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Kilocalorie (IT)/kilogram/K to Btu (IT)/pound/°F

Values for calculation

$ \{\text{Value}\in\mathbb{R}\} $
This set includes all real numbers.
$ \text{Value} $

Specific heat capacity converter

Specific heat capacity converter from kilocalorie (IT)/kilogram/K to Btu (IT)/pound/°F

$$ 1\ \mathrm{kcal\ (IT)/kg/K}=\cfrac{{1}\cdot{4186.8}}{4186.8}\ \mathrm{Btu\ (IT)/lb/°F}=1\ \mathrm{Btu\ (IT)/lb/°F} $$
$ 1\ \mathrm{Btu\ (IT)/lb/°F} $
Unrounded result
1
$1\ \mathrm{kcal\ (IT)/kg/K}=\cfrac{{1}\cdot{4186.8}}{4186.8}\ \mathrm{Btu\ (IT)/lb/°F}=1\ \mathrm{Btu\ (IT)/lb/°F}$
Equations in LaTeX code
1\ \mathrm{kcal\ (IT)/kg/K}=\cfrac{{1}\cdot{4186.8}}{4186.8}\ \mathrm{Btu\ (IT)/lb/°F}=1\ \mathrm{Btu\ (IT)/lb/°F}
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