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Calorie (IT)/gram/°F to joule/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 calorie (IT)/gram/°F to joule/gram/°C

$$ 1\ \mathrm{cal\ (IT)/g/°F}={{1}\cdot{4186.8}}\cdot{10^{-3}}\ \mathrm{J/(g\cdot\ °C)}=4.187\ \mathrm{J/(g\cdot\ °C)} $$
$ 4.1868\ \mathrm{J/(g\cdot\ °C)} $
Unrounded result
4.1868
$1\ \mathrm{cal\ (IT)/g/°F}={{1}\cdot{4186.8}}\cdot{10^{-3}}\ \mathrm{J/(g\cdot\ °C)}=4.187\ \mathrm{J/(g\cdot\ °C)}$
Equations in LaTeX code
1\ \mathrm{cal\ (IT)/g/°F}={{1}\cdot{4186.8}}\cdot{10^{-3}}\ \mathrm{J/(g\cdot\ °C)}=4.187\ \mathrm{J/(g\cdot\ °C)}
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