# 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"?

Jul 3, 2017

$\text{99.5 kJ}$, or $\text{4.61 kJ/mol}$, for what is apparently $\text{21.6 mols}$ of gold... How rich is this person?

Well, we should look up the melting point of gold first. It's ${1064}^{\circ} \text{C}$, 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:

${q}_{P} = m {C}_{P} \Delta T$,

where:

• ${q}_{P}$ is the heat flow at constant pressure.
• $m$ is the mass of the object in $\text{g}$ with specific heat capacity ${C}_{P}$ in $\text{J/g"^@ "C}$. You must already have the specific heat capacity of gold memorized, because you didn't put it in the question. :D
• $\Delta T$ is the change in temperature in the appropriate units. (What should we use, $\text{^@ "C}$ or $\text{K}$? 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

$\textcolor{b l u e}{{q}_{P}} = \left(4.25 \times {10}^{3} \text{g Au")("0.129 J/g"^@ "C")(197.0 - 15.5^ @"C}\right)$

$=$ $9.95 \times {10}^{4}$ $\text{J}$

$=$ $\textcolor{b l u e}{\text{99.5 kJ}}$

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

$\Delta \overline{H} = {q}_{P} / \left({n}_{\text{obj}}\right)$

where ${n}_{\text{obj}}$ 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"))

$=$ $\textcolor{b l u e}{\text{4.61 kJ/mol}}$ to three sig figs.