Answer
The original temperature of the cylinder was $~~873^{\circ}C$
Work Step by Step
In part (a), we found that $~~20,300~calories~~$ of energy was transferred to the water as heat.
In part (b), we found that $~~1110~calories~~$ of energy was transferred to the bowl as heat.
We can find the change in temperature of the cylinder:
$Q = cm\Delta T$
$\Delta T = \frac{Q}{cm}$
$\Delta T = \frac{20,300~cal+1110~cal}{(0.0923~cal/g\cdot K)(300~g)}$
$\Delta T = 773~C^{\circ}$
Since the equilibrium temperature is $100^{\circ}C$, the original temperature of the cylinder was $~~873^{\circ}C$
