Convert BTU / pound to BTU / pound
Learn how to convert
1
BTU / pound to
BTU / pound
step by step.
Calculation Breakdown
Set up the equation
\(1.0\left(\dfrac{BTU}{pound}\right)={\color{rgb(20,165,174)} x}\left(\dfrac{BTU}{pound}\right)\)
Define the base values of the selected units in relation to the SI unit \(\left(\dfrac{joule}{{\color{rgb(230,179,255)} kilo}gram}\right)\)
\(\text{Left side: 1.0 } \left(\dfrac{BTU}{pound}\right) = {\color{rgb(89,182,91)} 2109.0\left(\dfrac{joule}{{\color{rgb(230,179,255)} kilo}gram}\right)} = {\color{rgb(89,182,91)} 2109.0\left(\dfrac{J}{{\color{rgb(230,179,255)} k}g}\right)}\)
\(\text{Right side: 1.0 } \left(\dfrac{BTU}{pound}\right) = {\color{rgb(125,164,120)} \dfrac{1054.5}{0.45359237}\left(\dfrac{joule}{{\color{rgb(230,179,255)} kilo}gram}\right)} = {\color{rgb(125,164,120)} \dfrac{1054.5}{0.45359237}\left(\dfrac{J}{{\color{rgb(230,179,255)} k}g}\right)}\)
Insert known values into the conversion equation to determine \({\color{rgb(20,165,174)} x}\)
\(1.0\left(\dfrac{BTU}{pound}\right)={\color{rgb(20,165,174)} x}\left(\dfrac{BTU}{pound}\right)\)
\(\text{Insert known values } =>\)
\(1.0 \times {\color{rgb(89,182,91)} 2109.0} \times {\color{rgb(89,182,91)} \left(\dfrac{joule}{{\color{rgb(230,179,255)} kilo}gram}\right)} = {\color{rgb(20,165,174)} x} \times {\color{rgb(125,164,120)} {\color{rgb(125,164,120)} \dfrac{1054.5}{0.45359237}}} \times {\color{rgb(125,164,120)} \left(\dfrac{joule}{{\color{rgb(230,179,255)} kilo}gram}\right)}\)
\(\text{Or}\)
\(1.0 \cdot {\color{rgb(89,182,91)} 2109.0} \cdot {\color{rgb(89,182,91)} \left(\dfrac{J}{{\color{rgb(230,179,255)} k}g}\right)} = {\color{rgb(20,165,174)} x} \cdot {\color{rgb(125,164,120)} \dfrac{1054.5}{0.45359237}} \cdot {\color{rgb(125,164,120)} \left(\dfrac{J}{{\color{rgb(230,179,255)} k}g}\right)}\)
\(\text{Cancel SI units}\)
\(1.0 \times {\color{rgb(89,182,91)} 2109.0} \cdot {\color{rgb(89,182,91)} \cancel{\left(\dfrac{J}{{\color{rgb(230,179,255)} k}g}\right)}} = {\color{rgb(20,165,174)} x} \times {\color{rgb(125,164,120)} \dfrac{1054.5}{0.45359237}} \times {\color{rgb(125,164,120)} \cancel{\left(\dfrac{J}{{\color{rgb(230,179,255)} k}g}\right)}}\)
\(\text{Conversion Equation}\)
\(2109.0 = {\color{rgb(20,165,174)} x} \times \dfrac{1054.5}{0.45359237}\)
Switch sides
\({\color{rgb(20,165,174)} x} \times \dfrac{1054.5}{0.45359237} = 2109.0\)
Isolate \({\color{rgb(20,165,174)} x}\)
Multiply both sides by \(\left(\dfrac{0.45359237}{1054.5}\right)\)
\({\color{rgb(20,165,174)} x} \times \dfrac{1054.5}{0.45359237} \times \dfrac{0.45359237}{1054.5} = 2109.0 \times \dfrac{0.45359237}{1054.5}\)
\(\text{Cancel}\)
\({\color{rgb(20,165,174)} x} \times \dfrac{{\color{rgb(255,204,153)} \cancel{1054.5}} \times {\color{rgb(99,194,222)} \cancel{0.45359237}}}{{\color{rgb(99,194,222)} \cancel{0.45359237}} \times {\color{rgb(255,204,153)} \cancel{1054.5}}} = 2109.0 \times \dfrac{0.45359237}{1054.5}\)
\(\text{Simplify}\)
\({\color{rgb(20,165,174)} x} = \dfrac{2109.0 \times 0.45359237}{1054.5}\)
Solve \({\color{rgb(20,165,174)} x}\)
\({\color{rgb(20,165,174)} x} = 0.90718474\approx9.0718 \times 10^{-1}\)
\(\text{Conversion Equation}\)
\(1.0\left(\dfrac{BTU}{pound}\right)\approx{\color{rgb(20,165,174)} 9.0718 \times 10^{-1}}\left(\dfrac{BTU}{pound}\right)\)
