How much heat (in kj) is required to warm 11.0 g of ice, initially at - 12.0 ∘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?

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Chemistry

Dayanara

16 October, 06:18

How much heat (in kj) is required to warm 11.0 g of ice, initially at - 12.0 ∘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?

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Alonso Meza · Answer 1

Sensible heat (involving temperature change) is calculated as mc (deltaT), where c is the specific heat. Latent heat (involving phase change) is calculated as mH, where H is the latent heat of fusion or vaporization.

From ice at - 12 to ice at 0: (11.0 g) (2.09 J/g-degC) (0 - (-12) degC) = 275.88 J

From ice at 0 to water at 0: (11.0 g) (80 J/g) = 880 J

From water at 0 to water at 100: (11.0 g) (4.184 J/g-degC) (100 - 0 degC) = 4602.4 J

From water at 100 to steam at 100: (11.0 g) (540 J/g) = 5940 J

From steam at 100 to steam at 109: (11.0 g) (2.01 J/g-degC) (109 - 100 degC) = 198.99 J

Adding all the heats gives: 11897.27 J, which is equivalent to 11.89727 kJ.