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

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

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Answer 1 · 2017-07-03T20:02:25

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

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

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

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