔT in K is equivalent to ΔT in °C.
- No phase change– if melting, boiling or freezing occurs in the ΔT range, heat of change is needed.
- Specific heat variable with T– for large ΔT, consider interval averagingc.
- Losses to the environment– in real systems they may be significant. Enter the energy reserve or safety factor.
Table of engineering mini-cases
| Scenario | Data | Purpose of calculations | Key result |
|---|---|---|---|
| Heating the tank | m, c, ΔT | Determine the energy of the heater | Q = m·c·ΔT |
| Material test | Q, m, ΔT | Determine c material | c = Q/(m·ΔT) |
| Selection of heat accumulator | Q, c, ΔT | Determine the required mass | m = Q/(c·ΔT) |
| Superheat Analysis | Q, m, c | Estimate Temperature Rise | ΔT = Q/(m·c) |
How to use the calculator
- Select in the boxmodeunknown:Q, dT, morc.
- Enter known values in the appropriate fields:m, c, dT, Q.
- Set units:m_unit(kg/g),dT_unit(K),Q_unit(J/kJ).
- Setprecisionto match the output format.
- Click "Calculate". Read the appropriate result field:out_Q, out_dT, out_morout_c.
Practical applications
- Selection of heater power and heating times in industrial processes.
- Assessment of thermal inertia of electronic and mechanical components.
- Estimation of energy demand in heating and cooling systems.
- Didactic research on the thermal properties of materials.
- Preliminary energy balances before CFD/FEM simulations.
FAQ
Can I enter temperatures in °C?
Yes. The calculator operates on the temperature differenceΔTin K. If you have the initial and final temperatures in °C, enter their difference - in terms of value it corresponds to K.
What about melting or boiling?
The equationQ = m·c·ΔTdoes not take into account the heat of transformation. Add the termm·L(L – heat of fusion/vaporization) for the segments where a phase transition occurs.
Why is the result different from the measurement?
In real systems, there are losses to the environment, material inhomogeneity, non-linearc(T)and measurement errors. Account for adjustments and uncertainties.
How to select ΔT to limit overheating?
Increase the mass or use a material with a largerc. Alternatively, reduce the suppliedQor use power control.
Are kJ and kcal the same thing?
No. 1 kcal ≈ 4.1868 kJ. The calculator works in J/kJ; you can convert to kcal after obtaining the result.
Summary
The specific heat calculator simplifies the calculation of energy, mass, temperature difference or specific heat for homogeneous samples. It is based on a standard equation and consistent SI units, and flexibleQ/dT/m/cmodes allow you to solve a variety of engineering and educational tasks. With caution towards phase transitions and nonlinearitiesc(T)the tool provides reliable results for preliminary analyzes and design decisions.