How much heat (in kj) is required to warm 13.0 g of ice, initially at -15.0 ∘c, to steam at 114.0 ∘c? the heat capacity of ice is 2.09 j/g⋅∘c and that of steam is 2.01 j/g⋅∘c, the heat of fusion for water is 6.02 kj/mol, and the heat of vaporization for water is 40.7 kj/mol.

1)(temperature change)(mass)(Csolid)/1000=.40755kJ
(15)(13)(2.09)/1000
2)(Hfusion)(moles)=4.342952275kJ
(6.02)(13/18.02)
3)(temperature change)(mass)(Cliquid)/1000=2.613kJ
(100)(13)(2.01)/1000
4)(Hvap)(moles)=29.3618202kJ
(40.7)(13/18.02)
5)(temperature change)(mass)(Cgas)/1000=0.36582kJ
(14)(13)(2.01)/1000
Total=37.09114248kJ

Because a model would make the investigation easier to understand