Question

What amount of thermal energy (in kJ) is required to convert 138 g of ice at -18 °C completely to water vapour at 202 °C? The melting point of water is 0 °C and its normal boiling point is 100 °C.

The heat of fusion of water is 6.02 kJ mol-1

The heat of vaporization of water at its normal boiling point is 40.7 kJ mol-1

The specific heat capacity of ice is 2.09 J g-1 °C-1

The specific heat capacity of liquid water is 4.18 J g-1 °C-1

The specific heat capacity of water vapour is 2.01 J g-1 °C-1

Answer 1

Q = heat change for conversion of ice at -18 ^oC to ice at 0 ^oC + heat change for conversion of ice at 0^oC to water at 0^oC + heat change for conversion of water at 0^oC to water at 100 ^oC +heat change for conversion of water at 100 ^oC to vapour at 100 ^oC+ heat change for conversion of vapour at 100 ^oC to vapour at 202 ^oC

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

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

Where

m = mass of ice = 138 g

c” = Specific heat of steam = 2.01 J/g degree C

c' = Specific heat of water = 4.18 J/g degree C

c = Specific heat of ice= 2.09 J/g degree C

L’ = Heat of Vaporization of water = 40.7kJ/mol x (1000J/kJ) x (1mol/18g) = 2261.1 J/g

L= Heat of fusion of ice = 6.02 kJ/molx(1000J/kJ) x (1 mol/18g) = 334.4 J/g

dt’’ = 202-100 = 102^oC

dt' = 100 -0 =100 ^oC

dt = 0-(-18)=18 ^oC

Plug the values we get Q = m(cdt + L + c'dt '+ L' + c"dt" ) = 449.9x10³ J

                                  Q = 449.9 kJ