Converter of specific heat units
Specific heat capacity units converter - converts units based on various energy units (joules, calories, kilocalories), temperature units (Kelvins, Celsius or Fahrenheit degrees) and mass units (grams, kilograms).

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Inputs data - value and unit, which we're going to convert#

Value
Unit
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#

Popular (temperature in Kelvins)#

UnitSymbolSymbol
(plain text)
ValueNotes
joule / kilogram / KelvinShow sourceJkg×K\frac{J}{kg\times K}J / (kg × K)4189.9The basic specific heat unit in the SI system. The substance has specific heat of one joule per kilograme per kelwin (1 J/kg/K) if heating one kilograme sample (1 kg) by one kelvin (1 K) requires delivery of one joule of energy (1 J).
joule / gram / KelvinShow sourceJg×K\frac{J}{g\times K}J / (g × K)4.1899The basic specific heat unit in the SI system. The substance has specific heat of one joule per grame per kelwin (1 J/g/K) if heating one grame sample (1 g) by one kelvin (1 K) requires delivery of one joule of energy (1 J).1 Jg×K=1 J0.001 kg×K=1000 Jkg×K1\ \frac{J}{g \times K} = \frac{1\ J}{0.001\ kg \times K} = \frac{1000\ J}{kg \times K}
calorie (thermochemical) / kilogram / KelvinShow sourcecalthkg×K\frac{cal_{th}}{kg\times K}calth / (kg × K)1001.41013384The basic specific heat unit in the SI system. The substance has specific heat of one joule per kilograme per kelwin (1 calTH/kg/K) if heating one kilograme sample (1 kg) by one kelvin (1 K) requires delivery of one joule of energy (1 calTH).1 calTHkg×K=4.184 Jkg×K1\ \frac{cal_{TH}}{kg \times K} = \frac{4.184\ J}{kg \times K}
calorie (thermochemical) / gram / KelvinShow sourcecalthg×K\frac{cal_{th}}{g\times K}calth / (g × K)1.001410134The basic specific heat unit in the SI system. The substance has specific heat of one joule per grame per kelwin (1 calTH/g/K) if heating one grame sample (1 g) by one kelvin (1 K) requires delivery of one joule of energy (1 calTH).1 calTHg×K=1 calTH0.001 kg×K=1000 calTHkg×K1\ \frac{cal_{TH}}{g \times K} = \frac{1\ cal_{TH}}{0.001\ kg \times K} = \frac{1000\ cal_{TH}}{kg \times K}
calorie (International Table) / kilogram / KelvinShow sourcecalITkg×K\frac{cal_{IT}}{kg\times K}calIT / (kg × K)1000.74042228The basic specific heat unit in the SI system. The substance has specific heat of one joule per kilograme per kelwin (1 calIT/kg/K) if heating one kilograme sample (1 kg) by one kelvin (1 K) requires delivery of one joule of energy (1 calIT).1 calITkg×K=4.1868 Jkg×K1\ \frac{cal_{IT}}{kg \times K} = \frac{4.1868\ J}{kg \times K}
calorie (International Table) / gram / KelvinShow sourcecalITg×K\frac{cal_{IT}}{g\times K}calIT / (g × K)1.000740422The basic specific heat unit in the SI system. The substance has specific heat of one joule per grame per kelwin (1 calIT/g/K) if heating one grame sample (1 g) by one kelvin (1 K) requires delivery of one joule of energy (1 calIT).1 calITg×K=1 calIT0.001 kg×K=1000 calITkg×K1\ \frac{cal_{IT}}{g \times K} = \frac{1\ cal_{IT}}{0.001\ kg \times K} = \frac{1000\ cal_{IT}}{kg \times K}
kilocalorie / kilogram / KelvinShow sourcekcalkg×K\frac{kcal}{kg\times K}kcal / (kg × K)1.000740422The basic specific heat unit in the SI system. The substance has specific heat of one joule per kilograme per kelwin (1 kcal/kg/K) if heating one kilograme sample (1 kg) by one kelvin (1 K) requires delivery of one joule of energy (1 kcal).1 kcalkg×K=1000 calITkg×K1\ \frac{kcal}{kg \times K} = \frac{1000\ cal_{IT}}{kg \times K}
kilocalorie / gram / KelvinShow sourcekcalg×K\frac{kcal}{g\times K}kcal / (g × K)0.00100074The basic specific heat unit in the SI system. The substance has specific heat of one joule per grame per kelwin (1 kcal/g/K) if heating one grame sample (1 g) by one kelvin (1 K) requires delivery of one joule of energy (1 kcal).1 kcalg×K=1000 calIT0.001 kg×K=1 calITkg×K1\ \frac{kcal}{g \times K} = \frac{\cancel{1000}\ cal_{IT}}{\cancel{0.001}\ kg \times K} = 1\ \frac{cal_{IT}}{kg \times K}

Popular (temperature in Celsius degrees)#

UnitSymbolSymbol
(plain text)
ValueNotes
joule / kilogram / degree CelsiusShow sourceJkg×C\frac{J}{kg\times ^\circ C}J / (kg × °C)4189.9The non-SI specific heat unit. The substance has specific heat of one joule per kilograme per celsius degree (1 J/kg/F) if heating one kilograme sample (1 kg) by one Celsius degree (1 K) requires delivery of one joule of energy (1 J).1 Jkg×C=1 Jkg×K1\ \frac{J}{kg \times ^{\circ}C} = 1\ \frac{J}{kg \times K}
joule / gram / degree CelsiusShow sourceJg×C\frac{J}{g\times ^\circ C}J / (g × °C)4.1899The non-SI specific heat unit. The substance has specific heat of one joule per grame per celsius degree (1 J/g/F) if heating one grame sample (1 g) by one Celsius degree (1 K) requires delivery of one joule of energy (1 J).1 Jg×C=1 J0.001 kg×C=1000 Jkg×C1\ \frac{J}{g \times ^{\circ}C} = \frac{1\ J}{0.001\ kg \times ^{\circ}C} = \frac{1000\ J}{kg \times ^{\circ}C}
calorie (thermochemical) / kilogram / degree CelsiusShow sourcecalthkg×C\frac{cal_{th}}{kg\times ^\circ C}calth / (kg × °C)1001.41013384The non-SI specific heat unit. The substance has specific heat of one joule per kilograme per celsius degree (1 calTH/kg/F) if heating one kilograme sample (1 kg) by one Celsius degree (1 K) requires delivery of one joule of energy (1 calTH).1 calTHkg×C=4.184 Jkg×C1\ \frac{cal_{TH}}{kg \times ^{\circ}C} = \frac{4.184\ J}{kg \times ^{\circ}C}
calorie (thermochemical) / gram / degree CelsiusShow sourcecalthg×C\frac{cal_{th}}{g\times ^\circ C}calth / (g × °C)1.001410134The non-SI specific heat unit. The substance has specific heat of one joule per grame per celsius degree (1 calTH/g/F) if heating one grame sample (1 g) by one Celsius degree (1 K) requires delivery of one joule of energy (1 calTH).1 calTHg×C=1 calTH0.001 kg×C=1000 calTHkg×C1\ \frac{cal_{TH}}{g \times ^{\circ}C} = \frac{1\ cal_{TH}}{0.001\ kg \times ^{\circ}C} = \frac{1000\ cal_{TH}}{kg \times ^{\circ}C}
calorie (International Table) / kilogram / degree CelsiusShow sourcecalITkg×C\frac{cal_{IT}}{kg\times ^\circ C}calIT / (kg × °C)1000.74042228The non-SI specific heat unit. The substance has specific heat of one joule per kilograme per celsius degree (1 calIT/kg/F) if heating one kilograme sample (1 kg) by one Celsius degree (1 K) requires delivery of one joule of energy (1 calIT).1 calITkg×C=4.1868 Jkg×C1\ \frac{cal_{IT}}{kg \times ^{\circ}C} = \frac{4.1868\ J}{kg \times ^{\circ}C}
calorie (International Table) / gram / degree CelsiusShow sourcecalITg×C\frac{cal_{IT}}{g\times ^\circ C}calIT / (g × °C)1.000740422The non-SI specific heat unit. The substance has specific heat of one joule per grame per celsius degree (1 calIT/g/F) if heating one grame sample (1 g) by one Celsius degree (1 K) requires delivery of one joule of energy (1 calIT).1 calITg×C=1 calIT0.001 kg×C=1000 calITkg×C1\ \frac{cal_{IT}}{g \times ^{\circ}C} = \frac{1\ cal_{IT}}{0.001\ kg \times ^{\circ}C} = \frac{1000\ cal_{IT}}{kg \times ^{\circ}C}
kilocalorie / kilogram / degree CelsiusShow sourcekcalkg×C\frac{kcal}{kg\times ^\circ C}kcal / (kg × °C)1.000740422The non-SI specific heat unit. The substance has specific heat of one joule per kilograme per celsius degree (1 kcal/kg/F) if heating one kilograme sample (1 kg) by one Celsius degree (1 K) requires delivery of one joule of energy (1 kcal).1 kcalkg×C=1000 calITkg×C1\ \frac{kcal}{kg \times ^{\circ}C} = \frac{1000\ cal_{IT}}{kg \times ^{\circ}C}
kilocalorie / gram / degree CelsiusShow sourcekcalg×C\frac{kcal}{g\times ^\circ C}kcal / (g × °C)0.00100074The non-SI specific heat unit. The substance has specific heat of one joule per grame per celsius degree (1 kcal/g/F) if heating one grame sample (1 g) by one Celsius degree (1 K) requires delivery of one joule of energy (1 kcal).1 kcalg×C=1000 calIT0.001 kg×C=1 calITkg×C1\ \frac{kcal}{g \times ^{\circ}C} = \frac{\cancel{1000}\ cal_{IT}}{\cancel{0.001}\ kg \times ^{\circ}C} = 1\ \frac{cal_{IT}}{kg \times ^{\circ}C}

Popular (temperature in Fahrenheit degrees)#

UnitSymbolSymbol
(plain text)
ValueNotes
joule / kilogram / degree FahrenheitShow sourceJkg×F\frac{J}{kg\times ^\circ F}J / (kg × °F)2327.72222222The non-SI specific heat unit. The substance has specific heat of one joule per kilograme per Fahrenheit degree (1 J/kg/F) if heating one kilograme sample (1 kg) by one Fahrenheit degree (1 K) requires delivery of one joule of energy (1 J).1 Jkg×F=1.8 Jkg×K1\ \frac{J}{kg \times ^{\circ}F} = 1.8\ \frac{J}{kg \times K}
joule / gram / degree FahrenheitShow sourceJg×F\frac{J}{g\times ^\circ F}J / (g × °F)2.327722222The non-SI specific heat unit. The substance has specific heat of one joule per grame per Fahrenheit degree (1 J/g/F) if heating one grame sample (1 g) by one Fahrenheit degree (1 K) requires delivery of one joule of energy (1 J).1 Jg×F=1 J0.001 kg×F=1000 Jkg×F1\ \frac{J}{g \times ^{\circ}F} = \frac{1\ J}{0.001\ kg \times ^{\circ}F} = \frac{1000\ J}{kg \times ^{\circ}F}
calorie (thermochemical) / kilogram / degree FahrenheitShow sourcecalthkg×F\frac{cal_{th}}{kg\times ^\circ F}calth / (kg × °F)556.338963246The non-SI specific heat unit. The substance has specific heat of one joule per kilograme per Fahrenheit degree (1 calTH/kg/F) if heating one kilograme sample (1 kg) by one Fahrenheit degree (1 K) requires delivery of one joule of energy (1 calTH).1 calTHkg×F=4.184 Fkg×F1\ \frac{cal_{TH}}{kg \times ^{\circ}F} = \frac{4.184\ F}{kg \times ^{\circ}F}
calorie (thermochemical) / gram / degree FahrenheitShow sourcecalthg×F\frac{cal_{th}}{g\times ^\circ F}calth / (g × °F)0.556338963The non-SI specific heat unit. The substance has specific heat of one joule per grame per Fahrenheit degree (1 calTH/g/F) if heating one grame sample (1 g) by one Fahrenheit degree (1 K) requires delivery of one joule of energy (1 calTH).1 calTHg×F=1 calTH0.001 kg×F=1000 calTHkg×F1\ \frac{cal_{TH}}{g \times ^{\circ}F} = \frac{1\ cal_{TH}}{0.001\ kg \times ^{\circ}F} = \frac{1000\ cal_{TH}}{kg \times ^{\circ}F}
calorie (International Table) / kilogram / degree FahrenheitShow sourcecalITkg×F\frac{cal_{IT}}{kg\times ^\circ F}calIT / (kg × °F)555.966901266The non-SI specific heat unit. The substance has specific heat of one joule per kilograme per Fahrenheit degree (1 calIT/kg/F) if heating one kilograme sample (1 kg) by one Fahrenheit degree (1 K) requires delivery of one joule of energy (1 calIT).1 calITkg×F=4.1868 Jkg×F1\ \frac{cal_{IT}}{kg \times ^{\circ}F} = \frac{4.1868\ J}{kg \times ^{\circ}F}
calorie (International Table) / gram / degree FahrenheitShow sourcecalITg×F\frac{cal_{IT}}{g\times ^\circ F}calIT / (g × °F)0.555966901The non-SI specific heat unit. The substance has specific heat of one joule per grame per Fahrenheit degree (1 calIT/g/F) if heating one grame sample (1 g) by one Fahrenheit degree (1 K) requires delivery of one joule of energy (1 calIT).1 calITg×F=1 calIT0.001 kg×F=1000 calITkg×F1\ \frac{cal_{IT}}{g \times ^{\circ}F} = \frac{1\ cal_{IT}}{0.001\ kg \times ^{\circ}F} = \frac{1000\ cal_{IT}}{kg \times ^{\circ}F}
kilocalorie / kilogram / degree FahrenheitShow sourcekcalkg×F\frac{kcal}{kg\times ^\circ F}kcal / (kg × °F)0.555966901The non-SI specific heat unit. The substance has specific heat of one joule per kilograme per Fahrenheit degree (1 kcal/kg/F) if heating one kilograme sample (1 kg) by one Fahrenheit degree (1 K) requires delivery of one joule of energy (1 kcal).1 kcalkg×F=1000 calITkg×F1\ \frac{kcal}{kg \times ^{\circ}F} = \frac{1000\ cal_{IT}}{kg \times ^{\circ}F}
kilocalorie / gram / degree FahrenheitShow sourcekcalg×F\frac{kcal}{g\times ^\circ F}kcal / (g × °F)0.000555967The non-SI specific heat unit. The substance has specific heat of one joule per grame per Fahrenheit degree (1 kcal/g/F) if heating one grame sample (1 g) by one Fahrenheit degree (1 K) requires delivery of one joule of energy (1 kcal).1 kcalg×F=1000 calIT0.001 kg×F=1 calITkg×F1\ \frac{kcal}{g \times ^{\circ}F} = \frac{\cancel{1000}\ cal_{IT}}{\cancel{0.001}\ kg \times ^{\circ}F} = 1\ \frac{cal_{IT}}{kg \times ^{\circ}F}

Some facts#

  • ⓘ Remember: Specific heat is a physical quantity determining the amount of energy (heat) that should be supplied (transferred) to one unit of substance mass to raise the temperature by one unit.
  • The basic specific heat unit in the SI system is joule per kilogram per kelvin:
    Jkg×K\dfrac{J}{kg × K}
  • Specific heat is a property of substance (the so-called material constant). Also, it depends on external conditions: pressure and temperature.
  • The specific heat tells us how difficult it is to heat the given body. Substances with low specific heat change their temperature easily, whereas high ones require much more energy delivered to achieve identical effect.
  • To calculate the specific heat of the selected substance, we can use the following formula:
    c=ΔQm×ΔTc = \dfrac{\Delta Q}{m \times \Delta T}
    where:
    • cc - specific heat of the substance,
    • ΔQ\Delta Q - amount of heat delivered,
    • mm - mass of the heated sample,
    • ΔT\Delta T - noted temperature change.
  • Due to the high compressibility of gases, two values of specific heat are given in their case:
    • CpC_p - specific heat under constant pressure,
    • CvC_v - specific heat under constant volume.
    *) Compressibility is the property of a substance meaning that substance changes volume significantly under pressure.

How to convert#

  • Enter the number to field "value" - enter the NUMBER only, no other words, symbols or unit names. You can use dot (.) or comma (,) to enter fractions.
    Examples:
    • 1000000
    • 123,23
    • 999.99999
  • Find and select your starting unit in field "unit". Some unit calculators have huge number of different units to select from - it's just how complicated our world is...
  • And... you got the result in the table below. You'll find several results for many different units - we show you all results we know at once. Just find the one you're looking for.

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