homework help 5

##3030kJ##

Your method is 100% correct, but you made a small calculation error in converting the lb to g, by a factor of 10, resulting in you final answer being out by a factor of 10.

The amount of energy required to melt ##1kg## of ice at ##0^@C## is equal to the specific latent heat of fusion for ice and has value ##L_f=334kJ//kg##.

So for ##20xx0.4536=9.072kg##, we get that energy required to melt it is
##W=mL_f=9.072kgxx334000J//kg##

##=3.03MJ##

##=3030kJ##

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