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Btu (IT)/pound/°F to calorie (IT)/gram/°C

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 Btu (IT)/pound/°F to calorie (IT)/gram/°C

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