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How much heat (in kJ) is required to warm 13.0 g of ice?
How much heat (in kJ) is required to warm 13.0 g of ice, initially at -14.0 ∘C, to steam at 109.0 ∘C? The heat capacity of ice is 2.09 J/g⋅∘C and that of steam is 2.01 J/g⋅∘C.
1 Answer
- Roger the MoleLv 74 years agoFavorite Answer
In order to answer this question you also need to know three additional constants for various kinds of heat transfer in water. The source below gives them all.
(2.09 J/g⋅∘C) x (13.0 g) x (0 - (-14.0)) ∘C = 380.38 J to warm the ice to its melting point
(333.6 J/g) x (13.0 g) = 4336.8 J to melt the ice
(4.184 J/g·°C) x (13.0 g) x (100 - 0) ∘C = 5439.2 J to warm the melted ice to its boiling point
(2257 J/g) x (13.0 g) = 29341 J to vaporize the water
(2.01 J/g⋅∘C) x (13.0 g) x (109.0 - 100) ∘C = 235.17 J to heat the steam to 109.0 ∘C
380.38 J + 4336.8 J + 5439.2 J + 29341 J + 235.17 J = 39733 J = 39.7 kJ total
