Answer
$1110~calories~~$ of energy was transferred to the bowl as heat.
Work Step by Step
We can find the amount of energy transferred to the bowl as heat:
$Q = cm\Delta T$
$Q = (0.0923~cal/g\cdot K)(150~g)(80.0~C^{\circ})$
$Q = 1110~cal$
$1110~calories~~$ of energy was transferred to the bowl as heat.
