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) ?

Question

Chemistry

Larry

21 November, 00:31

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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Isaac Ferguson · Answer 1

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