Question #d5e47

1 Answer
Jan 11, 2016

Answer:

#3030kJ#

Explanation:

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#