Answer
$4.7\times 10^{5}\,J$
Work Step by Step
We find:
$m=(1.0\,g/mL)(1.50\times10^{3}\,mL)=1500\,g$
$c= 4.18\,J/g\cdot\,^{\circ}C$
$\Delta T= 100.0^{\circ}C-25.0^{\circ}C=75.0^{\circ}C$
$q=mc\Delta T$
$=(1500\,g)(4.18\,J/g\cdot\,^{\circ}C)(75.0^{\circ}C)$
$= 4.7\times 10^{5}\,J$