# 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?

## EDIT: Should be $\text{6.02 kJ/mol}$, as in, $\text{6020 J/mol}$! - Truong-Son

Jun 14, 2017

The heat of phase transition is given by the expression $Q = n \Delta {H}_{f}$.

$Q =$ heat transferred in $\text{J}$

$n =$ $\text{mols}$ of substance receiving or giving off heat

$\Delta {H}_{f}$ = Enthalpy of Fusion

For a 75.0 gram block of ice at ${0}^{\circ} \text{C}$, the heat needed to convert the ice into liquid water is $Q = n \Delta {H}_{f}$:

$n = \left(\text{75 g")/("18 g/mol}\right)$ = $\text{4.17 mols}$ ${H}_{2} O$

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