kilocalorie (IT)/kilogram/K (kcal/(kg*K)) = Btu (th)/pound/°F (Btu (th)/(lb*°F)) / 1.0006698877
To get Kilocalorie it per kilogram per kelvin specific heat capacity, simply divide Btu th per pound per fahrenheit by 1.0006698877. With the help of this specific heat capacity converter, we can easily convert Btu th per pound per fahrenheit to Kilocalorie it per kilogram per kelvin. Here you are provided with the converter, proper definitions,relations in detail along with the online tool to convert Btu (th)/pound/°F (Btu (th)/(lb*°F)) to kilocalorie (IT)/kilogram/K (kcal/(kg*K)).
1 Btu (th)/pound/°F (Btu (th)/(lb*°F)) is 0.99933056073906 kilocalorie (IT)/kilogram/K (kcal/(kg*K)).
Btu (th)/pound/°F (Btu (th)/(lb*°F)) to kilocalorie (IT)/kilogram/K (kcal/(kg*K)) converter is the specific heat capacity converter from one unit to another. It is required to convert the unit of specific heat capacity from Btu th per pound per fahrenheit to Kilocalorie it per kilogram per kelvin, in specific heat capacity. This is the very basic unit conversion, which you will learn in primary classes. It is one of the most widely used operations in a variety of mathematical applications. In this article, let us discuss how to convert Btu (th)/pound/°F (Btu (th)/(lb*°F)) to kilocalorie (IT)/kilogram/K (kcal/(kg*K)), and the usage of a tool that will help to convert one unit from another unit, and the relation between Btu th per pound per fahrenheit and Kilocalorie it per kilogram per kelvin with detailed explanation.
A thermochemical British thermal unit per pound per degree Fahrenheit (Btu(th)/lb·°F) is a unit of specific heat capacity in the US Customary Units and British Imperial Units. A material has the heat capacity of 1 Btu(th)/lb·°F if heat energy of one thermochemical British thermal unit is required to raise the temperature of one pound of this material by one degree Fahrenheit.
An international kilocalorie per kilogram per kelvin (kcal(IT)/kg·K) is a metric unit of specific heat capacity. A material has the heat capacity of 1 kcal(IT)/kg·K if heat energy of one international kilocalorie is required to raise the temperature of one kilogram of this material by one kelvin.