Question

How much heat (in kJ) is required to warm 11.0 g of ice, initially at -13.0 ∘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.

Answer 1

Given mass of ice, m = 11.0g

specific heat of ice, S(ice) = 2.09 J/g⋅∘C

specific heat of water, S(water) = 4.184 J/g⋅∘C

specific heat of steam, S(steam) = 2.01 J/g⋅∘C

Latent heat of fusion of ice, Lf = 334 J / g

Latent heat of vaporization of water, Lv = 2257 J / g

When ice at - 13 DegC is converted to steam at 113 DegC, it undergoes the following changes.

1. ice at - 13 DegC to ice at 0 DegC (DeltaH1)

2.  Ice at 0 DegC to water at 0 DegC (DeltaH2)

3. water at 0 DegC to water at 100 DegC (DeltaH3)

4. water at 100 DegC to steam at 100 DegC (DeltaH4)

5. steam at 100 DegC to steam at 113 DegC (DeltaH5)

Hence total heat required, H = H1 + H2 + H2 +H4 + H5

= mxS(ice)x[0 - (-13)] + mxLf + mxS(water)x[100 - 0] + mxLv + mxS(steam)x[113 - 100]

= 11.0gx(2.09 J/g⋅∘C)x13 + 11.0gx334J/g + 11.0gx(4.184 J/g⋅∘C)x100 + 11.0gx2257 J / g

+ 11.0gx(2.01 J/g⋅∘C)x13

= 298.87J + 3674J + 4602.4 J + 24827 J + 287.43J

= 33689.7 J

= 33689.7 J (1KJ / 1000J) = 33.7 KJ (answer)