Question

How much heat (in kJ) is required to warm 13.0 g of ice, initially at -13.0 ∘C, to steam at 114.0 ∘C? The heat capacity of ice is 2.09 J/g⋅∘C and that of steam is 2.01 J/g⋅∘C.

Answer 1

Q = m Cp T

here 5 conversions are there

1) -13.0 ∘C   to 0 ∘C   (ice)

Q = m Cp T

    = 13 x 2.09 x (0-(-13))

    = 353.21 J

2) at 0⁰C   (Latent heat of fusion )

L = latent heat of fusion of water = 334 J/g

Q= m L

    = 13 g x 334 J/g

     = 4342 J

3) 0⁰C   to 100⁰C    (water )

Q = m Cp T

     = 13 x 4.18 x (100-0)

     = 5434 J

4 ) at 100⁰C (latent heat of vapourisation :)

   the heat of vaporization of water is L = 2270 J/g

Q = m L

     = 13 x 2270

       = 29510 J

5) 100⁰C to 114⁰C      

Q = m Cp T

    = 13 x 2.01 x (114-100)

     = 365.82 J

now add 1 to 5 Q values

total heat = (353.21 + 4342 + 5434 + 29510 + 365.82 ) J

                = 40 kJ

heat required = 40 kJ

.