How much heat (in kj) is required to warm 13.0 g of ice, initially at - 12.0 ∘c, to steam at 113.0 ∘c? the heat capacity of ice is 2.09 j/g⋅∘c and that of steam is 2.01 j/g⋅∘

c.

The total amount of heat required is the sum of all the sensible heat and latent heats involved in bringing the ice to a desired temperature and state. The latent heat of fusion and vaporization of water 333.55 J/g and 2260 J/g, respectively. Solving for the total amount of heat,

total amount of heat = 13.0 g (2.09 J/gC) (12) + 13 (333.55 J/g) + 13.0 g (4.18 J/gC) (100 - 0) + (13.0 g) (2260 J/g) + (13 g) (2.01 J/g) (113-100)

= 39815.88 J

= 39.82 kJ