Answer
$20,300~calories~~$ of energy was transferred to water as heat.
Work Step by Step
We can find the amount of energy transferred to water as heat:
$Q = cm\Delta T+L_v~m'$
$Q = (1.00~cal/g\cdot K)(220~g)(80.0~C^{\circ})+(539~cal/g)(5.00~g)$
$Q = 20,300~cal$
$20,300~calories~~$ of energy was transferred to water as heat.
