Answer
(a) $0C^{\circ}$
(b) $1.8C^{\circ}$
Work Step by Step
(a) We know that $Q_{lem}=m_{lem}c_w\Delta T$.
We plug in the known values to obtain:
$Q_{lem}=(2.00Kg)(4186J/Kg.K)(1.00C^{\circ}-0C^{\circ})$
$Q_{lem}=8372J$
Now $m_{mel}=\frac{Q_{lem}}{L_f}$
We plug in the known values to obtain:
$m_{mel}=\frac{8372J}{33.5\times 10^4J/Kg}=0.025Kg$
From the above result, we can see that the mass of the melted ice cube is less than that of the total mass of ice cube at $0C^{\circ}$; thus the ice cube is not melted completely and therefore, the final temperature of the system is $T=0C^{\circ}$.
(b) We know that
$T_i=\frac{m_{ice}L_f}{m_{lem}c_w}$
We plug in the known values to obtain:
$T_i=\frac{(0.0450Kg)(33.5\times 10^4J/Kg)}{(2.00Kg)(4186J/Kg.K)}$
$T_i=1.8C^{\circ}$
