How much heat is evolved in converting 1.00 mol of steam at 135.0 ∘c to ice at -45.0 ∘c? the heat capacity of steam is 2.01 j/(g⋅∘c) and of ice is 2.09 j/(g⋅∘c)?

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21-01-2018
Chemistry

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How much heat is evolved in converting 1.00 mol of steam at 135.0 ∘c to ice at -45.0 ∘c? the heat capacity of steam is 2.01 j/(g⋅∘c) and of ice is 2.09 j/(g⋅∘c)?

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jamuuj jamuuj · Answer 1 · 2018-02-02T13:14:33+01:00

1 mole of steam (H2O) = 18g
Heat evolved = MCФ where m is the mass and c s specific heat capacity while Ф is change in temperature. Latent heat of fusion is 334 kj/kg while latent heat of vaporization is 2260 kj/kg and specific heat of water is 4.2 j/g/c
= 18 ×2.01 × (135-100) = 1266.3 J
0.018 × 334000 = 6012 J (change of state from gas (steam) to liquid (water)
18 × 4.186× (100 -0) = 7534.8 J
0.018 × 2260000 = 40680 J (change of state from liquid to solid ice)
18 × 2.09 × (0--45) = 1692.9 J
The total heat evolved is therefore 57186 J or 57.186 kJ