Answer
$7.23\times10^{-20}J$
Work Step by Step
We can write,
The average energy, given to a = $\frac{Q}{N}$
single water molecule (E)
Also, according to equation 12.5 we can write,
$Q=mL_{v}$, Therefore,
$E=\frac{mL_{v}}{N}=\frac{m_{water molecule}NL_{v}}{N}$
Let's plug known values into this equation.
$E=(18\space u)(\frac{1.66\times10^{-27}kg}{1\space u})(2.42\times10^{6}J/kg)=7.23\times10^{-20}J$
