# If #4.25 xx 10^3# #"g"# of gold was heated from #15.5^@ "C"# to #197.0^@ "C"#, how much heat was involved, and what is the molar enthalpy in #"kJ/mol"#? #C_P = "0.129 J/g"^@ "C"#?

##### 1 Answer

Well, we should look up the melting point of gold first. It's

The **heat flow** equation is to be used at constant atmospheric pressure:

#q_P = mC_PDeltaT# ,where:

#q_P# is theheat flowat constant pressure.#m# is themassof the object in#"g"# withspecific heat capacity#C_P# in#"J/g"^@ "C"# . You must already have the specific heat capacity of gold memorized, because you didn't put it in the question. :D#DeltaT# is the change intemperaturein the appropriate units. (What should we use,#""^@ "C"# or#"K"# ? Does it matter for changes in temperature?)

Thus, if we assume the specific heat capacity does not change in this temperature range, the heat absorbed is

#color(blue)(q_P) = (4.25 xx 10^3 "g Au")("0.129 J/g"^@ "C")(197.0 - 15.5^ @"C")#

#=# #9.95 xx 10^4# #"J"#

#=# #color(blue)("99.5 kJ")#

At constant pressure, the heat flow is also related to the molar enthalpy:

#DeltabarH = q_P/(n_"obj")# where

#n_"obj"# is just the mols of the object.

Therefore, the **enthalpy for this heating process at constant pressure** is:

#color(blue)(DeltabarH) = ("99.5 kJ")/(4.25 xx 10^3 cancel"g Au" xx "1 mol"/(196.96657 cancel"g Au"))#

#=# #color(blue)("4.61 kJ/mol")# to three sig figs.