joule per kilogram

SI coherent derived unit whose name and symbol includes an SI coherent derived unit with a special name and symbol

Name	Symbol	Quantity	Base units
joule per kilogram	J/kg	specific energy	m² s⁻²
The joule per kilogram, symbol J/kg, is the SI coherent derived unit of specific energy. One joule per kilogram is equal to the specific energy of a substance that has one joule of energy associated with one kilogram of the substance.
Definition			c²
$1 \mspace{4mu} \text{J} / \text{kg} \mspace{6mu} = \dfrac{1}{(299 \mspace{4mu} 792 \mspace{4mu} 458)^2} \mspace{6mu} c^2\\ \\ \\ 1 \mspace{4mu} \text{J} / \text{kg} \mspace{6mu} \approx 1.112 \mspace{4mu} 650 \mspace{4mu} 056 \mspace{4mu} 053 \mspace{4mu} 618 \mspace{4mu} 432 \times10^{-17} \mspace{6mu} c^2$

The joule is equal to kg m² s^-2, when expressed in terms of SI base units. It follows that the joule per kilogram is equal to (kg m² s^-2)/kg.

This can be simplified to give an expression for the joule per kilogram in terms of just two base units – the metre and the second:

$1 \mspace{6mu} \text{J/kg} \mspace{6mu} = 1 \mspace{6mu} ( \text{kg} \mspace{4mu} \text{m}^2 \mspace{4mu} \text{s}^{-2}) / \text{kg} \\ \\ 1 \mspace{6mu} \text{J/kg} \mspace{6mu} = 1 \mspace{6mu} \text{m}^2 \mspace{4mu} \text{s}^{-2} \\ \\ 1 \mspace{6mu} \text{J/kg} \mspace{6mu} = 1 \mspace{6mu} \text{(m} \mspace{4mu} \text{s}^{-1} )^2$

The fixed numerical value of the speed of light in vacuum, c, is defined exactly, when expressed in the unit m s⁻¹ :

$c \mspace{6mu} = 299 \mspace{4mu} 792 \mspace{4mu} 458 \mspace{6mu} \text{m} \mspace{4mu} \text{s}^{-1}$

Inverting this relation gives an exact expression for the metre per second in terms of the speed of light in vacuum, c :

$1 \mspace{6mu} \text{m} \mspace{6mu} \text{s}^{-1} \mspace{6mu} = \dfrac{1}{299 \mspace{4mu} 792 \mspace{4mu} 458} \mspace{6mu} c$

It follows that the joule per kilogram can be expressed exactly in terms of the SI defining constant, the speed of light in vacuum, c :

$1 \mspace{6mu} \text{J/kg} \mspace{6mu} = 1 \mspace{6mu} ( \text{m} \mspace{4mu} \text{s}^{-1})^2 \\ \\ 1 \mspace{6mu} \text{J/kg} \mspace{6mu} = \dfrac{1}{(299 \mspace{4mu} 792 \mspace{4mu} 458)^2} \mspace{6mu} c^2$