Question

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"`?

Answer 1

`"99.5 kJ"`, or `"4.61 kJ/mol"`, for what is apparently `"21.6 mols"` of gold... How rich is this person?

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Well, we should look up the melting point of gold first. It's `1064^@ "C"`, so we can safely assume that only heating is involved (and not melting).

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

`q_P = mC_PDeltaT`,

where:

• `q_P` is the heat flow at constant pressure.
• `m` is the mass of the object in `"g"` with specific 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 in temperature in 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.