How is #DeltaH_(fusion)# used to calculate the mass of solid that 1kJ of energy will melt?
1 Answer
With no need of a formula. Is
Given the enthalpy of fusion, simple unit conversion gives:
#overbrace("kJ"/"mol")^(DeltabarH_"tr") xx overbrace("mol"/"g" xx "g")^(m_"solid"//M_"solid") = overbrace("kJ")^(q)# where
#DeltabarH_"tr"# is the molar enthalpy of phase transition in#"kJ/mol"# ,#q# is heat flow in#"kJ"# ,#m# is mass in#"g"# , and#M# is molar mass in#"g/mol"# .
Thus:
#"g" = overbrace(cancel"kJ")^(q) xx overbrace(cancel"mol"/cancel"kJ")^(DeltabarH_"fus"^(-1)) xx overbrace("g"/cancel"mol")^(M_"solid"^(-1))#
And yet, this still refers to a real equation that we could have invoked only at constant pressure:
#q = nDeltabarH_"tr"# where
#n# is the mols of substance and#DeltabarH_"tr"# is its molar enthalpy of phase transition.
