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.