Convert ton of coal to watt • minute
Learn how to convert
1
ton of coal to
watt • minute
step by step.
Calculation Breakdown
Set up the equation
\(1.0\left(ton \text{ } of \text{ } coal\right)={\color{rgb(20,165,174)} x}\left(watt \times minute\right)\)
Define the base values of the selected units in relation to the SI unit \(\left(joule\right)\)
\(\text{Left side: 1.0 } \left(ton \text{ } of \text{ } coal\right) = {\color{rgb(89,182,91)} 2.9288 \times 10^{10}\left(joule\right)} = {\color{rgb(89,182,91)} 2.9288 \times 10^{10}\left(J\right)}\)
\(\text{Right side: 1.0 } \left(watt \times minute\right) = {\color{rgb(125,164,120)} 60.0\left(joule\right)} = {\color{rgb(125,164,120)} 60.0\left(J\right)}\)
Insert known values into the conversion equation to determine \({\color{rgb(20,165,174)} x}\)
\(1.0\left(ton \text{ } of \text{ } coal\right)={\color{rgb(20,165,174)} x}\left(watt \times minute\right)\)
\(\text{Insert known values } =>\)
\(1.0 \times {\color{rgb(89,182,91)} 2.9288 \times 10^{10}} \times {\color{rgb(89,182,91)} \left(joule\right)} = {\color{rgb(20,165,174)} x} \times {\color{rgb(125,164,120)} {\color{rgb(125,164,120)} 60.0}} \times {\color{rgb(125,164,120)} \left(joule\right)}\)
\(\text{Or}\)
\(1.0 \cdot {\color{rgb(89,182,91)} 2.9288 \times 10^{10}} \cdot {\color{rgb(89,182,91)} \left(J\right)} = {\color{rgb(20,165,174)} x} \cdot {\color{rgb(125,164,120)} 60.0} \cdot {\color{rgb(125,164,120)} \left(J\right)}\)
\(\text{Cancel SI units}\)
\(1.0 \times {\color{rgb(89,182,91)} 2.9288 \times 10^{10}} \cdot {\color{rgb(89,182,91)} \cancel{\left(J\right)}} = {\color{rgb(20,165,174)} x} \times {\color{rgb(125,164,120)} 60.0} \times {\color{rgb(125,164,120)} \cancel{\left(J\right)}}\)
\(\text{Conversion Equation}\)
\(2.9288 \times 10^{10} = {\color{rgb(20,165,174)} x} \times 60.0\)
Switch sides
\({\color{rgb(20,165,174)} x} \times 60.0 = 2.9288 \times 10^{10}\)
Isolate \({\color{rgb(20,165,174)} x}\)
Multiply both sides by \(\left(\dfrac{1.0}{60.0}\right)\)
\({\color{rgb(20,165,174)} x} \times 60.0 \times \dfrac{1.0}{60.0} = 2.9288 \times 10^{10} \times \dfrac{1.0}{60.0}\)
\(\text{Cancel}\)
\({\color{rgb(20,165,174)} x} \times {\color{rgb(255,204,153)} \cancel{60.0}} \times \dfrac{1.0}{{\color{rgb(255,204,153)} \cancel{60.0}}} = 2.9288 \times 10^{10} \times \dfrac{1.0}{60.0}\)
\(\text{Simplify}\)
\({\color{rgb(20,165,174)} x} = \dfrac{2.9288 \times 10^{10}}{60.0}\)
Solve \({\color{rgb(20,165,174)} x}\)
\({\color{rgb(20,165,174)} x}\approx488133333.33\approx4.8813 \times 10^{8}\)
\(\text{Conversion Equation}\)
\(1.0\left(ton \text{ } of \text{ } coal\right)\approx{\color{rgb(20,165,174)} 4.8813 \times 10^{8}}\left(watt \times minute\right)\)
