Question

The heat capacity of liquid water is 4.18 J/g

Answer 1

Answer: Amount of heat provided = heat lost by water to convert into vapour

                 = Heating the water from 67⁰C to 100⁰C + converting to vapour.

                 = m₁ x Cw xt + m₂ x Cv ------------------ equation 1

                where, m₁= mass of water, m₁ =1.00g, Cw = specific heat of water , Cw = 4.18J/g/^oC,

            t = rise in temperature = T₂ -T₁t = T₂ -T₁ = (100 + 273) - ( 67 +273) = 373 -340 = 33 K

           m₂= mass of steam, m₂ = 1.00g,   Cv= heat of vapourisation = 40.7KJ/mol = 40.7 x1000 J/mol.

           substituting all the values in equation 1 we get,

           Amount of heat to be provided is = (1.00)(4.18)(33) + (1.00)(40.7 x 10^3)

                                                          = 137.94+40700 = 40837.94 J

                                                          = 40.8379 KJ

      therefore amount of heat to be provided is 40.8379 KJ