Answer
$2.88\times 10^6J$
Work Step by Step
We can find the required heat as follows:
$Q_{ice}=mc\Delta T$
We plug in the known values to obtain:
$Q_{ice}=(8.50976Kg)(2093J/kg.C^{\circ})(2C^{\circ})$
$Q_{ice}=35621.85536J$
Heat required to melt the ice without temperature is given as
$Q_{ice}^{\prime}=m_{ice}L_f$
$\implies Q_{ice}^{\prime}=(8.50976Kg)(33.5\times 10^4J/Kg.C^{\circ})$
$Q_{ice}^{\prime}=2850769.6J$
Now $Q=Q_{ice}+Q_{ice}^{\prime}$
We plug in the known values to obtain:
$Q=35621.85536J+2850769.6J$
$Q=2.88\times 10^6J$
