Specific Heat Calculator (2024)

1. Determine whether you want to warm up the sample (give it some thermal energy) or cool it down (take some thermal energy away).

2. Insert the amount of energy supplied as a positive value. If you want to cool down the sample, insert the subtracted energy as a negative value. For example, say that we want to reduce the sample's thermal energy by 63,000 J. Then $Q = -63,\!000 \ \text{J}$Q=−63,000J.

   See Also

   Natuurkunde.nl - Q =m*C* delta T- Rekenen, formules en warmtepomp feitenFormules: q=m*c*delta t - WetenschapsforumNatuurkunde.nl - Formules snelheden: wanneer delta-t en wanneer t

3. Decide the temperature difference between the initial and final state of the sample and type it into the heat capacity calculator. If the sample is cooled down, the difference will be negative, and if warmed up - positive. Let's say we want to cool the sample down by 3 degrees. Then $\Delta T = -3 \ \text{K}$ΔT=−3K. You can also select the Show initial and final temperatures checkbox to type the initial and final values of temperature manually.

4. Determine the mass of the sample. We will assume $m = 5 \ \text{kg}$m=5kg.

5. Calculate specific heat as $c = \frac{Q}{m \Delta T}$c=mΔTQ​. In our example, it will be equal to:

   $c = \mathrm{\frac{-63,000 \ J}{5 \ kg \cdot \ -3 \ K}} = \mathrm{4,\!200 \ \frac{J}{kg \cdot K}}$c=5kg⋅−3K−63,000J​=4,200kg⋅KJ​

   This is the typical heat capacity of water.

You don't need to use the heat capacity calculator for most common substances. The values of specific heat for some of the most popular ones are listed below.

• Ice: $\mathrm{2,\!100 \ {J}/{kg\! \cdot\! K}}$2,100J/kg⋅K
• Water: $\mathrm{4,\!200 \ {J}/{kg\! \cdot\! K}}$4,200J/kg⋅K
• Water vapor: $\mathrm{2,\!000 \ {J}/{kg\! \cdot\! K}}$2,000J/kg⋅K
• Basalt: $\mathrm{840 \ {J}/{kg\! \cdot\! K}}$840J/kg⋅K
• Granite: $\mathrm{790 \ {J}/{kg\! \cdot\! K}}$790J/kg⋅K
• Aluminum: $\mathrm{890 \ {J}/{kg\! \cdot\! K}}$890J/kg⋅K
• Iron: $\mathrm{450 \ {J}/{kg\! \cdot\! K}}$450J/kg⋅K
• Copper: $\mathrm{380 \ {J}/{kg\! \cdot\! K}}$380J/kg⋅K
• Lead: $\mathrm{130 \ {J}/{kg\! \cdot\! K}}$130J/kg⋅K

The specific heat capacity is the heat or energy required to change one unit mass of a substance of a constant volume by 1 °C. The formula is Cv = Q / (ΔT × m).

The formula for specific heat capacity, C, of a substance with mass m, is C = Q /(m × ΔT). Where Q is the energy added and ΔT is the change in temperature. The specific heat capacity during different processes, such as constant volume, Cv and constant pressure, Cp, are related to each other by the specific heat ratio, ɣ= Cp/Cv, or the gas constant R = Cp - Cv.

Specific heat capacity is measured in J/kg·K or J/kg·°C, as it is the heat or energy required during a constant volume process to change the temperature of a substance of unit mass by 1 °C or 1 K.

The specific heat of water at 25 °C is 4,181.3 J/kg·K, the amount of heat required to raise the temperature of 1 kg of water by 1 Kelvin.

Specific heat is measured in BTU / lb °F in imperial units and in J/kg·K in SI units.

The specific heat of copper is 385 J/kg·K. You can use this value to estimate the energy required to heat a 100 g of copper by 5 °C, i.e., Q = m × Cp × ΔT = 0.1 × 385 × 5 = 192.5 J.

The specific heat of aluminum is 897 J/kg K. This value is almost 2.3 times of the specific heat of copper. You can use this value to estimate the energy required to heat a 500 g of aluminum by 5 °C, i.e., Q = m × Cp × ΔT = 0.5 × 897 × 5 = 2242.5 J.