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Metals - Specific Heats

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The specific heat of metals and metalloids (semimetals) are given in the table below.

See also tabulated values for gases, food and foodstuff, common liquids and fluids, common solids and other common substances as well as values of molar specific heat for common organic substances and inorganic substances.

Metals - Specific Heats
MetalSpecific Heat
- cp -
(kJ/(kg K))



Aluminum 0.91
Antimony 0.21
Barium 0.20
Beryllium 1.83
Bismuth 0.13
Cadmium 0.23
Calsium 0.63
Carbon Steel 0.49
Cast Iron 0.46
Cesium 0.24
Chromium 0.46
Cobalt 0.42
Copper 0.39
Gallium 0.37
Germanium 0.32
Gold 0.13
Hafnium 0.14
Indium 0.24
Iridium 0.13
Iron 0.45
Lanthanum 0.195
Lead 0.13
Lithium 3.57
Lutetium 0.15
Magnesium 1.05
Manganese 0.48
Mercury 0.14
Molybdenum 0.25
Nickel 0.44
Niobium (Columbium) 0.27
Osmium 0.13
Palladium 0.24
Platinum 0.13
Plutonium 0.13
Potassium 0.75
Rhenium 0.14
Rhodium 0.24
Rubidium 0.36
Ruthenium 0.24
Scandium 0.57
Selenium 0.32
Silicon 0.71
Silver 0.23
Sodium 1.21
Strontium 0.30
Tantalum 0.14
Thallium 0.13
Thorium 0.13
Tin 0.21
Titanium 0.54
Tungsten 0.13
Uranium 0.12
Vanadium 0.39
Yttrium 0.30
Zinc 0.39
Zirconium 0.27
Wrought Iron 0.50

Metalloids - also known as semimetals - are elements containing properties similar and midway between metals and nonmetals.

  • 1 J/(kg K) = 2.389x10-4 kcal/(kg oC) = 2.389x10-4 Btu/(lbm oF)
  • 1 kJ/(kg K) = 0.2389 kcal/(kg oC) = 0.2389 Btu/(lbm oF) = 103 J/(kg oC) = 1 J/(g oC)
  • 1 Btu/(lbm oF) = 4186.8 J/ (kg K) = 1 kcal/(kg oC)
  • 1 kcal/(kg oC) = 4186.8 J/ (kg K) = 1 Btu/(lbm oF)

For conversion of units, use the Specific heat online unit converter.

See also tabulated values for Gases, Food and foodstuffCommon liquids and fluids, Common solids and other Common substances as well as values of molar specific heat for common organic substances and inorganic substances.

Heating Energy

The energy required to heat a product can be calculated as

q = cp m dt                                           (1)

where

q = heat required (kJ)

cp = specific heat (kJ/kg K, kJ/kg C°)

dt = temperature difference (K, C°)

Example - Heating Carbon Steel

2 kg of carbon steel is heated from 20 oC to 100 oC. The specific heat of carbon steel is 0.49 kJ/kgC° and the heat required can be calculated as

q = (0.49 kJ/kg oC) (2 kg) ((100 oC) - (20 oC)) 

     = 78.4 (kJ)

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