The molar heat of fusion for water is 6.01 J/mol. How much energy must be added to a 75.0-g block of ice at 0°C to change it to a 75.0 g of liquid water at 0°C?

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EDIT: Should be #"6.02 kJ/mol"#, as in, #"6020 J/mol"#!
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Answer 1 · 2017-06-14T00:44:45

The heat of phase transition is given by the expression #Q = nDelta H_f#.

#Q = # heat transferred in #"J"#

#n = # #"mols"# of substance receiving or giving off heat

#Delta H_f# = Enthalpy of Fusion

For a 75.0 gram block of ice at #0^@ "C"#, the heat needed to convert the ice into liquid water is #Q = nDelta H_f#:

#n = ("75 g")/("18 g/mol")# = #"4.17 mols"# #H_2O#

#=> Q = nDeltaH_f = ("4.17 mols")("6.02 kJ/mol") = "25.1 kJ"# to 3 sig. figs.