Convert calorie / gram to BTU / pound
Learn how to convert
1
calorie / gram to
BTU / pound
step by step.
Calculation Breakdown
Set up the equation
\(1.0\left(\dfrac{calorie}{gram}\right)={\color{rgb(20,165,174)} x}\left(\dfrac{BTU}{pound}\right)\)
Define the base values of the selected units in relation to the SI unit \(\left(\dfrac{joule}{{\color{rgb(230,179,255)} kilo}gram}\right)\)
\(\text{Left side: 1.0 } \left(\dfrac{calorie}{gram}\right) = {\color{rgb(89,182,91)} \dfrac{4.1868}{10^{-3}}\left(\dfrac{joule}{{\color{rgb(230,179,255)} kilo}gram}\right)} = {\color{rgb(89,182,91)} \dfrac{4.1868}{10^{-3}}\left(\dfrac{J}{{\color{rgb(230,179,255)} k}g}\right)}\)
\(\text{Right side: 1.0 } \left(\dfrac{BTU}{pound}\right) = {\color{rgb(125,164,120)} \dfrac{1055.05585262}{0.5}\left(\dfrac{joule}{{\color{rgb(230,179,255)} kilo}gram}\right)} = {\color{rgb(125,164,120)} \dfrac{1055.05585262}{0.5}\left(\dfrac{J}{{\color{rgb(230,179,255)} k}g}\right)}\)
Insert known values into the conversion equation to determine \({\color{rgb(20,165,174)} x}\)
\(1.0\left(\dfrac{calorie}{gram}\right)={\color{rgb(20,165,174)} x}\left(\dfrac{BTU}{pound}\right)\)
\(\text{Insert known values } =>\)
\(1.0 \times {\color{rgb(89,182,91)} \dfrac{4.1868}{10^{-3}}} \times {\color{rgb(89,182,91)} \left(\dfrac{joule}{{\color{rgb(230,179,255)} kilo}gram}\right)} = {\color{rgb(20,165,174)} x} \times {\color{rgb(125,164,120)} {\color{rgb(125,164,120)} \dfrac{1055.05585262}{0.5}}} \times {\color{rgb(125,164,120)} \left(\dfrac{joule}{{\color{rgb(230,179,255)} kilo}gram}\right)}\)
\(\text{Or}\)
\(1.0 \cdot {\color{rgb(89,182,91)} \dfrac{4.1868}{10^{-3}}} \cdot {\color{rgb(89,182,91)} \left(\dfrac{J}{{\color{rgb(230,179,255)} k}g}\right)} = {\color{rgb(20,165,174)} x} \cdot {\color{rgb(125,164,120)} \dfrac{1055.05585262}{0.5}} \cdot {\color{rgb(125,164,120)} \left(\dfrac{J}{{\color{rgb(230,179,255)} k}g}\right)}\)
\(\text{Cancel SI units}\)
\(1.0 \times {\color{rgb(89,182,91)} \dfrac{4.1868}{10^{-3}}} \cdot {\color{rgb(89,182,91)} \cancel{\left(\dfrac{J}{{\color{rgb(230,179,255)} k}g}\right)}} = {\color{rgb(20,165,174)} x} \times {\color{rgb(125,164,120)} \dfrac{1055.05585262}{0.5}} \times {\color{rgb(125,164,120)} \cancel{\left(\dfrac{J}{{\color{rgb(230,179,255)} k}g}\right)}}\)
\(\text{Conversion Equation}\)
\(\dfrac{4.1868}{10^{-3}} = {\color{rgb(20,165,174)} x} \times \dfrac{1055.05585262}{0.5}\)
Switch sides
\({\color{rgb(20,165,174)} x} \times \dfrac{1055.05585262}{0.5} = \dfrac{4.1868}{10^{-3}}\)
Isolate \({\color{rgb(20,165,174)} x}\)
Multiply both sides by \(\left(\dfrac{0.5}{1055.05585262}\right)\)
\({\color{rgb(20,165,174)} x} \times \dfrac{1055.05585262}{0.5} \times \dfrac{0.5}{1055.05585262} = \dfrac{4.1868}{10^{-3}} \times \dfrac{0.5}{1055.05585262}\)
\(\text{Cancel}\)
\({\color{rgb(20,165,174)} x} \times \dfrac{{\color{rgb(255,204,153)} \cancel{1055.05585262}} \times {\color{rgb(99,194,222)} \cancel{0.5}}}{{\color{rgb(99,194,222)} \cancel{0.5}} \times {\color{rgb(255,204,153)} \cancel{1055.05585262}}} = \dfrac{4.1868 \times 0.5}{10^{-3} \times 1055.05585262}\)
\(\text{Simplify}\)
\({\color{rgb(20,165,174)} x} = \dfrac{4.1868 \times 0.5}{10^{-3} \times 1055.05585262}\)
Rewrite equation
\(\dfrac{1.0}{10^{-3}}\text{ can be rewritten to }10^{3}\)
\(\text{Rewrite}\)
\({\color{rgb(20,165,174)} x} = \dfrac{10^{3} \times 4.1868 \times 0.5}{1055.05585262}\)
Solve \({\color{rgb(20,165,174)} x}\)
\({\color{rgb(20,165,174)} x}\approx1.9841603597\approx1.9842\)
\(\text{Conversion Equation}\)
\(1.0\left(\dfrac{calorie}{gram}\right)\approx{\color{rgb(20,165,174)} 1.9842}\left(\dfrac{BTU}{pound}\right)\)
