Calculate the enthalpy change on freezing one mode of water at `10^(@)C` to ice at temperature` -10^(0)C``, Delta fus H=6.03KJmol^(-1)` At `0^(0)`,`Cp H_(2)O_((l))` =`75.3Jmol^(-1)K^(-1)`and` Cp H_(2)O_((s)) =36.8Jmol`

Calcualte the enthalpy change on freezing of 1.0 mole of water at 10.0^(@)C to ice at -10^(@) C. Delta_(fs)H=6.03 kJ mol^(-1) at 0^(@)C . C_(p)[H_(2)O(l)] = 75.3 J mol^(-1) K^(-1), C_(P)[H_(2)O(s)] = 36.8 Jmol^(-1)K^(-1)

The enthalpy change of freezing of 1 mol of water at 5^(@)C to ice at -5^(@)C is (Given Delta_(fus)H=6kJ"mol"^(-1) at 0^(@)C , (C_(p)(H_(2)O,s)=36.8J"mol"^(-1)K^(-1))

What is the enthalpy change when 1.0 g of water is frozen at 0^(@)C ? Delta_(fus) = H = 6.02 kJ "mol"^(-1) .

Calculate the change in enthalpy for the following process at 1 atm H_(2)O(1, 50^(@)C) rarr H_(2)O (g, 150^(@)C) given that Delta H_(v) at 100^(@)C is 40.7 kJ mol^(-1) C_(p)(H_(2)O, l) = 75.0 J mol^(-1)K^(-1) C_(p)(H_(2)O, g) = 33.3 J mol^(-1)K^(-1)

The enthalpy change for the formation of 3.6 kg water is _____ . H_(2(g))+1/2 O_(2(g)) to H_2 O_((l)), Delta H=-284.5 kJ mol^-1

Calculate the entropy change accompanying the following change of state H_(2)O (s, 10^(@)C, 1 atm) rarr H_(2)O(l, 10^(@)C, 1atm) C_(P) for ice = 9 cla deg^(-1) mol^(-1) C_(P) for H_(2)O = 18 cal deg^(-1) mol^(-1) Latent heat of fustion of ice = 1440 cal mol^(-1) at 0^(@)C .