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).

Beta version#

BETA TEST VERSION OF THIS ITEM
This online calculator is currently under heavy development. It may or it may NOT work correctly.
You CAN try to use it. You CAN even get the proper results.
However, please VERIFY all results on your own, as the level of completion of this item is NOT CONFIRMED.
Feel free to send any ideas and comments !

Symbolic algebra

ⓘ Hint: This calculator supports symbolic math. You can enter numbers, but also symbols like a, b, pi or even whole math expressions such as (a+b)/2. If you still don't sure how to make your life easier using symbolic algebra check out our another page: Symbolic calculations

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

Value
Unit
Decimals

$4189.9$ (joule / kilogram / Kelvin) is equal to:#

Popular (temperature in Kelvins)#

Unit	Symbol	Symbol (plain text)	Value as symbolic	Value as numeric	Notes	Unit conversion formula
joule / kilogram / Kelvin	Show source $\frac{J}{kg\times K}$	J / (kg × K)	Show source $\text{...}$	-	The 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).	Show source $...$
joule / gram / Kelvin	Show source $\frac{J}{g\times K}$	J / (g × K)	Show source $\text{...}$	-	The 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). $1\ \frac{J}{g \times K} = \frac{1\ J}{0.001\ kg \times K} = \frac{1000\ J}{kg \times K}$	Show source $...$
calorie (thermochemical) / kilogram / Kelvin	Show source $\frac{cal_{th}}{kg\times K}$	cal_th / (kg × K)	Show source $\text{...}$	-	The basic specific heat unit in the SI system. The substance has specific heat of one joule per kilograme per kelwin (1 cal_TH/kg/K) if heating one kilograme sample (1 kg) by one kelvin (1 K) requires delivery of one joule of energy (1 cal_TH). $1\ \frac{cal_{TH}}{kg \times K} = \frac{4.184\ J}{kg \times K}$	Show source $...$
calorie (thermochemical) / gram / Kelvin	Show source $\frac{cal_{th}}{g\times K}$	cal_th / (g × K)	Show source $\text{...}$	-	The basic specific heat unit in the SI system. The substance has specific heat of one joule per grame per kelwin (1 cal_TH/g/K) if heating one grame sample (1 g) by one kelvin (1 K) requires delivery of one joule of energy (1 cal_TH). $1\ \frac{cal_{TH}}{g \times K} = \frac{1\ cal_{TH}}{0.001\ kg \times K} = \frac{1000\ cal_{TH}}{kg \times K}$	Show source $...$
calorie (International Table) / kilogram / Kelvin	Show source $\frac{cal_{IT}}{kg\times K}$	cal_IT / (kg × K)	Show source $\text{...}$	-	The basic specific heat unit in the SI system. The substance has specific heat of one joule per kilograme per kelwin (1 cal_IT/kg/K) if heating one kilograme sample (1 kg) by one kelvin (1 K) requires delivery of one joule of energy (1 cal_IT). $1\ \frac{cal_{IT}}{kg \times K} = \frac{4.1868\ J}{kg \times K}$	Show source $...$
calorie (International Table) / gram / Kelvin	Show source $\frac{cal_{IT}}{g\times K}$	cal_IT / (g × K)	Show source $\text{...}$	-	The basic specific heat unit in the SI system. The substance has specific heat of one joule per grame per kelwin (1 cal_IT/g/K) if heating one grame sample (1 g) by one kelvin (1 K) requires delivery of one joule of energy (1 cal_IT). $1\ \frac{cal_{IT}}{g \times K} = \frac{1\ cal_{IT}}{0.001\ kg \times K} = \frac{1000\ cal_{IT}}{kg \times K}$	Show source $...$
kilocalorie / kilogram / Kelvin	Show source $\frac{kcal}{kg\times K}$	kcal / (kg × K)	Show source $\text{...}$	-	The 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). $1\ \frac{kcal}{kg \times K} = \frac{1000\ cal_{IT}}{kg \times K}$	Show source $...$
kilocalorie / gram / Kelvin	Show source $\frac{kcal}{g\times K}$	kcal / (g × K)	Show source $\text{...}$	-	The 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). $1\ \frac{kcal}{g \times K} = \frac{\cancel{1000}\ cal_{IT}}{\cancel{0.001}\ kg \times K} = 1\ \frac{cal_{IT}}{kg \times K}$	Show source $...$

Popular (temperature in Celsius degrees)#

Unit	Symbol	Symbol (plain text)	Value as symbolic	Value as numeric	Notes	Unit conversion formula
joule / kilogram / degree Celsius	Show source $\frac{J}{kg\times ^\circ C}$	J / (kg × °C)	Show source $\text{...}$	-	The 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). $1\ \frac{J}{kg \times ^{\circ}C} = 1\ \frac{J}{kg \times K}$	Show source $...$
joule / gram / degree Celsius	Show source $\frac{J}{g\times ^\circ C}$	J / (g × °C)	Show source $\text{...}$	-	The 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). $1\ \frac{J}{g \times ^{\circ}C} = \frac{1\ J}{0.001\ kg \times ^{\circ}C} = \frac{1000\ J}{kg \times ^{\circ}C}$	Show source $...$
calorie (thermochemical) / kilogram / degree Celsius	Show source $\frac{cal_{th}}{kg\times ^\circ C}$	cal_th / (kg × °C)	Show source $\text{...}$	-	The non-SI specific heat unit. The substance has specific heat of one joule per kilograme per celsius degree (1 cal_TH/kg/F) if heating one kilograme sample (1 kg) by one Celsius degree (1 K) requires delivery of one joule of energy (1 cal_TH). $1\ \frac{cal_{TH}}{kg \times ^{\circ}C} = \frac{4.184\ J}{kg \times ^{\circ}C}$	Show source $...$
calorie (thermochemical) / gram / degree Celsius	Show source $\frac{cal_{th}}{g\times ^\circ C}$	cal_th / (g × °C)	Show source $\text{...}$	-	The non-SI specific heat unit. The substance has specific heat of one joule per grame per celsius degree (1 cal_TH/g/F) if heating one grame sample (1 g) by one Celsius degree (1 K) requires delivery of one joule of energy (1 cal_TH). $1\ \frac{cal_{TH}}{g \times ^{\circ}C} = \frac{1\ cal_{TH}}{0.001\ kg \times ^{\circ}C} = \frac{1000\ cal_{TH}}{kg \times ^{\circ}C}$	Show source $...$
calorie (International Table) / kilogram / degree Celsius	Show source $\frac{cal_{IT}}{kg\times ^\circ C}$	cal_IT / (kg × °C)	Show source $\text{...}$	-	The non-SI specific heat unit. The substance has specific heat of one joule per kilograme per celsius degree (1 cal_IT/kg/F) if heating one kilograme sample (1 kg) by one Celsius degree (1 K) requires delivery of one joule of energy (1 cal_IT). $1\ \frac{cal_{IT}}{kg \times ^{\circ}C} = \frac{4.1868\ J}{kg \times ^{\circ}C}$	Show source $...$
calorie (International Table) / gram / degree Celsius	Show source $\frac{cal_{IT}}{g\times ^\circ C}$	cal_IT / (g × °C)	Show source $\text{...}$	-	The non-SI specific heat unit. The substance has specific heat of one joule per grame per celsius degree (1 cal_IT/g/F) if heating one grame sample (1 g) by one Celsius degree (1 K) requires delivery of one joule of energy (1 cal_IT). $1\ \frac{cal_{IT}}{g \times ^{\circ}C} = \frac{1\ cal_{IT}}{0.001\ kg \times ^{\circ}C} = \frac{1000\ cal_{IT}}{kg \times ^{\circ}C}$	Show source $...$
kilocalorie / kilogram / degree Celsius	Show source $\frac{kcal}{kg\times ^\circ C}$	kcal / (kg × °C)	Show source $\text{...}$	-	The 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). $1\ \frac{kcal}{kg \times ^{\circ}C} = \frac{1000\ cal_{IT}}{kg \times ^{\circ}C}$	Show source $...$
kilocalorie / gram / degree Celsius	Show source $\frac{kcal}{g\times ^\circ C}$	kcal / (g × °C)	Show source $\text{...}$	-	The 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). $1\ \frac{kcal}{g \times ^{\circ}C} = \frac{\cancel{1000}\ cal_{IT}}{\cancel{0.001}\ kg \times ^{\circ}C} = 1\ \frac{cal_{IT}}{kg \times ^{\circ}C}$	Show source $...$

Popular (temperature in Fahrenheit degrees)#

Unit	Symbol	Symbol (plain text)	Value as symbolic	Value as numeric	Notes	Unit conversion formula
joule / kilogram / degree Fahrenheit	Show source $\frac{J}{kg\times ^\circ F}$	J / (kg × °F)	Show source $\text{...}$	-	The 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). $1\ \frac{J}{kg \times ^{\circ}F} = 1.8\ \frac{J}{kg \times K}$	Show source $...$
joule / gram / degree Fahrenheit	Show source $\frac{J}{g\times ^\circ F}$	J / (g × °F)	Show source $\text{...}$	-	The 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). $1\ \frac{J}{g \times ^{\circ}F} = \frac{1\ J}{0.001\ kg \times ^{\circ}F} = \frac{1000\ J}{kg \times ^{\circ}F}$	Show source $...$
calorie (thermochemical) / kilogram / degree Fahrenheit	Show source $\frac{cal_{th}}{kg\times ^\circ F}$	cal_th / (kg × °F)	Show source $\text{...}$	-	The non-SI specific heat unit. The substance has specific heat of one joule per kilograme per Fahrenheit degree (1 cal_TH/kg/F) if heating one kilograme sample (1 kg) by one Fahrenheit degree (1 K) requires delivery of one joule of energy (1 cal_TH). $1\ \frac{cal_{TH}}{kg \times ^{\circ}F} = \frac{4.184\ F}{kg \times ^{\circ}F}$	Show source $...$
calorie (thermochemical) / gram / degree Fahrenheit	Show source $\frac{cal_{th}}{g\times ^\circ F}$	cal_th / (g × °F)	Show source $\text{...}$	-	The non-SI specific heat unit. The substance has specific heat of one joule per grame per Fahrenheit degree (1 cal_TH/g/F) if heating one grame sample (1 g) by one Fahrenheit degree (1 K) requires delivery of one joule of energy (1 cal_TH). $1\ \frac{cal_{TH}}{g \times ^{\circ}F} = \frac{1\ cal_{TH}}{0.001\ kg \times ^{\circ}F} = \frac{1000\ cal_{TH}}{kg \times ^{\circ}F}$	Show source $...$
calorie (International Table) / kilogram / degree Fahrenheit	Show source $\frac{cal_{IT}}{kg\times ^\circ F}$	cal_IT / (kg × °F)	Show source $\text{...}$	-	The non-SI specific heat unit. The substance has specific heat of one joule per kilograme per Fahrenheit degree (1 cal_IT/kg/F) if heating one kilograme sample (1 kg) by one Fahrenheit degree (1 K) requires delivery of one joule of energy (1 cal_IT). $1\ \frac{cal_{IT}}{kg \times ^{\circ}F} = \frac{4.1868\ J}{kg \times ^{\circ}F}$	Show source $...$
calorie (International Table) / gram / degree Fahrenheit	Show source $\frac{cal_{IT}}{g\times ^\circ F}$	cal_IT / (g × °F)	Show source $\text{...}$	-	The non-SI specific heat unit. The substance has specific heat of one joule per grame per Fahrenheit degree (1 cal_IT/g/F) if heating one grame sample (1 g) by one Fahrenheit degree (1 K) requires delivery of one joule of energy (1 cal_IT). $1\ \frac{cal_{IT}}{g \times ^{\circ}F} = \frac{1\ cal_{IT}}{0.001\ kg \times ^{\circ}F} = \frac{1000\ cal_{IT}}{kg \times ^{\circ}F}$	Show source $...$
kilocalorie / kilogram / degree Fahrenheit	Show source $\frac{kcal}{kg\times ^\circ F}$	kcal / (kg × °F)	Show source $\text{...}$	-	The 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). $1\ \frac{kcal}{kg \times ^{\circ}F} = \frac{1000\ cal_{IT}}{kg \times ^{\circ}F}$	Show source $...$
kilocalorie / gram / degree Fahrenheit	Show source $\frac{kcal}{g\times ^\circ F}$	kcal / (g × °F)	Show source $\text{...}$	-	The 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). $1\ \frac{kcal}{g \times ^{\circ}F} = \frac{\cancel{1000}\ cal_{IT}}{\cancel{0.001}\ kg \times ^{\circ}F} = 1\ \frac{cal_{IT}}{kg \times ^{\circ}F}$	Show source $...$

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:

$\dfrac{J}{kg × K}$
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:

$c = \dfrac{\Delta Q}{m \times \Delta T}$
where:
- $c$ - specific heat of the substance,
- $\Delta Q$ - amount of heat delivered,
- $m$ - mass of the heated sample,
- $\Delta T$ - noted temperature change.

Due to the high compressibility of gases, two values of specific heat are given in their case:
- $C_p$ - specific heat under constant pressure,
- $C_v$ - 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.