Question

How much heat is released when 85.0 g of steam at 100.0°C is cooled to ice at -15.0°C? The enthalpy of vaporization of water is 40.67 kJ/mol, the enthalpy of fusion for water is 6.01 kJ/mol, the molar heat capacity of liquid water is 75.4 J/(mol ⋅ °C), and the molar heat capacity of ice is 36.4 J/(mol ⋅ °C).

Answer 1

The amount of heat released is Q = heat change for conversion of steam at 100 ^oC to water at 100 ^oC + heat change for conversion of water at 100^oC to water at 0 ^oC + heat change for conversion of water at 0^oC to ice at 0^oC + heat change for conversion of ice at 0 ^oC to ice at -15 ^oC   

Amount of heat released , Q = mL + mcdt + mL' + mc'dt'

                                              = m(L + cdt + L' + c'dt')

Where

m = mass of water = 85.0 g

c = Specific heat of water = molar heat capacity of water / molar mass

     = 74.5x1000 J /(mol oC ) / 18 (g/mol)

     = 4.18 J/g ^oC

c' = Specific heat of ice = molar heat capacity of ice / molar mass of water

    = 36.4 x1000 J / ( mol ^oC ) / 18 (g/mol)

    = 2.02 J/g ^oC

L = Heat of Vaporization of water = enthalpy of vaporization of water / molar mass of water

     = 40.67 x1000 J/mol / 18 (g/mol)

     = 2260 J/g

L'= Heat of fusion of ice = enthalpy of fusion for water / molar mass of water

    = 6.01x1000 (J/mol ) / 18 (g/mol)

    = 333.9 J/g

dt = 100 - 0 = 100 ^oC

dt' = 0-(-15) = 15 ^oC

Plug the values we get Q = m(L + cdt + L' + c'dt')

                                          = 85.0 [ 2260 + (4.18x100) + 333.9 + (2.02x15)]

                                         = 258587 J

                                         = 258.6 kJ

Therefore the heat released is 258.6 kJ